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NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Ex 6.3

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NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Ex 6.3

Get Free NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Ex 6.3 PDF in Hindi and English Medium. Sets Class 12 Maths NCERT Solutions are extremely helpful while doing your homework. Application of Derivatives Exercise 6.3 Class 12 Maths NCERT Solutions were prepared by Experienced LearnCBSE.in Teachers. Detailed answers of all the questions in Chapter 5 Class 12 Application of Derivatives Ex 6.3 provided in NCERT Textbook.

Free download NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Ex 6.3 PDF in Hindi Medium as well as in English Medium for CBSE, Uttarakhand, Bihar, MP Board, Gujarat Board, BIE, Intermediate and UP Board students, who are using NCERT Books based on updated CBSE Syllabus for the session 2019-20.

The topics and sub-topics included in the Applications of Derivatives chapter are the following:

Section NameTopic Name
6Applications of Derivatives
6.1Introduction
6.2Rate of Change of Quantities
6.3Increasing and Decreasing Functions
6.4Tangents and Normals
6.5Approximations
6.6Maxima and Minima
6.7Maximum and Minimum Values of a Function in a Closed Interval
6.8Summary

NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Ex 6.3

Ex 6.3 Class 12 Maths Question 1.
Find the slope of the tangent to the curve y = 3x4 – 4x at x = 4.
Solution:
The curve is y = 3x4 – 4x
\frac { dy }{ dx } = 12x3 – 4
∴Req. slope = { \left( \frac { dy }{ dx } \right) }_{ x=4 }
= 12 x 43 – 4 = 764.

Ex 6.3 Class 12 Maths Question 2.
Find the slope of the tangent to the curve y=\frac { x-1 }{ x-2 } ,x\neq 2 at x = 10.
Solution:
The curve is y=\frac { x-1 }{ x-2 }
NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives 2

Ex 6.3 Class 12 Maths Question 3.
Find the slope of the tangent to curve y = x3 – x + 1 at the point whose x-coordinate is 2.
Solution:
The curve is y = x3 – x + 1
\frac { dy }{ dx } = 3x² – 1
∴slope of tangent = { \left( \frac { dy }{ dx } \right) }_{ x=2 }
= 3 x 2² – 1
= 11

Ex 6.3 Class 12 Maths Question 4.
Find the slope of the tangent to the curve y = x3 – 3x + 2 at the point whose x-coordinate is 3.
Solution:
The curve is y = x3 – 3x + 2
\frac { dy }{ dx } = 3x² – 3
∴slope of tangent = { \left( \frac { dy }{ dx } \right) }_{ x=3 }
= 3 x 3² – 3
= 24

Ex 6.3 Class 12 Maths Question 5.
Find the slope of the normal to the curve x = a cos3 θ, y = a sin3 θ at θ = \frac { \pi }{ 4 } .
Solution:
\frac { dx }{ d\theta } =-3a\quad { cos }^{ 2 }\theta sin\theta ,\frac { dy }{ d\theta } =3a\quad { sin }^{ 2 }\theta cos\theta
NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives 5

Ex 6.3 Class 12 Maths Question 6.
Find the slope of the normal to the curve x = 1 – a sin θ, y = b cos² θ at θ = \frac { \pi }{ 2 }
Solution:
\frac { dx }{ d\theta } =-a\quad cos\theta \quad \& \quad \frac { dy }{ d\theta } =2b\quad cos\theta (-sin\theta )
NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives 6

Ex 6.3 Class 12 Maths Question 7.
Find points at which the tangent to the curve y = x3 – 3x2 – 9x + 7 is parallel to the x-axis.
Solution:
Differentiating w.r.t. x; \frac { dy }{ dx } = 3 (x – 3) (x + 1)
Tangent is parallel to x-axis if the slope of tangent = 0
or \frac { dy }{ dx }=0
⇒3(x + 3)(x + 1) = 0
⇒x = -1, 3
when x = -1, y = 12 & When x = 3, y = – 20
Hence the tangent to the given curve are parallel to x-axis at the points (-1, -12), (3, -20)

Ex 6.3 Class 12 Maths Question 8.
Find a point on the curve y = (x – 2)² at which the tangent is parallel to the chord joining the points (2,0) and (4,4).
Solution:
The equation of the curve is y = (x – 2)²
Differentiating w.r.t x
\frac { dy }{ dx }=2(x-2)
The point A and B are (2,0) and (4,4) respectively.
byjus class 12 maths Chapter 6 Application of Derivatives 8
Slope of AB = \frac { { y }_{ 2 }-{ y }_{ 1 } }{ { x }_{ 2 }-{ x }_{ 1 } } =\frac { 4-0 }{ 4-2 } =\frac { 4 }{ 2 } = 2 …(i)
Slope of the tangent = 2 (x – 2) ….(ii)
from (i) & (ii) 2 (x – 2)=2
∴ x – 2 = 1 or x = 3
when x = 3,y = (3 – 2)² = 1
∴ The tangent is parallel to the chord AB at (3,1)

Ex 6.3 Class 12 Maths Question 9.
Find the point on the curve y = x3 – 11x + 5 at which the tangent is y = x – 11.
Solution:
Here, y = x3 – 11x + 5
\frac { dy }{ dx } = 3x² – 11
The slope of tangent line y = x – 11 is 1
∴ 3x² – 11 = 1
⇒ 3x² = 12
⇒ x² = 4, x = ±2
When x = 2, y = – 9 & when x = -2,y = -13
But (-2, -13) does not lie on the curve
∴ y = x – 11 is the tangent at (2, -9)

Ex 6.3 Class 12 Maths Question 10.
Find the equation of all lines having slope -1 that are tangents to the curve y=\frac { 1 }{ x-1 }, x≠1
Solution:
Here
y=\frac { 1 }{ x-1 }
\frac { dy }{ dx } =\frac { -1 }{ { (x-1) }^{ 2 } }
NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives 10

Ex 6.3 Class 12 Maths Question 11.
Find the equation of ail lines having slope 2 which are tangents to the curve y=\frac { 1 }{ x-3 }, x≠3.
Solution:
Here
y=\frac { 1 }{ x-3 }
\frac { dy }{ dx } ={ (-1)(x-3) }^{ -2 }=\frac { -1 }{ { (x-3) }^{ 2 } }
∵ slope of tangent = 2
\frac { -1 }{ { (x-3) }^{ 2 } } =2\Rightarrow { (x-3) }^{ 2 }=-\frac { 1 }{ 2 }
Which is not possible as (x – 3)² > 0
Thus, no tangent to y=\frac { 1 }{ x-3 } has slope 2.

Ex 6.3 Class 12 Maths Question 12.
Find the equations of all lines having slope 0 which are tangent to the curve y=\frac { 1 }{ { x }^{ 2 }-2x+3 }
Solution:
Let the tangent at the point (x1, y1) to the curve
byjus class 12 maths Chapter 6 Application of Derivatives 12

Ex 6.3 Class 12 Maths Question 13.
Find points on the curve \frac { { x }^{ 2 } }{ 9 } +\frac { { y }^{ 2 } }{ 16 } =1 at which the tangents are
(a) parallel to x-axis
(b) parallel to y-axis
Solution:
The equation of the curve is \frac { { x }^{ 2 } }{ 9 } +\frac { { y }^{ 2 } }{ 16 } =1…(i)
NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives 13

Ex 6.3 Class 12 Maths Question 14.
Find the equations of the tangent and normal to the given curves at the indicated points:
(i) y = x4 – 6x3 + 13x2 – 10x + 5 at (0,5)
(ii) y = x4 – 6x3 + 13x2 – 10x + 5 at (1,3)
(iii) y = x3 at (1, 1)
(iv) y = x2 at (0,0)
(v) x = cos t, y = sin t at t = \frac { \pi }{ 4 }
Solution:
\frac { dy }{ dx } ={ 4x }^{ 3 }-18{ x }^{ 2 }+26x-10
Putting x = 0, \frac { dy }{ dx } at (0,5) = – 10
NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives 14
NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives 14.1

Ex 6.3 Class 12 Maths Question 15.
Find the equation of the tangent line to the curve y = x2 – 2x + 7 which is
(a) parallel to the line 2x – y + 9 = 0
(b) perpendicular to the line 5y – 15x = 13.
Solution:
Equation of the curve is y = x² – 2x + 7 …(i)
\frac { dy }{ dx } = 2x – 2 = 2(x – 1)
(a) Slope of the line 2x – y + 9 = 0 is 2
⇒ Slope of tangent = \frac { dy }{ dx } = 2(x – 1) = 2
byjus class 12 maths Chapter 6 Application of Derivatives 15

Ex 6.3 Class 12 Maths Question 16.
Show that the tangents to the curve y = 7x3 + 11 at the points where x = 2 and x = – 2 are parallel.
Solution:
Here, y = 7x3 + 11
=> x \frac { dy }{ dx } = 21 x²
Now m1 = slope at x = 2 is { \left( \frac { dy }{ dx } \right) }_{ x=2 } = 21 x 2² = 84
and m2 = slope at x = -2 is { \left( \frac { dy }{ dx } \right) }_{ x=-2 } = 21 x (-2)² = 84
Hence, m1 = m2 Thus, the tangents to the given curve at the points where x = 2 and x = – 2 are parallel

Ex 6.3 Class 12 Maths Question 17.
Find the points on the curve y = x3 at which the slope of the tangent is equal to the y-coordinate of the point
Solution:
Let P (x1, y1) be the required point.
The given curve is: y = x3
NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives 17

Ex 6.3 Class 12 Maths Question 18.
For the curve y = 4x3 – 2x5, find all the points at which the tangent passes through the origin.
Solution:
Let (x1, y1) be the required point on the given curve y = 4x3 – 2x5, then y1 = 4x13 – 2x15 …(i)
byjus class 12 maths Chapter 6 Application of Derivatives 18
NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives 18.1

Ex 6.3 Class 12 Maths Question 19.
Find the points on the curve x2 + y2 – 2x – 3 = 0 at which the tangents are parallel to the x-axis.
Solution:
Here, x2 + y2 – 2x – 3 = 0
=> \frac { dy }{ dx } =\frac { 1-x }{ y }
Tangent is parallel to x-axis, if \frac { dy }{ dx }=0 i.e.
if 1 – x = 0
⇒ x = 1
Putting x = 1 in (i)
⇒ y = ±2
Hence, the required points are (1,2), (1, -2) i.e. (1, ±2).

Ex 6.3 Class 12 Maths Question 20.
Find the equation of the normal at the point (am2, am3) for the curve ay2 = x3.
Solution:
Here, ay2 = x3
2ay\frac { dy }{ dx } ={ 3x }^{ 2 }\Rightarrow \frac { dy }{ dx } =\frac { { 3x }^{ 2 } }{ 2ay }
NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives 20

Ex 6.3 Class 12 Maths Question 21.
Find the equation of the normal’s to the curve y = x3 + 2x + 6 which are parallel to the line x + 14y + 4 = 0.
Solution:
Let the required normal be drawn at the point (x1, y1)
The equation of the given curve is y = x3 + 2x + 6 …(i)
NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives 21
NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives 21.1

Ex 6.3 Class 12 Maths Question 22.
Find the equations of the tangent and normal to the parabola y² = 4ax at the point (at²,2at).
Solution:
NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives 22

Ex 6.3 Class 12 Maths Question 23.
Prove that the curves x = y² and xy = k cut at right angles if 8k² = 1.
Solution:
The given curves are x = y² …(i)
and xy = k …(ii)
byjus class 12 maths Chapter 6 Application of Derivatives 23

Ex 6.3 Class 12 Maths Question 24.
Find the equations of the tangent and normal to the hyperbola \frac { { x }^{ 2 } }{ { a }^{ 2 } } -\frac { { y }^{ 2 } }{ { b }^{ 2 } } =1 at the point (x0 ,y0).
Solution:
NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives 24
NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives 24.1

Ex 6.3 Class 12 Maths Question 25.
Find the equation of the tangent to the curve y=\sqrt { 3x-2 } which is parallel to the line 4x – 2y + 5 = 0.
Solution:
Let the point of contact of the tangent line parallel to the given line be P (x1, y1) The equation of the curve is y=\sqrt { 3x-2 }
NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives 25

Choose the correct answer in Exercises 26 and 27.

Ex 6.3 Class 12 Maths Question 26.
The slope of the normal to the curve y = 2x² + 3 sin x at x = 0 is
(a) 3
(b) \frac { 1 }{ 3 }
(c) -3
(d) -\frac { 1 }{ 3 }
Solution:
(d) ∵ y = 2x² + 3sinx
\frac { dy }{ dx }=4x+3cosx at
x = 0,\frac { dy }{ dx }=3
∴ slope = 3
⇒ slope of normal is = \frac { 1 }{ 3 }

Ex 6.3 Class 12 Maths Question 27.
The line y = x + 1 is a tangent to the curve y² = 4x at the point
(a) (1,2)
(b) (2,1)
(c) (1,-2)
(d) (-1,2)
Solution:
(a) The curve is y² = 4x,
\frac { dy }{ dx } =\frac { 4 }{ 2y } =\frac { 2 }{ y }
Slope of the given line y = x + 1 is 1 ∴ \frac { 2 }{ y }=1
y = 2 Putting y= 2 in y² = 4x 2² = 4x
⇒ x = 1
∴ Point of contact is (1,2)

NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Hindi Medium Ex 6.3

NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.3 AOD
NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.3 in English medium
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The post NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Ex 6.3 appeared first on Learn CBSE.


How To Improve Your Aim In PUBG Mobile

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Aiming can contribute up to forty percent of your chicken dinner because stealth and strategy can only get you so far – when push comes to shove, the person with the better aim wins.

Aiming is one of the more important parts of PUBG Mobile. Know how to aim would dramatically improve your rate of survival, as everything can be decided in just a single moment of weakness. This guide will provide you with the tips and tricks needed to get better aim – to take down the enemies before they act.

Know your weapon

PUBG Mobile Know Your Weapon
The training mode in which you can test every weapon in the game – at every distance

Firstly, you need to adjust your aim based on whatever weapons you are using, as each one has their own quirk and features. Always check your weapon’s stats, especially range, firing modes, recoil, spray, and bullet speed. After getting your hand on a weapon, you might even fire a few shots to test it out. However, you would have to make sure that there are no enemies nearby first.

Actually, it is best to learn this outside of the matches, as you can test just about every gun in the weapon Practice Range. There are also stationary and moving targets, so it is just as effective as in a normal match.

Pubg Mobile : How To AIM Like A PRO! And Control The Shakiness/Recoil Of The Gun For A Perfect Shot!

Dealing with moving targets

PUBG Mobile Dealing with moving targets
Know your weapon is crucial, as some weapon like the AK is very hard to use

Follow a moving target by tracking them with your cursor – this would also require a degree of prediction, as you have to keep up with enemy movements without knowing their trajectory.

Snipers and DMRs can have different bullet speeds and drops
Snipers and DMRs can have different bullet speeds and drops

If you are new to track, it might be better to just aim for the body, as it is a bigger target. Getting a headshot is of course recommended, but it might be tricky sometimes. In a melee, tracking is much easier, as you can just full auto on your enemies to take them down.

Reading your enemies is important – sometimes you should not fire immediately as soon as an enemy enters your sight from a safe distance. Spend like a few seconds to ascertain the situation, check and predict their movement. This momentarily delay might be crucial, or even a deciding factor between a hit and a miss. For example, if an enemy is moving to the right, you can place your cursor on his way and only fire when he has entered your cursor’s area.

Taking bullet drops and bullet speed into consideration

Aiming at long range requires you to take a lot of factors into consideration
Aiming at long range requires you to take a lot of factors into consideration

Some longer-range weapons such as the sniper rifles have slow bullet speed – which means the bullet would take longer to hit the designated location. To use those kinds of guns effectively, you have to circumvent their weaknesses by predicting the enemy movements.

Adjust your aim to compensate for the bullet drops by aiming a bit higher than your intended target – if you are gunning for a headshot, aim a bit higher.

Customize Your Settings to Aim Better

SouL MortaL's Settings for the best PUBG Experience
SouL MortaL’s Settings for the best PUBG Experience

Change the game’s settings to your own preference would assist you heavily in aiming and moving around in-game. Every small detail is important, even the cursor colors.

Changing your device’s sensitivity can assist heavily in quite a few situations, as lower sensitivity would be the best setting for close quarter combats while higher sensitivity is best for long-distance. The reason behind this is that you would need a stable cursor for the moving enemies in close range – on longer-range however, you would have to be quick to be able to follow enemies movement.

Pick one of the two shooting options – the second one is easier to use, as you can both shoot and aim with just one finger.

Tracking is Best to Hit Enemies in Close Combat

Tracking is Best to Hit Enemies in Close Combat

The closer the enemy, the easier it is to follow their movements as they’ll be bigger targets. Turn on your burst fire or full auto mode to gun them down quickly!

Top 10 PUBG Weapons

How to Move + Shoot + Aim + Peek simultaneously Using Two Thumbs | No need for claw anymore! PubgM

Articles on PUBG Mobile

How To Set Up A Successful Ambush In PUBG Mobile

Camping and set up ambushes are the best ways to get a Chicken Dinner in PUBG Mobile – this guide would list out various useful tips and tricks for you to recreate this strategy.

Setting up a trap and flank from multiple directions are amongst the core strategies for taking out the enemy squads in PUBG Mobile – or even opponents in duo or solo mode. Ambushes are not hard to set up, however, it would require some map knowledge and coordination between you and your teammates. While ambushes in solo mode are possible – the strategy is way more effective with an extra hand or two in Duos or Squads.

Setting up a trap is one of the best ways to improve your KD score with minimal risk
Setting up a trap is one of the best ways to improve your KD score with minimal risk

Our basic guide below would give you a general idea of how to create an ambush and gain advantages in combat.

Firstly, you would have to pick a location. There are a lot of good camping locations in PUBG – and many more that you can create by yourself. However, there are a few simple rules that could make a successful ambush.

Holding out on a vantage point over the bridge

You have to put yourself in the enemies’ shoes: where would they go first when approaching a building? Where would the enemies go when the blue zone is spawned so far from this location. With that knowledge in mind, you can either use a jeep to block the bridge for a bridge camp or stand on top of the stairs waiting for enemies to poke their head up. Remember that the hunter can also become the hunted – if you picked a large building with many exits, there would be times that the enemy slips through your traps and flank you.

Drop a weapon or some item to distract the enemy
Drop a weapon or some item to distract the enemy

Secondly, you can make use of their psychological weakness: Greed. Dropdown a gun or a medkit and wait nearby for unsuspecting enemies to fall into your trap. In Duo or Squad mode, the airdrops might also be a great place for an ambush as well, as the enemies would be too focused to get their hand on the sweet loot instead of looking out for a trap. You would need to prepare yourself for a fight, however, as many enemies would be attracted to this as well.

 

 

Structures with only one way to get in are prime targets to set up an ambush
Structures with only one way to get in are prime targets to set up an ambush

Lastly, be patient. PUBG is a Survival Battle Royale game – you can just play by yourself until much later in the match when the safe zone shrinks. Sometimes you don’t even have to get the loots, as killing your enemies and picking up the gears from their cold dead body is way easier. Pick a place to lie down and set up your camp – you would be on the top sooner or later.

Top 10 PUBG Mobile Players in India

Top 10 PUBG Mobile Players in India

Receiving 9 awards and being nominated over 25 times, PlayerUnknown’s Battlegrounds or PUBG by PUBG Corporation of South Korean developer Bluehole Inc., has taken the world by storm. This critically acclaimed multiplayer online battle royale video game designed by Brendan Greene has turned the gaming industry upside down for good.

Making over US$4.6 billion approximately within the first 6 months of it’s launch on March 2017, PUBG became one of the most watched and played video games on platforms like Twich and YouTube. Soon on 9 February 2018, PUBG mobile was launched by PUBG corporation and Tencent Games.

This video game brought a new era of gaming and live streaming with opportunities for thousands of streamers around the world. While everyone is having fun with this game, how could have India stayed any behind having so many die-hard PUBG fans? With excellent gameplays and entertaining interaction skills, Indian PUBGM streamers have won hearts of millions. We have listed top 10 players/streamers focused on playing PUBG mobile.

Top 10 Best & Beloved PUBG Mobile Players in India

MortaL

Who doesn’t know the heart of Soul clan, MortaL aka Naman Mathur. He started his YouTube channel on 15 September, 2016. The first videos that he posted on his channel were of his gameplays of Mini Militia. He soon got viral from his absolute epic gameplay video clips of PUBGM.
MortaL plays PUBGM on tab and those of you who are not aware enough, you might need to watch out if this guy is put in the same server as yours for how stunningly excellent he is at what he does. Mathur undoubtedly is one of the most respected PUBGM streamers on YouTube in India.

Poor gamer clan

Poor gamer aka Mayank Yadav from Lucknow started his YouTube gaming journey on 25 December, 2017. The first few videos he uploaded was of his gameplay of the video game Rules of Survival. He soon then started playing PUBGM and within a year, Yadav gained enormous fame and affection because of his gameplay, down-to-earth nature, positive attitude and sense of humour.

Today beloved Padulal is one of the most successful and loved streamers on YouTube India.

CarryisLive

“Toh kaise hain aap log”? This man needs no introduction and if you don’t know him yet, you’re missing out on some great YouTube gem. Carry Minati aka Ajey Nagar from Faridabad, India is one of the biggest Youtubers in the nation. He started his channel CarryisLive on 9 Jan, 2017. He is one of the earliest progressive video game streaming channels in India. Call him a pioneer.

With a witty sense of humour Carry has got his audience rolling on the floor. He is one of the most renowned Youtubers and streamers in India today.

Dynamo Gaming

“Patt se headshot”. Does that sound any familiar to you? We all love Dynamo, don’t we? Aditya Sawant aka Dynamo is one of the biggest gaming content creators on YouTube India.
Sawant again is one of the earliest gaming streamers in India. He took his first step into gaming on 24 March, 2015. He started playing PUBG on 2017 and since then there has been no turning back him for him.

The RawKnee Games

The RawKnee Show is a comedic roast channel on YouTube owned by The RawKnee aka Rony Dasgupta from Mumbai. He created his gaming channel called The RawKnee Games and stepped into the world of YouTube gaming on 4 April, 2017 playing FIFA.
Starting from video games like CSGO to games like Mad Max, RawKnee is no behind when it comes to trying out different types of video games. His mad gameplay, energetic humour and absolute unique commentary makes him stand out and is one the best Indian gaming streamers on YouTube.

BeastBoyShub

BeastBoyShub started his channel on 24 September, 2016. He did not start as a gaming channel exactly, but he soon began posting videos of his gameplays of various video games. He set his foot into PUBG PC by uploading his gameplay videos on 2017 after which he gained a wide range of followers.
He doesn’t just plays PUBG or PUBGM, but he also other experiments with video games of different genres like Granny, Resident Evil, Evil Within, Walking Dead and Tomb Raider.
BeastBoyShub is definitely one of the biggest gaming channels on YouTube in India and the funny Hindi commentaries just keeps adding to the goodness.

Kronten Gaming

Chetan Chandgude aka Kronten started his streaming career on 4 March, 2018. The first video he streamed was of him playing PUBG. Within no time Kronten rose to fame because of his great interaction skills with his audience and his very well known ‘rush gameplay’. This streamer with an uncountable number of ‘chicken dinners’ has a massive fan following and is undoubtedly one of the most supported gamers in India.

Cosmic YT

Inderpal Singh aka Cosmic YT is again one of the most well known PUBGM streaming gamer on YouTube in India.
Singh started his channel on 5 March, 2017. His first video was of him live streaming his gameplay of Grand Theft Auto V or GTA V. He played games like GTA and Resident Evil 7 for the first few months, soon then he started streaming PUBGM which caught more and more eyes for him everyday which just keeps growing and growing. His killer gameplay at PUBG mobile and his polite nature are some of the examples of his many good traits that makes his fans absolutely love him.

MDisCrazy

MDisCrazy is one of the few famous Indian PUBGM streamers who was absolutely adored by his fans within no time. Starting his channel not too long ago, he surely is one of the most promising YouTube gamers in India. He uploaded his first video of his PUBGM gameplay highlights on 6 July 2018 which got a lot of attention for of his rad FPP/First Person Perspective gameplay.
He often plays with some other Indian YouTube streamers like Ron Gaming and Cosmic YT.

Ron Gaming

Ron Gaming from Gandhinagar, Gujrat is one of the most well known and versatile gaming channels on YouTube in India. He started his channel on 30 July 2016. The first video he uploaded was of his GTA V gameplay. Initially, he played video games like Rainbow Six Siege, Destiny 2, Far Cry, Dead By Daylight, Ark Survival and of course GTA V. He started streaming PUBG in the year 2018 and today with a large fan following Ron is one of the most down to earth and prestigious gamers on YouTube.

Note: Dates mentioned are according to the oldest uploaded video on their channels.

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NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Ex 6.4

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NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Ex 6.4

Get Free NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Ex 6.4 PDF in Hindi and English Medium. Sets Class 12 Maths NCERT Solutions are extremely helpful while doing your homework. Application of Derivatives Exercise 6.4 Class 12 Maths NCERT Solutions were prepared by Experienced LearnCBSE.in Teachers. Detailed answers of all the questions in Chapter 5 Class 12 Application of Derivatives Ex 6.4 provided in NCERT Textbook.

Free download NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Ex 6.4 PDF in Hindi Medium as well as in English Medium for CBSE, Uttarakhand, Bihar, MP Board, Gujarat Board, BIE, Intermediate and UP Board students, who are using NCERT Books based on updated CBSE Syllabus for the session 2019-20.

The topics and sub-topics included in the Applications of Derivatives chapter are the following:

Section NameTopic Name
6Applications of Derivatives
6.1Introduction
6.2Rate of Change of Quantities
6.3Increasing and Decreasing Functions
6.4Tangents and Normals
6.5Approximations
6.6Maxima and Minima
6.7Maximum and Minimum Values of a Function in a Closed Interval
6.8Summary

NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Ex 6.4

Ex 6.4 Class 12 Maths Question 1.
Using differentials, find the approximate value of each of the following up to 3 places of decimal.
(i) \sqrt { 25.3 }
(ii) \sqrt { 49.5 }
(iii) \sqrt { 0.6 }
(iv) { \left( 0.009 \right) }^{ \frac { 1 }{ 3 } }
(v) { \left( 0.999 \right) }^{ \frac { 1 }{ 10 } }
(vi) { \left( 15 \right) }^{ \frac { 1 }{ 4 } }
(vii) { \left( 26 \right) }^{ \frac { 1 }{ 3 } }
(viii) { \left( 255 \right) }^{ \frac { 1 }{ 4 } }
(ix) { \left( 82 \right) }^{ \frac { 1 }{ 4 } }
(x) { \left( 401 \right) }^{ \frac { 1 }{ 2 } }
(xi) { \left( 0.0037 \right) }^{ \frac { 1 }{ 2 } }
(xii) { \left( 26.57 \right) }^{ \frac { 1 }{ 3 } }
(xiii) { \left( 81.5 \right) }^{ \frac { 1 }{ 4 } }
(xiv) { \left( 3.968 \right) }^{ \frac { 3 }{ 2 } }
(xv) { \left( 32.15 \right) }^{ \frac { 1 }{ 5 } }
Solution:
(i) y + ∆y = \sqrt { 25.3 }
= \sqrt { 25+0.3 }
= \sqrt { x+\Delta x }
∴ x = 25
∆x = 0.3
⇒ y = √x
NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives 1
NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives 1.1
NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives 1.2
tiwari academy class 12 maths Chapter 6 Application of Derivatives 1.3
NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives 1.4

Ex 6.4 Class 12 Maths Question 2.
Find the approximate value of f (2.01), where f (x) = 4x² + 5x + 2
Solution:
f(x+∆x) = f(2.01), f(x) = f (2) = 4.2² + 5.2 + 2 = 28,
f’ (x) = 8x + 5 Now, f(x + ∆x) = f(x) + ∆f(x)
= f(x) + f’ (x) • ∆x = 28 + (8x + 5) ∆x
= 28 + (16 + 5) x 0.01
= 28 + 21 x 0.01
= 28 + 0.21
Hence,f(2 x 01)
= 28 x 21.

Ex 6.4 Class 12 Maths Question 3.
Find the approximate value of f (5.001), where f(x) = x3 – 7x2 +15.
Solution:
Let x + ∆x = 5.001, x = 5 and ∆x = 0.001,
f(x) = f(5) = – 35
f(x + ∆x) = f(x) + ∆f(x) = f(x) + f'(x).∆x
= (x3 – 7x² + 15) + (3x² – 14x) × ∆x
f(5.001) = – 35 + (3 × 5² – 14 × 5) × 0.001
⇒ f (5.001) = – 35 + 0.005
= – 34.995.

Ex 6.4 Class 12 Maths Question 4.
Find the approximate change in the volume V of a cube of side x metres caused by increasing the side by 1%.
Solution:
The side of the cube = x meters.
Increase in side = 1% = 0.01 × x = 0.01 x
Volume of cube V= x3
∴ ∆v =\frac { dv }{ dx } × ∆x
= 3x² × 0.01 x
= 0.03 x3 m3

Ex 6.4 Class 12 Maths Question 5.
Find the approximate change in the surface area of a cube of side x metres caused by decreasing the side by 1%.
Solution:
The side of the cube = x m;
Decrease in side = 1% = 0.01 x
Increase in side = ∆x = – 0.01 x
Surface area of cube = 6x² m² = S
\frac { ds }{ dx } × ∆x = 12x × (- 0.01 x)
= – 0.12 x² m².

Ex 6.4 Class 12 Maths Question 6.
If the radius of a sphere is measured as 7m with an error of 0.02 m, then find the approximate error in calculating its volume.
Solution:
Radius of the sphere = 7m : ∆r = 0.02 m.
Volume of the sphere V = \frac { 4 }{ 3 } \pi { r }^{ 3 }
\Delta V=\frac { dV }{ dr } \times \Delta r=\frac { 4 }{ 3 } .\pi .3{ r }^{ 2 }\times \Delta r
= 4π × 7² × 0.02
= 3.92 πm³

Ex 6.4 Class 12 Maths Question 7.
If the radius of a sphere is measured as 9 m with an error of 0.03 m, then find the approximate error in calculating its surface area.
Solution:
Radius of the sphere = 9 m: ∆r = 0.03m
Surface area of sphere S = 4πr²
∆s = \frac { ds }{ dr } × ∆r
= 8πr × ∆r
= 8π × 9 × 0.03
= 2.16 πm².

Ex 6.4 Class 12 Maths Question 8.
If f (x) = 3x² + 15x + 5, then the approximate value of f (3.02) is
(a) 47.66
(b) 57.66
(c) 67.66
(d) 77.66
Solution:
(d) x + ∆x = 3.02, where x=30, ∆x=.02,
∆f(x) = f(x + ∆x) – f(x)
⇒ f(x + ∆x) = f(x) + ∆f(x) = f(x) + f’ (x)∆x
Now f(x) = 3×2 + 15x + 5; f(3) = 77, f’ (x) = 6x + 15
f’ (3) = 33
∴ f (3.02) = 87 + 33 x 0 02 = 77.66

Ex 6.4 Class 12 Maths Question 9.
The approximate change in the volume of a cube of side x metres caused by increasing the side by 3% is
(a) 0.06 x³ m³
(b) 0.6 x³ m³
(c) 0.09 x³ m³
(d) 0.9 x³ m³
Solution:
(c) Side of a cube = x meters
Volume of cube = x³,
for ∆x. ⇒ 3% of x = 0.03 x
Let ∆v be the change in v0l. ∆v = \frac { dv }{ dx } x ∆x = 3x² × ∆x
But, ∆x = 0.03 x
⇒ ∆v = 3x² x 0.03 x
= 0.09 x³m³

NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Hindi Medium Ex 6.4

NCERT Solutions for Class 12 Maths Exercise 6.4 AOD
12 Maths ex. 6.4
approximation class 12 maths
6.4 class 12
12 Maths 6.4 AOD
AOD 6.4 maths
NCERT Solutions for Class 12 Maths Exercise 6.4 AOD in English Medium
12 Maths Exercise 6.4
12 Maths Exercise 6.4 in Hindi Medium

Class 12 Maths NCERT Solutions

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Scholarships for Students

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Scholarships for Students

State Wise Scholarships

Top Scholarships for Class 1 to Class 10

Popular Scholarships in India

Best Scholarships in India

  • NTSE – National Talent Search Examination
  • KVPY – Kishore Vaigyanik Protsahan Yojana 2019
  • NSP – National Scholarship Portal

State Level National Talent Search Exam (NTSE) 2019 – 2020

Top Scholarships for Class 1 to Class 10

Pre Matric Scholarship for SC, ST & General Students, Uttar Pradesh 2019-20

Eligibility
  • Be a domicile of Uttar Pradesh
  • Be studying in class 9 and 10
  • Belong to ST/SC/General Category
  • Have a family income of not more than INR 1 Lakh per annum
Important Dates
  • Last date to apply under fresh application: 10 September 2019
  • Last date to submit an application for renewal: 10 August 2019
Important Links

Pre Matric Scholarship for OBC Students, Uttar Pradesh 2019-20

Eligibility
  • Be a domicile of Uttar Pradesh
  • Belong to OBC category
  • Have a family income less than INR 1 Lac per annum from all sources
Important Dates
  • Last date to apply under fresh application: 10 September 2019
  • Last date to submit an application for renewal: 10 August 2019
Important Links

Pre Matric Scholarship for Minority Students, Uttar Pradesh 2019-20

Eligibility
  • Be a domicile of Uttar Pradesh
  • Belong to the Minority community
  • Have the family income of less than INR 1 Lakh per annum from all sources
  • Be studying in class 9 and 10
Important Dates
  • Last date to apply under fresh application: 10 September 2019
  • Last date to submit application for renewal: 10 August 2019
Important Links

West Bengal Pre Matric Scholarship

Eligibility
  • Be a domicile of West Bengal
  • Be studying in class 1 to 10 at a school or institution which is affiliated with a Council/Educational Board/University of State or Central Government
  • Have scored at least 50% marks in his/her last qualifying exam
  • Have a family income of less than INR 2 Lakhs per annum
  • Belong to minority section of West Bengal

Note – Students pursuing their studies outside West Bengal are not eligible for this scholarship

Important Dates
  • Application deadline – 15th September 2019
Important Links

National Scholarship Exam (NSE) 2019

Eligibility
  • Students in Class 5 to 12 (SSC/ICSE/CBSE board)
  • Students enrolled in a Diploma course (any stream and any year)
  • Students enrolled in a Degree (any stream and any year)
Important Dates
  • Registrations open: 1 April 2019
  • Last date to apply: 30 September 2019
  • Date of exam: 15th December 2019
Important Links

Kind Scholarship for Meritorious Students

Kind Scholarship for Meritorious Students is an initiative to support academically meritorious students coming from under-privileged section of society to build a foundation for their bright future.  The scholarship is funded by  donations received from Individuals & professionals who want to support meritorious students in need.

Scholarship award varies as per the need of candidates. However, the standard scholarship awards are the following:-

  • For Class 9 & 10 students: INR 6,000 per year
  • For Class 11 & 12 students: INR 12,000 per year
  • For Polytechnic / ITI / Diploma / Graduation & Others: INR 18,000 per year
Eligibility
  • Be studying in any Indian school (Class 9 to 12) or college (Graduation; ITI, Polytechnic, Vocational & Professional Courses)
  • Have an annual household income less than INR 4 Lakhs
  • Have cleared the previous examination with at least 60% marks

Note: Preference will be given to candidates having high merit, single parent, orphan or wards of Indian Armed Forces died on duty.

Important Dates
  • Starting date: July 1, 2019
  • End date: September 30, 2019

Begum Hazrat Mahal National Scholarship Scheme for Minorities Girls 2019-20

Eligibility
  • Only girl students belonging to Muslim, Christian, Sikh, Buddhist, Jain, and Parsi communities are eligible for this scholarship.
  • They must be studying in class 9 to 12.
  • The annual family income of the student should be less than INR 2 Lakhs.
  • The students must have secured a minimum of 50% marks (in aggregate) in the previous class.
Important Dates
  • Last date to submit the application: 30th September 2019
Important Links

24th Anuvrat Essay Writing Competition 2019

Eligibility
  • Junior Group: Class 1 to 8
  • Senior Group: Class 9 to 12

What are the benefits?

  • 3 graded prizes for 1st, 2nd  and 3rd winning essays in each group
  • Some consolation prizes for each participating group
  • Honor for best participants
  • Merit/appreciation certificate for commendable essay writers
Important Dates
  • Deadline for essay submission – 30th September 2019
  • Result declaration – December 2019
Important Links

The Gaud Saraswat Brahman (G.S.B.) Scholarship 2019

The scholarship offers the following-

  • Free scholarship- Awarded to students from Class 5 to Graduate level
  • Foreign loan scholarship- Up to INR 17 lakhs (INR 1.7 million) with a loan scholarship of INR 2 lakhs per student
Eligibility
  • Applicants must belong to the Gaud Saraswat Brahman community.
  • Students of class 5th to undergraduate courses will be eligible for this scholarship scheme.
  • All Students enrolled in courses like diploma /B.A/B.Sc/B.Com/B.Arch/MBBS/B.Pharma/B.E can also apply.
Important Dates
  • Application deadline for students of Class 5 to 10 – 15th July 2019
  • Application deadline for students of PUC I & II/Jr. College – 15th August 2019
  • Application deadline for students of Degree (BA/BSc/B.Com) & Diploma – 15th September 2019
  • Application deadline for students of BE/BArch/BPharm/MBBS etc. – 15th October 2019
Important Links

Pre Matric Scholarships Scheme for Minorities 2019-20

Eligibility
  • Must be studying in class 1 to 10
  • Must have secured at least 50% marks in the previous final examination
  • Must have a family income of not more than INR 1 lakh per annum from all sources
  • Must belong to the minority community (Muslims, Sikhs, Christians, Buddhists, Jains and Zoroastrians/Parsis)
Important Dates
  • Scholarship announcement date: 15 July 2019
  • Scholarship closing date: 15 October 2019
  • Last date for defect verification: 31 October 2019
  • Last date for Institute verification: 31 October 2019
Important Links

Pre-Matric Scholarship for Students with Disabilities 2019-20

Eligibility
  • Be a regular full-time student of class 9 or class 10 in government or recognised school.
  • Have more than 40% disability and a valid certificate for the same.
  • Belong to the family where the annual income is not more than INR 2.50 Lakhs from all the sources.
Important Dates
  • Scholarship announcement date: 15 July 2019
  • Scholarship closing date: 15 October 2019
  • Last date for defect verification: 31 October 2019
  • Last date for Institute verification: 31 October 2019
Important Links

Financial Assistance for Education of the Wards of Beedi/Cine/IOMC/LSDM Workers – Post-Matric 2019-20

The selected scholar will receive the following benefits:

  • Girls studying in PUC I & II will receive INR 2,440
  • Boys studying in PUC I & II will receive INR 2,000
  • Girls and boys enrolled in ITI courses will receive INR 10,000
  • Girls and boys enrolled in Degree courses will receive INR 3,000
  • Girls and boys enrolled in Professional courses such as BE/MBBS/BSc-Agri. will receive INR 15,000
Eligibility
  • One or both the parents of the applicants should be working as a Beedi worker/ Iron Ore Manganese & Chrome Ore Mines (IOMC) worker/ Limestone & Dolomite Mines (LSDM) worker/ Cine worker for at least last six months.
  • The monthly income of the family must not exceed INR 10,000 from all the sources except for Cine workers where the income must not exceed INR 8,000 per month or INR 1,00,000 per annum.
  • The student must have passed the last qualifying examination in the first attempt.
  • The student must have enrolled in a regular course of general or technical education, including medical, engineering and agricultural studies.

Note: This scholarship is also for the children of Contract/ Gharkhataworkers.

Important Dates
  • Scholarship announcement date: 15 July 2019
  • Scholarship closing date: 31 October 2019
  • Last date for defective verification: 15 November 2019
  • Last date for institute verification: 15 November 2019
Important Links
  • please visit NSP website https://scholarships.gov.in/

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NEC Scholarship 2019 | Dates, Eligibility, Rewards, Application, Selection Process

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NEC Scholarship 2019: The Directorate of Technical Education (DTE), Assam will release the notification for North Eastern Council (NEC) Scholarship 2019. The NEC Scholarship offers support to economically backward class students of the North Eastern Region for pursuing higher education. Students should pursue engineering and professional courses from the approved list by NEC. The application for NEC Scholarship releases in the official website of DTE, Assam on October 2019. Candidates can fill the application and apply for this scholarship till the mid of November 2019. NEC Scholarship provides financial help to the disadvantaged students of the North Eastern Region every year.

Candidates studying in several disciplines within or outside North-East but within the country can avail the benefits of NEC Scholarship. Shortlisted candidates will get NEC Scholarship amount of Rs. 30,000/- from the organization. The article below provides more information that is the date, eligibility, and so forth about NEC Scholarship 2019.

NEC Scholarship 2019

NEC Scholarship 2019 is the initiatives of DTE, Assam to encourage outstanding disadvantaged students for pursuing higher education. DTE, Assam offers NEC scholarship to students who desire to study in diploma, degree, Ph.D. and M.Phil courses. DTE, Assam organizes NEC Scholarship only for the students of North-Eastern Region. NEC Merit scholarship helps lower-income class students of Assam to complete engineering and professional courses without financial hurdle.

NEC Scholarship Overview

ParticularsDetails
Organization NameDirectorate of Technical Education, Assam
Scholarship NameNEC Merit Scholarship
Amount of ScholarshipRs. 30,000 per annum to each scholar
Applicable StateAssam
Courses OfferedDiploma / Certificate courses
Bachelor Degree
Post Graduation Degree
Ph.D.
M.Phil
Official Websitewww.dte.assam.gov.in

Refer to the below list of courses covered in diploma, degree and postgraduate level as per Annexure I, II and III in the official notification.

Scholarships for Students

List of Courses Covered in Diploma Level As per Annexure – I

Name of SubjectCourses
Textile TechnologyDiploma in Textile Technology
Computer Science & EngineeringBCA, DCA, PGDCA, Diploma in Computer Science, Computer Engineering
ElectronicsDiploma in Electronics Engineering
Telecom & Communication EngineeringDiploma in Telecom Engineering
Automobile EngineeringDiploma in Automobile Engineering
Hotel/Travel & Tourism ManagementDiploma in Hotel Management for 3 years, PG Diploma in Travel & Tourism Management (1 year)
The Teaching of Physically/ Mentally HandicappedDiploma in Teaching of Physically/ Mentally Handicapped
Ceramic & Glass TechnologyDiploma in Glass & or Ceramic Technology
Physical Education and SportsDiploma in Physical Education
Fashion Technology & DesignDiploma in Fashion Technology & Design
Para-Medical including General NursingDiploma in Pharmacy, General Nursing & Psychiatric Nursing
Foreign Trade ManagementDiploma in Foreign Trade Management
ForestryB.Sc. (Forestry) 3 years course.
Ecology & EnvironmentB.Sc. (Ecology & Environment) 3 years course.
GemologyDiploma in Gemology
Tea TechnologyDiploma in Tea Technology
Leather TechnologyDiploma in Leather Technology
Resin TechnologyDiploma in Resin Technology
Plastic TechnologyDiploma in Plastic Technology
Mass CommunicationBA/B.Sc. in Mass Communication
Interior DecorationDiploma in Interior Decoration
Indian System of MedicineDiploma in Indian System of Medicine
Bio-TechnologyDiploma in Bio-Tech, 3 years course
Business ManagementBBA/BBM

List of Courses Covered in Degree Level As per Annexure-II

Name of SubjectCourses
Dairy TechnologyB.Tech./B.E.
Food TechnologyB.Tech./B.E.
Ecology & EnvironmentB.Tech./M.Sc.
ArchitectureB. Arch.
Mining/Metallurgical Engg.B.Tech./B.E.
Textile TechnologyB.Tech./B.E.
ElectronicsB.Tech./B.E./M.Sc. (Electronics)
Computer ScienceB.Tech./B.E./M.Sc. (Computer Science)
Computer EngineeringB.Tech./B.E.
Computer ApplicationM.C.A.
Foreign Trade ManagementDegree in Foreign Trade Management
Physical Education & SportsB.P. ED., P.G. Diploma in Physical Education & sports.
Chemical TechnologyB.Tech./B.E.
Bio-TechnologyB.Tech. (Bio-Tech)/M.Sc. (Bio-Tech)
Fashion Technology & Design
Mass Communication & JournalismM.Sc. /P.G. Diploma
Petroleum EngineeringB.Tech./B.E.
ForestryB.Tech. (Forestry)/M.Sc. (Forestry)
Medical ScienceMBBS/BDS/BHMS/BAMS/BUMS/ B.PHARM.
Business AdministrationMBA(PGDM/PGDBM)
Information TechnologyB.Tech./B.E.
Physiotherapy(BPT – 4 years Course)
NursingB.Sc. (Nursing).

List of Courses Covered in Post Graduation Level As per Annexure – III

Name of SubjectCourses
Computer EngineeringM.Tech./M.E.
Computer ApplicationM.Tech.
Computer ScienceM.Tech.
Agriculture & Allied subjectM.Sc.(Agri)/M.V.Sc./M.F.Sc.
Civil EngineeringM.Tech./M.E.
Mechanical EngineeringM.Tech./M.E.
Electrical EngineeringM.Tech./M.E.
ElectronicsM.Tech./M.E.
ArchitectureM. Arch.
Medical SciencesMD/MS/PGD/Diploma/M.Pharm.
Bio-TechnologyM.Tech. (Bio-Tech)
Foreign TradeM.A. (Foreign Trade)
Physical Education & SportsM.A. (Physical Education)
Fine ArtsMFA
Food TechnologyM.Tech.
Information TechnologyM.Tech (IT)
ForestryM.Tech. (Forestry)
Environmental SciencesM.Tech.
Genetic TechnologyM.Tech.
Energy ManagementM.Tech.
NursingM.Sc. (Nursing)

NEC Scholarship 2019 – Important Dates

EventsImportant Dates
Start of NEC Scholarship ApplicationOctober 2019
Deadline to Submit NEC Scholarship ApplicationMid of November 2019
Application Scrutinization and Shortlisting of CandidatesDecember 2019
Declaration of NEC Scholarship ResultJanuary 2020
NEC Scholarship Disbursement to Shortlisted CandidatesMarch 31, 2020

NEC Scholarship 2019 – Eligibility Criteria

Candidates must fulfill the following eligibility criteria to apply for NEC Scholarship 2019:

  • Candidate must be a permanent resident of any of the states in the North-Eastern Region.
  • Candidates must possess the relevant qualifications for the chosen course from a recognized Institute/ University.
  • General candidates should secure at least 70% marks in the last qualifying examination for the chosen courses. Courses which includes diploma/ degree/ post-graduate/ Ph.D. / M.Phil.
  • SC/ ST candidates should secure at least 60% marks in the last qualifying examination for the chosen courses.
  • Candidates family income should be less than 4.5 Lakh per annum from all sources.
  • Candidate must not be employed.
  • Candidate should not be getting any other financial assistance from any other sources.

NEC Scholarship 2019 – Rewards

The rewards details for shortlisted candidates are given below

Level of Courses of Approved List by NECRewards (In Rs.)Book-grant Amount
DiplomaRs. 900/- per monthRs. 1000/- per annum
DegreeRs. 1000/- per monthRs. 1400/- per annum
Post-graduateRs. 1200/- per monthRs. 2000/- per annum
Ph.D. / M.Phil.Rs. 1500/- per monthRs. 3000/- per annum
  • M.Phil. rewards are sanctioned for a period of one and half year and can extend up to 6 months.
  • Ph.D. rewards and book-grants are generally sanctioned for a period of 3 years.
  • It can be extended for 1 more year based on the Head of the Institutions recommendation.

NEC Scholarship 2019 – Selection Process

The selection process for the NEC Scholarship are given below:

  • The NEC Scholarship is a merit-based scholarship. So, Selections for this scholarship are made on the basis of Merit.
  • This scholarship considers academic merit prescribed for each chosen courses for final selection.
  • State Govt. is responsible for the selection of candidates after the proper advertisement.
  • The selection procedure also needs to be followed as outlined at Annexure V in the official notification.
  • State Govt. will pay the stipend directly at stipulated rates for diploma level and above to the respective institute.
  • First, DTE scrutinizes the candidate’s application in December.
  • Later on, prepares the list of shortlisted candidates by January every year.
  • After selection, the sanctioned amount will be drawn within 31st March 2020.
  • Then, shortlisted candidates will be paid as per the sanctioned fund.
  • Direct Beneficiary Account transfer through ECS will be used to make payment to shortlisted candidates.

NEC Scholarship 2019 – Application Process

The application for the NEC Scholarship will be available online in October 2019. Candidates can fill and submit the application in the mid of November 2019. Refer to the below points to apply for NEC Scholarship 2019.

  • Candidates should visit the official website scholarships.gov.in.
  • Click on the “State Schemes” appeared on the home page.
  • Then, click on “NEC Merit Scholarship Tripura” appeared under the “Tripura” link.
  • Click on the “Guidelines” to download the NEC Merit Scholarship application form. The NEC application form will be available in pdf form.
  • Read all the NEC Scholarship instructions carefully with entire details.
  • After reading, go back to the previous page and click on “Apply”.
  • Login with email id and password to fill the online application.
  • Candidates are required to fill all the essential and mandatory details in capital letters.
  • Candidates must upload a scanned copy of their recent passport size photo with signature.
  • Candidates must attach self-attested copies of the required documents.
  • Candidates need to pay the application fee through proper manner.
  • Recheck the application thoroughly to avoid rejection by DTE during scrutinization.
  • Submit the completed application to the Head of the institutions on or before the due date.
  • The Head of the institutions verifies and fill the required details in the application. Then, forwards the print out of the application along with all relevant documents to DTE on 30th November 2019.

NEC Scholarship 2019 – Checklist of Documents

The below-mentioned documents must be uploaded along with the application to apply for NEC Scholarship 2019.

  • Recommendation by the Head of the institute
  • Permanent residence certificate
  • Undertaking not availed any other scholarship
  • Undertaking to refund stipend in case of furnishing false information
  • Attested copies of all the relevant certificates and mark Sheets
  • Letter of recognition of the course and institute where studying
  • In the case of Ph.D. and M.Phil. enclose synopsis
  • Candidates annual family income certificate and affidavit for any additional income
  • Attested copies of caste certificate and domicile certificate
  • Age proof (birth certificate or class 10th board certificate)
  • Residential proof (electricity bill, PAN card, voter Id, aadhar card, and so forth.)
  • The hard copy of the first page of the Bank passbook

NEC Scholarship 2019 – Renewal

Candidates must apply for the renewal of scholarship every year. Renewal of scholarship is generally for a period of 1 year from 1st April to 31st March of the next year. Each candidate can get the renewal form from the State Govt. and fill it up in the prescribed format. Thereon, the candidate must get a recommendation from the Head of the institute to renew this scholarship. The recommendation strictly based on the satisfactory progress of the candidate. Failure in any academic year will result in the discontinuation of the stipend for that entire period. Candidates future renewal should be recommended only from the date of clearance to the next higher class/semester. The sanction order along with the statement for the renewal of stipend should be endorsed to NEC. Refer the below points to Renew the NEC Scholarship application form.

  • Candidates should visit the official website of scholarships.gov.in to renew the application.
  • Login with email id and password to renew the existing application. Edit the old application form correctly which exists in the National Scholarship portal.
  • Ph.D./ M.Phil candidates must enclose the latest progress report from the guide attested by the Registrar.
  • Then, submit the NEC Renewal application form to the Head of the institution for the recommendation.
  • The Head of the Institution recommends that whether the application renewed to next year or not forwarded for renewal.

NEC Scholarship 2019 – Result

The candidate’s application along with documents will be scrutinized and shortlisted in December 2019 by DTE. After that, the result for NEC Scholarship will be published on the official website of DTE, Assam in January 2020. DTE prepares the final selection list of candidates by January every year. Candidates can check the list of selected candidates name on the DTE, Assam portal. Candidates can also search their name in the selection list to confirm their chances of receiving the NEC scholarship. After releasing the result, the sanctioned amount for NEC Scholarship will be drawn within 31st March 2020. Then, the sanctioned fund will be disbursed only to the shortlisted candidates by DTE, Assam on March 31, 2020.

NEC Scholarship 2019 – Contact Details

In case of any queries related to NEC Scholarship or any assistance contact on the below-mentioned details.

Phone Number+91-361-2382276
AddressDirectorate of Technical Education (DTE), Kahilipara, Guwahati, Assam – 781019
Email IDcoedte.assam@gov.in

FAQ’s on NEC Scholarship 2019

Question 1.
How is the NEC Scholarship helpful to North Eastern Region students?

Answer:
The NEC Scholarship offers support to economically backward class students of the North Eastern Region for pursuing higher education. Students should pursue engineering and professional courses from the approved list by NEC.

Question 2.
What are the courses offered under the NEC Scholarship?

Answer:
The following courses offered under the NEC Scholarship:

  • Diploma / Certificate courses
  • Bachelor Degree
  • Post Graduation Degree
  • Ph.D
  • M.Phil

Question 3.
What are the NEC Scholarship application starting and closing date?

Answer:
The NEC Scholarship application starting date is October 2019. Whereas the closing date is in the mid of November 2019.

Question 4.
What is the date for NEC Scholarship application scrutinization and selection of candidates?

Answer:
The date for Scholarship application scrutinization and selection of candidates is December 2019.

Question 5.
What is the date for NEC Scholarship Disbursement to shortlisted candidates?

Answer:
The date for NEC Scholarship Disbursement to shortlisted candidates is 31st March 2020.

Hope this article will help you to get more information about NEC Scholarship 2019. If you have queries related to NEC Scholarship, then leave it in the comment box to get in touch with us.

The post NEC Scholarship 2019 | Dates, Eligibility, Rewards, Application, Selection Process appeared first on Learn CBSE.

NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Ex 6.5

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NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Ex 6.4

Get Free NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Ex 6.5 PDF in Hindi and English Medium. Sets Class 12 Maths NCERT Solutions are extremely helpful while doing your homework. Application of Derivatives Exercise 6.5 Class 12 Maths NCERT Solutions were prepared by Experienced LearnCBSE.in Teachers. Detailed answers of all the questions in Chapter 6 Class 12 Application of Derivatives Ex 6.5 provided in NCERT Textbook.

Free download NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Ex 6.5 PDF in Hindi Medium as well as in English Medium for CBSE, Uttarakhand, Bihar, MP Board, Gujarat Board, BIE, Intermediate and UP Board students, who are using NCERT Books based on updated CBSE Syllabus for the session 2019-20.

The topics and sub-topics included in the Applications of Derivatives chapter are the following:

Section NameTopic Name
6Applications of Derivatives
6.1Introduction
6.2Rate of Change of Quantities
6.3Increasing and Decreasing Functions
6.4Tangents and Normals
6.5Approximations
6.6Maxima and Minima
6.7Maximum and Minimum Values of a Function in a Closed Interval
6.8Summary

NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Ex 6.5

Ex 6.5 Class 12 Maths Question 1.
Find the maximum and minimum values, if any, of the following functions given by
(i) f(x) = (2x – 1)² + 3
(ii) f(x) = 9x² + 12x + 2
(iii) f(x) = – (x – 1)² + 10
(iv) g(x) = x3 + 1
Solution:
(i) Minimum value of (2x – 1)² is zero.
Minimum value of (2x – 1)² + 3 is 3
Clearly it does not have maximum value,
(ii) f(x) = 9x² + 12x + 2
⇒ f(x) = (3x + 2)² – 2
Minimum value of (3 + 2)² is zero.
∴ Min.value of (3x + 2)² – 2 = 9x² + 12x + 2 is – 2
f (x) does not have finite maximum value
(iii) f(x) = – (x – 1)² + 10
Maximum value of – (x – 1)² is zero
Maximum valuer f f(x) = – (x – 1)² + 10 is 10
f (x) does not have finite minimum value.
(iv) As x—»∞,g(x)—»∞;Also x—»-∞,g(x)—»-∞
Thus there is no maximum or minimum value of f(x)

Ex 6.5 Class 12 Maths Question 2.
Find the maximum and minimum values, if any, of the following functions given by
(i) f(x) = |x + 2| – 1
(ii) g(x) = -|x + 1| + 3
(iii) h (x) = sin 2x + 5
(iv) f(x) = |sin(4x + 3)|
(v) h(x) = x + 1,x∈(-1,1)
Solution:
(i) We have :f(x) = |x + 2 |-1 ∀x∈R
Now |x + 2|≥0∀x∈R
|x + 2| – 1 ≥ – 1 ∀x∈R ,
So -1 is the min. value of f(x)
now f(x) = -1
⇒ |x + 2|-1
⇒ |x + 2| = 0
⇒ x = – 2
(ii) We have g(x) = -|x + 1| + 3 ∀x∈R
Now | x + 1| ≥ 0 ∀x∈R
-|x+ 1| + 3 ≤3 ∀x∈R
So 3 is the minimum value of f(x).
Now f(x) = 3
⇒ -|x+1| + 3
⇒ |x+1| = 0
⇒ x = – 1.
(iii) Thus maximum value of f(x) is 6 and minimum value is 4.
(iv) Let f(x) = |sin4x + 3|
Maximum value of sin 4x is 1
∴ Maximum value of |sin(4x+3)| is |1+3| = 4
Minimum value of sin 4x is -1
∴ Minimum value of f(x) is |-1+3| = |2|= 2
(v) Greatest value of f (x) is 2 and least value is 0.

Ex 6.5 Class 12 Maths Question 3.
Find the local maxima and local minima, if any, of the following functions. Find also the local maximum and the local minimum values, as the case may be:
(i) f(x) = x2
(ii) g(x) = x3 – 3x
(iii) h(x) = sinx+cosx,0<x<\frac { \pi }{ 4 }
(iv) f(x) = sin4x + cos4x,0<x<\frac { \pi }{ 2 }
(v) f(x) = x– 6x2 + 9x:+15
(vi) g(x) = \frac { x }{ 2 } +\frac { 2 }{ x } , x>0
(vii) g(x) = \frac { 1 }{ { x }^{ 2 }+2 } , x>0
(viii) f(x) = x\sqrt { 1-x } , x>0
Solution:
(i) Let f(x) = x² ⇒ f’(x) = 2x
Now f'(x) = 0 ⇒ 2x = 0 i.e., x = 0
At x = 0; When x is slightly < 0, f’ (x) is -ve When x is slightly > 0, f(x) is +ve
∴ f(x) changes sign from -ve to +ve as x increases through 0.
⇒ f’ (x) has a local minimum at x = 0 local minimum value f(0) = 0.
NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives 3
NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives 3.1
NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives 3.2
NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives 3.3
NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives 3.4
vedantu class 12 maths Chapter 6 Application of Derivatives 3.5

Ex 6.5 Class 12 Maths Question 4.
Prove that the following functions do not have maxima or minima:
(i) f(x) = ex
(ii) f(x) = log x
(iii) h(x) = x3 + x2 + x + 1
Solution:
(i) f'(x) = ex;
Since f’ (x) ≠ 0 for any value of x.
So f(x) = ex does not have a max. or min.
(ii) f’ (x) = \frac { 1 }{ x }; Clearly f’ (x) ≠ 0 for any value of x.
So,f’ (x) = log x does not have a maximum or a minimum.
(iii) We have f(x) = x3 + x2 + x + 1
⇒f’ (x) = 3x2 + 2x + 1
Now, f’ (x) = 0 => 3x2 + 2x + 1 = 0
x=\frac { -2\pm \sqrt { 4-12 } }{ 6 } =\frac { -1+\sqrt { -2 } }{ 3 }
i.e. f'(x) = 0 at imaginary points
i.e. f'(x) ≠ 0 for any real value of x
Hence, there is neither max. nor min.

Ex 6.5 Class 12 Maths Question 5.
Find the absolute maximum value and the absolute minimum value of the following functions in the given intervals:
(i) f(x) = x3, x∈ [-2,2]
(ii) f(x) = sin x + cos x, x ∈ [0, π]
(iii) f(x) = 4x-\frac { 1 }{ 2 } { x }^{ 2 },x\in \left[ -2,\frac { 9 }{ 2 } \right]
(iv) f(x) = { (x-1) }^{ 2 }+3,x\in \left[ -3,1 \right]
Solution:
(i) We have f’ (x) = x3 in [ -2,2]
∴ f'(x) = 3x²; Now, f’ (x) = 0 at x = 0, f(0) = 0
Now, f(-2) = (-2)3 = – 8; f(0) = (0)² = 0 and f(0) = (2) = 8
Hence, the absolute maximum value of f (x) is 8 which it attained at x = 2 and absolute minimum value of f(x) = – 8 which is attained at x = -2.
(ii) We have f (x) = sin x + cos x in [0, π]
f’ (x) = cos x – sin x for extreme values f’ (x) = 0
NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives 5

Ex 6.5 Class 12 Maths Question 6.
Find the maximum profit that a company can make, if the profit function is given by p(x) = 41 – 24x – 18x²
Solution:
Profit function in p(x) = 41 – 24x – 18x²
∴ p'(x) = – 24 – 36x = – 12(2 + 3x)
for maxima and minima, p'(x) = 0
Now, p'(x) = 0
⇒ – 12(2 + 3x) = 0
⇒ x = -\frac { 2 }{ 3 },
p'(x) changes sign from +ve to -ve.
⇒ p (x) has maximum value at x = -\frac { 2 }{ 3 }
Maximum Profit = 41 + 16 – 8 = 49.

Ex 6.5 Class 12 Maths Question 7.
Find both the maximum value and the minimum value of 3x4 – 8x3 + 12x2 – 48x + 25 on the interval [0,3].
Solution:
Let f(x) = 3x4 – 8x3 + 12x2 – 48x + 25
∴f'(x) = 12x3 – 24x2 + 24x – 48
= 12(x2 + 2)(x – 2)
For maxima and minima, f'(x) = 0
⇒ 12(x2 + 2)(x – 2) = 0
⇒ x = 2
Now, we find f (x) at x = 0,2 and 3, f (0) = 25,
f (2) = 3 (24) – 8 (23) + 12 (22) – 48 (2) + 25 = – 39
and f (3) = (34) – 8 (33) + 12 (32) – 48 (3) + 25
= 243 – 216 + 108 – 144 + 25 = 16
Hence at x = 0, Maximum value = 25
at x = 2, Minimum value = – 39.

Ex 6.5 Class 12 Maths Question 8.
At what points in the interval [0,2π], does the function sin 2x attain its maximum value?
Solution:
We have f (x) = sin 2x in [0,2π], f’ (x) = 2 cos 2 x
For maxima and minima f’ (x) = 0 => cos 2 x = 0
vedantu class 12 maths Chapter 6 Application of Derivatives 8

Ex 6.5 Class 12 Maths Question 9.
What is the maximum value of the function sin x + cos x?
Solution:
Consider the interval [0, 2π],
Let f(x) = sinx + cosx,
f’ (x) = cosx – sinx
For maxima and minima, f’ (x) = 0
NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives 9

Ex 6.5 Class 12 Maths Question 10.
Find the maximum value of 2x3 – 24x + 107 in the interval [1,3]. Find the maximum value of the same function in [-3, -1].
Solution:
∵ f(x) = 2x3 – 24x + 107
∴f(x) = 6x2 – 24 ,
For maxima and minima f'(x) = 0;⇒ x = ±2
For the interval [ 1,3], we find the values of f (x)
at x = 1,2,3; f(1) = 85, f(2) = 75, f(3) = 89
Hence, maximum f (x) = 89 at x = 3
For the interval [-3, -1], we find the values of f(x) at x = – 3, – 2, – 1;
f(-3) = 125;
f(-2) = 139
f(-1) = 129
∴ max.f(x) = 139 at x = – 2.

Ex 6.5 Class 12 Maths Question 11.
It is given that at x = 1, the function x4 – 62x2 + ax + 9 attains its maximum value, on the interval [0,2]. Find the value of a.
Solution:
∵ f(x) = x4 – 62x2 + ax + 9
∴ f’ (x) = 4x3 – 124x + a
Now f’ (x) = 0 at x = 1
⇒ 4 – 124 + a = 0
⇒ a = 120
Now f” (x) = 12x2 – 124:
At x = 1 f” (1) = 12 – 124 = – 112 < 0
⇒ f(x) has a maximum at x = 1 when a = 120.

Ex 6.5 Class 12 Maths Question 12.
Find the maximum and minimum values of x + sin 2x on [0,2π]
Solution:
∴f(x) = x + sin2x on[0,2π]
∴f’ (x) = 1+2 cos2x
For maxima and minima f’ (x) = 0
NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives 12

Ex 6.5 Class 12 Maths Question 13.
Find two numbers whose sum is 24 and whose product is as large as possible.
Solution:
Let the required numbers hex and (24-x)
∴Their product,p = x(24 – x) = 24x – x²
Now \frac { dp }{ dx } = 0 ⇒24 – 2x = 0 ⇒ x = 12
Also \frac { { d }^{ 2 }p }{ { dx }^{ 2 } } = -2<0: ⇒ p is max at x = 12
Hence, the required numbers are 12 and (24-12)i.e. 12.

Ex 6.5 Class 12 Maths Question 14.
Find two positive numbers x and y such that x + y = 60 and xy3 is maximum.
Solution:
We have x + y = 60
⇒ y = 60 – x …(i)
NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives 14
Hence, the req. numbers are 15 and (60 -15) i.e. 15 and 45.

Ex 6.5 Class 12 Maths Question 15.
Find two positive numbers x and y such that their sum is 35 and the product x2 y5 is a maximum.
Solution:
We have x + y = 35 ⇒ y = 35 – x
Product p = x2 y5
= x2 (35 – x)5
vedantu class 12 maths Chapter 6 Application of Derivatives 15

Ex 6.5 Class 12 Maths Question 16.
Find two positive numbers whose sum is 16 and the sum of whose cubes is minimum.
Solution:
Let two numbers be x and 16 – x
NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives 16
Hence, the required numbers are 8 and (16-8) i.e. 8 and 8.

Ex 6.5 Class 12 Maths Question 17.
A square piece of tin of side 18 cm is to be made into a box without top, by cutting a square from each corner and folding up the flaps to form the box. What should be the side of the square to be cut off so that the volume of the box is the maximum possible.
Solution:
Let each side of the square to be cut off be x cm.
∴ for the box length = 18 – 2x: breadth = 18 – 2x and height = x
NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives 17
NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives 17.1

Ex 6.5 Class 12 Maths Question 18.
A rectangular sheet of tin 45 cm by 24 cm is to be made into a box without top, by cutting off square from each comer and folding up the flaps. What should be the side of the square to be cut off so that the volume of the box is maximum?
Solution:
Let each side of the square cut off from each comer be x cm.
∴ Sides of the rectangular box are (45 – 2x), (24 – 2x) and x cm.
NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives 18

Ex 6.5 Class 12 Maths Question 19.
Show that of all the rectangles inscribed in a given fixed circle, the square has the maximum area.
Solution:
Let the length and breadth of the rectangle inscribed in a circle of radius a be x and y respectively.
∴ x² + y² = (2a)² => x² + y² = 4a² …(i)
∴ Perimeter = 2 (x + y)
vedantu class 12 maths Chapter 6 Application of Derivatives 19
NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives 19.1

Ex 6.5 Class 12 Maths Question 20.
Show that the right circular cylinder of given surface and maximum volume is such that its height is equal to the diameter of the base.
Solution:
Let S be the given surface area of the closed cylinder whose radius is r and height h let v be the its Volume. Then
NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives 20

Ex 6.5 Class 12 Maths Question 21.
Of all the closed cylindrical cans (right circular), of a given volume of 100 cubic centimeters, find the dimensions of the can which has the minimum surface area ?
Solution:
Let r be the radius and h be the height of cylindrical can.
NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives 21

Ex 6.5 Class 12 Maths Question 22.
A wire of length 28 m is to be cut into two pieces. One of the pieces is to be made into a square and the other into a circle. What should be the length of the two pieces so that the combined area of the square and the circle is minimum ?
Solution:
Let one part be of length x, then the other part = 28 – x
Let the part of the length x be converted into a circle of radius r.
NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives 22
vedantu class 12 maths Chapter 6 Application of Derivatives 22.1
NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives 22.2
NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives 22.3

Ex 6.5 Class 12 Maths Question 23.
Prove that the volume of the largest cone that can be inscribed in a sphere of radius R is \frac { 8 }{ 27 } of the volume of the sphere.
Solution:
Let a cone. VAB of greatest volume be inscribed in the sphere let AOC = θ
∴ AC, radius of the base of the cone = R sin θ
and VC = VO + OC = R(1 +cosθ)
= R + Rcosθ
= height of the cone.,
V, the volume of the cone.
NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives 23
NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives 23.1
NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives 23.2

Ex 6.5 Class 12 Maths Question 24.
Show that die right circular cone of least curved surface and given volume has an altitude equal to √2 time the radius of the base.
Solution:
Let r and h be the radius and height of the cone.
NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives 24

Ex 6.5 Class 12 Maths Question 25.
Show that the semi-vertical angle of the cone of the maximum volume and of given slant height is tan-1 √2.
Solution:
Let v be the volume, l be the slant height and 0 be the semi vertical angle of a cone.
vedantu class 12 maths Chapter 6 Application of Derivatives 25
NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives 25.1

Ex 6.5 Class 12 Maths Question 26.
Show that semi-vertical angle of right circular cone of given surface area and maximum volume is { sin }^{ -1 }\left( \frac { 1 }{ 3 } \right)
Solution:
Let r be radius, l be the slant height and h be the height of the cone of given surface area s.Then
NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives 26
NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives 26.1

Choose the correct answer in the Exercises 27 and 29.

Ex 6.5 Class 12 Maths Question 27.
The point on die curve x² = 2y which is nearest to the point (0,5) is
(a) (2 √2,4)
(b) (2 √2,0)
(c) (0,0)
(d) (2,2)
Solution:
(a) Let P (x, y) be a point on the curve The other point is A (0,5)
Z = PA² = x² + y² + 25 – 10y [∵ x² = 2y]
NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives 27

Ex 6.5 Class 12 Maths Question 28.
For all real values of x, the minimum value of \frac { 1-x+{ x }^{ 2 } }{ 1+x+{ x }^{ 2 } }
(a) 0
(b) 1
(c) 3
(d) \frac { 1 }{ 3 }
Solution:
(d) Let y=\frac { 1-x+{ x }^{ 2 } }{ 1+x+{ x }^{ 2 } }
NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives 28
vedantu class 12 maths Chapter 6 Application of Derivatives 28.1

Ex 6.5 Class 12 Maths Question 29.
The maximum value of { \left[ x\left( x-1 \right) +1 \right] }^{ \frac { 1 }{ 3 } },0\le x\le 1 is
(a) { \left( \frac { 1 }{ 3 } \right) }^{ \frac { 1 }{ 3 } }
(b) \frac { 1 }{ 2 }
(c) 1
(d) 0
Solution:
(c) Let y = { \left[ x\left( x-1 \right) +1 \right] }^{ \frac { 1 }{ 3 } },0\le x\le 1
NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives 29

NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Hindi Medium Ex 6.5

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NCERT Solutions for Class 12 Maths Chapter 8 Application of Integrals Ex 8.2

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NCERT Solutions for Class 12 Maths Chapter 8 Application of Integrals

Get Free NCERT Solutions for Class 12 Maths Chapter 8 Application of Integrals Ex 8.2 PDF in Hindi and English Medium. Sets Class 12 Maths NCERT Solutions are extremely helpful while doing your homework. Application of Integrals Exercise 8.2 Class 12 Maths NCERT Solutions were prepared by Experienced LearnCBSE.in Teachers. Detailed answers of all the questions in Chapter 8 Class 12 Application of Integrals Ex 8.2 provided in NCERT Textbook.

Free download NCERT Solutions for Class 12 Maths Chapter 8 Application of Integrals Ex 8.2 PDF in Hindi Medium as well as in English Medium for CBSE, Uttarakhand, Bihar, MP Board, Gujarat Board, BIE, Intermediate and UP Board students, who are using NCERT Books based on updated CBSE Syllabus for the session 2019-20.

NCERT Solutions for Class 12 Maths Chapter 8 Application of Integrals Ex 8.2

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Call of Duty Mobile launching on October 1: PUBG MOBILE rival to feature new game modes

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Call of Duty Mobile launching on October 1: PUBG MOBILE rival to feature new game modes

Call of Duty Mobile is releasing on Android and iOS on October 1. The game will feature all popular Call of Duty playing modes from the PC versions.

PUBG MOBILE Call of Duty Mobile launching on October 1

HIGHLIGHTS

  • Call of Duty Mobile launching on October 1 across the globe.
  • The game will be available on both Android and iOS platforms.
  • Call of Duty Mobile will have many gameplay modes from the PC versions.

Earlier this year, Tencent Games released another online shooter title called Call of Duty Mobile for Android devices. The game was released in a closed beta for selected participants and it brought all the popular Call of Duty game modes from the PC versions in the past. We tried it out during the closed beta and it came close to displacing PUBG Mobile as the ultimate mobile shooter game you can play. However, once the beta was pulled off, many have been left wondering when the stable version will hit the servers. Well now, we have an answer.

The official page for Call of Duty Mobile has announced that the stable version of the game will be live from October 1. And the game will be available for both Android and iOS platform this time. The game is developed by Tencent’s Timi Studio and similar to the beta, it will be free-to-play for everyone.

For those in the unknown, the Call of Duty series is one of the most popular battle games from Activision. The mobile version aims to bring similar gameplay style to players around the globe along with all the classic Call of Duty gameplay modes and maps.

“We are delivering the definitive, first-person action experience on mobile with signature Call of Duty gameplay in the palms of your hands,” said Chris Plummer, VP, Mobile at Activision. “We are bringing together some of the best the franchise has to offer, including Modern Warfare maps like Crash and Crossfire, Black Ops maps like Nuketown and Hijacked, and many more, into one epic title.

The battle royale mode will be similar to the Classic mode matches in PUBG MOBILE, allowing 100 players at a time to battle it out in an open map. Similar to PUBG Mobile, Call of Duty Mobile can let you play in various modes like solo, duo and quad-player competitions. The maps will include battles across land, sea and air, with players able to rely on vehicles such as an ATV, helicopter and raft. Players can even choose first or third-person modes. And in typical Call of Duty style, the players can choose between any one of the six classes.

The Call of Duty Mobile will be a change for those players who have been playing PUBG MOBILE for long. In our short time with the beta, we found that Call of Duty Mobile fared better in many ways than PUBG MOBILE, especially with regards to gameplay style, graphics and more. In fact, PUBG MOBILE lifted off the PVP mode from COD Mobile’s Team Deathmatch mode.

Articles on PUBG Mobile

Call of Duty Mobile: Here’s how you can download it on your Android and iOS smartphone

Call of Duty Mobile

Activision’s Call of Duty Mobile is out now on Android and iOS in a closed beta. Closed beta restricts the distribution of the game but there is a way to install and play it on your phone right now.

HIGHLIGHTS

  • Call of Duty Mobile is now available in beta for Android and iOS users.
  • The game offers classic Call of Duty gameplay modes from the PC version and a new Battle Royale Mode.
  • Call of Duty Mobile will be a direct competitor to PUBG MOBILE.

For a long time, PUBG MOBILE has been sitting as the dominant game in the mobile world, increasingly gathering more popularity as the days pass by. Many other games have tried to steal PUBG of its limelight but the popular battle royale online fighter has maintained itself at the top of the charts. However, that hasn’t stopped other studios from trying and Activision is the latest one with its renowned Call of Duty series for Android and iOS.

Call of Duty Mobile is the latest title under Tencent’s umbrella and has been introduced as a new game that offers the thrills of the PC versions of the game while also dipping its foot in the viral battle royale mode. I even played it for a while and it seems like a breath of fresh air after being into PUBG for so long. The graphics are great and the gameplay does offer a variety of options. While it’s too early to say, it seems that Call of Duty Mobile could be the next big thing in the world of mobile games.

If you have been tempted to try it out, the game is available to the world in a closed beta form. However, Activision has been taking the registration for the game for almost a month now and those who were the early ones now have a chance to download it and try out the beta. If you didn’t register for the beta, here’s how you can do that from your Android and iOS device.

Call of Duty Mobile Guide: Everything you need to know about its class system

Call of Duty Mobile Guide

Call of Duty Mobile offers the class system from its PC counterpart to players. If you haven’t played Call of Duty before, here’s a brief guide to the class system and how you can take advantage of it.

HIGHLIGHTS

  • Call of Duty Mobile offers anew class-based system while playing as a team.
  • Those who play as clowns get a very special power – summon zombies.
  • As a Medic, the player is essentially the one responsible for looking after the team members.

It has been a few weeks since Call of Duty rolled out for the mobile platforms. And at the very least, the game competes with PUBG MOBILE in terms of player involvement. In fact, we did a comparison with PUBG MOBILE a few days ago and found that the Activision’s Call of Duty Mobile is ahead of PUBG MOBILE in certain areas, especially gameplay and graphics. However, like PUBG MOBILE, Call of Duty Mobile requires the player to understand some of its features to enjoy the game.

In essence, for you to play Call of Duty Mobile like an ace, you need to have a basic idea about the franchise and how its previous games used to work. And if you are like me who hasn’t played Call of Duty ever, then it must be a struggle to get along with the game, especially with the class system. Unlike PUBG MOBILE, Call of Duty Mobile requires you to choose a class while playing in the multiplayer mode.

If this class system is confusing to you, then here’s a brief explanation that might help you in a massive way as a novice.

Call of Duty Mobile: Class system explained

Scout:

As a scout, the player gets a massive power give you an idea about hostile positions in the vicinity. The player gets to use Sensor Darts that can give you an idea about which areas nearby you run a huge risk. Scouts can also track footprints of enemies nearby with the tracker ability.

Clown:

This is nothing like the name suggests. In fact, as a clown, the player gets a very special power summon zombies. Yes, zombies!

With the toy bomb, a clown can summon zombies to protect him/her from the enemies nearby. Zombies will attack the enemy and giving the player time to escape or run for safety. The clown can also limit enemy zombies in their aggression.

Mechanic:

As the name suggests, a mechanic gets to use all the cool gadgets in the team. As a mechanic, the player can summon an EMP drone that will halt all electrical operations on the enemy team.

Apart from the EMP drone, a mechanic can also get to use the Engineer ability that will help him/her get an augmented view of vehicles, traps set up by enemies and other items.

Medic:

As a Medic, the player is essentially the one responsible for looking after the team members. The Medic gets to heal players nearby. Additionally, the rate at which the Medic can heal himself/herself is faster than other players.

Call of Duty Mobile vs PUBG MOBILE: Two splendid battle royale games but is the Chicken Dinner stale now?

Call of Duty Mobile vs PUBG MOBILE

Activision’s popular battle title Call of Duty is now on both the mobile platforms – Android and iOS. Published by Tencent, it sits in the same space as PUBG MOBILE with a dedicated battle royale mode. So, is this the new game of the year and does it make the Chicken Dinner stale?

HIGHLIGHTS

  • Call of Duty Mobile offers regular mission based game modes and battle royale mode.
  • Both PUBG MOBILE and Call of Duty Mobile have great console-quality graphics.
  • Both PUBG MOBILE and Call of Duty Mobile have an impressively detailed roster of ammunition

Since PUBG stepped into the mobile gaming arena, mobile games haven’t been the same since then. New titles keep coming out every few weeks with graphics and gameplay that can challenge proper console games any given day. However, the battle royale formula is superhit and every PC gaming franchise wants to be a part of this – there are talks that EA could bring Apex Legends to smartphones soon. Activision, who’s responsible for the worldwide-hit franchise Call of Duty (COD), has worked with Tencent to bring it to the mobile platform and it’s already available in beta now.

Simply titled Call of Duty Mobile, it aims to bring the familiar COD gameplay to the mobile for fans of the franchise. In addition, to tackle the competition, it also brings in the viral battle royale mode that challenges you to win as the last man standing. Sounds familiar? On paper, the concept sounds very similar to PUBG MOBILE’s gameplay. And when I tried it in person, it did feel very much like PUBG in a lot of ways. And that makes it interesting – a proper PUBG MOBILE rival is finally here after a long time.

So after playing it for a couple of hours, here’s how I feel it compares to PUBG MOBILE.

Gameplay: Classic Call of Duty experience with a PUBG twist

PUBG has immortalized the battle royale genre in the world of mobile games. Ever since PUBG MOBILE, the idea of fighting it out with up to 100 real-life players in a desolated map sprinkled with weapons has appealed to a larger crowd and pulled in more people in the world of games. Call of Duty carries that idea forward with its own twist to make it slightly different from what we are used to in PUBG MOBILE.

The pace of the game in PUBG MOBILE seems fast

The pace of the game in PUBG MOBILE seems fast, even in the Classic mode matches. In comparison, Call of Duty multiplayer mode matches have a more relaxed pace. I often found spending more time looking at the environment during the gameplay – something that I have always struggled to do in a PUBG match. Of course, you have to keep in mind that PUBG is themed around hardcore battle royale-style survival while Call of Duty Mobile has a more mission-based approach to its gameplay (excluding the Battle Royale mode). The missions require you to work on the strategies and focus more on ammunition management. This is classic Call of Duty gameplay.

In the Battle Royale mode, there’s not really much difference between Call of Duty and PUBG. Both the games have the same concept – you need to jump in a big map, collect weapons, protect your team members, move around using vehicles, kill enemies using smart strategies and be the last man (or team) standing to win the match. However, in PUBG MOBILE, a player is just part of a team and has no superiority over others, apart from the playing skills. Call of Duty goes for a different approach by offering each player to play a role with a special weapon or be a medic to look after your team.

What I love about Call of Duty is its approach to make new players more comfortable. For example, the game offers you an option to either let you fire the weapon on your own or let the game fire for you once you aim. For new players like me, it was a great way to increase the involvement in a match and have a better chance in playing with the professionals. PUBG MOBILE has always gone for a more rough approach by not offering any training or tips on how to play.

Graphics: Great graphics aren’t just restricted to PC anymore

PUBG MOBILE has led the mobile gaming industry in terms of graphics quality for a while until Gameloft dropped Asphalt 9: Legends. That said, the game still looks good on high graphics and each update has been improving the visual effects – the latest one added rain to two maps.

Call of Duty Mobile in comparison has noticeably better graphics. If you were impressed with the details in PUBG MOBILE, then Call of Duty Mobile will simply blow you away. The detailing on the character’s dresses are impeccable – all the pockets on the dress, the weapons, the stitching and the colour patterns – this is on par with what you get in PC games now.

The detailing also extends beyond the characters. The lighting effects are impeccable and the reflections on puddles surely look close to what you see in the console games. The textures are quite detailed for a mobile game and map design is very clever.

Weapons: Both are equally good

Both PUBG MOBILE and Call of Duty Mobile have an impressively detailed roster of ammunition with extreme attention to details. Players need to learn about weapons and how they work in order to have a better shot at killing enemies. Call of Duty has a slightly serious approach to the weapon design and modification whereas PUBG goes for a slightly more fun approach.

Early impressions: Does the Chicken Dinner still hold its relevance over “the Duty”?

Not very much. Since Call of Duty Mobile dropped in, I have mostly noticed many PUBG players spending more time on the new game instead of logging hours in PUBG MOBILE. However, a major reason for that is that Call of Duty is new and will draw people’s fancy for the first few days. In comparison, PUBG MOBILE hasn’t changed much, even with the new Zombie modes.

However, Call of Duty Mobile has a better chance at scooping out a lot of PUBG followers in its favour. Those who have been yearning for a more serious military-based gameplay will find Call of Duty Mobile appealing. The graphics is also good and the game offers a lot many modes to get hooked to it. The gameplay is slightly relaxed but holds the classic Call of Duty vibe, which in itself will help its cause. And this is a great thing considering the game is still in the beta stage.

If you have registered for the closed beta, go ahead and download it. When this comes out, PUBG MOBILE will have serious competition and it would be interesting to see whether gamers embrace Call of Duty like PUBG or ignore it like Fortnite.

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NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations Ex 9.2

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NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations Ex 9.2

Get Free NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations Ex 9.2 PDF in Hindi and English Medium. Sets Class 12 Maths NCERT Solutions are extremely helpful while doing your homework. Differential Equations Exercise 9.2 Class 12 Maths NCERT Solutions were prepared by Experienced LearnCBSE.in Teachers. Detailed answers of all the questions in Chapter 9 Class 12 Differential Equations Ex 9.2 provided in NCERT Textbook.

Free download NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations Ex 9.2 PDF in Hindi Medium as well as in English Medium for CBSE, Uttarakhand, Bihar, MP Board, Gujarat Board, BIE, Intermediate and UP Board students, who are using NCERT Books based on updated CBSE Syllabus for the session 2019-20.

NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations Ex 9.2

Verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation

Ex 9.2 Class 12 Maths Question 1.
y={ e }^{ x }+1:{ y }^{ II }-{ y }^{ I }=0
Solution:
y={ e }^{ x }+1:{ y }^{ II }-{ y }^{ I }=0
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations 1

Ex 9.2 Class 12 Maths Question 2.
y=x^{ 2 }+2x+c:{ y }^{ I }-2x-2=0
Solution:
y=x^{ 2 }+2x+c:{ y }^{ I }-2x-2=0
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations 2

Ex 9.2 Class 12 Maths Question 3.
y=cosx+c:{ y }^{ I }+sinx=0
Solution:
y=cosx+c:{ y }^{ I }+sinx=0
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations 3

Ex 9.2 Class 12 Maths Question 4.
y=\sqrt { 1+{ x }^{ 2 } } :{ y }^{ I }=\frac { xy }{ 1+{ x }^{ 2 } }
Solution:
y=\sqrt { 1+{ x }^{ 2 } } :{ y }^{ I }=\frac { xy }{ 1+{ x }^{ 2 } }
vedantu class 12 maths Chapter 9 Differential Equations 4

Ex 9.2 Class 12 Maths Question 5.
y=Ax:x{ y }^{ I }=y(x\neq 0)
Solution:
y=Ax:x{ y }^{ I }=y(x\neq 0)
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations 5

Ex 9.2 Class 12 Maths Question 6.
y=x\quad sinx;{ xy }^{ I }=y+x\sqrt { { x }^{ 2 }-{ y }^{ 2 } } (x\neq 0\quad and\quad x>y\quad or\quad x<-y)
Solution:
y=x\quad sinx;{ xy }^{ I }=y+x\sqrt { { x }^{ 2 }-{ y }^{ 2 } } (x\neq 0\quad and\quad x>y\quad or\quad x<-y)
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations 6

Ex 9.2 Class 12 Maths Question 7.
xy = logy + C,
UP Board Solutions for Class 12 Maths Chapter 9 Differential Equations 7
Solution:
xy = logy + C,
UP Board Solutions for Class 12 Maths Chapter 9 Differential Equations 7
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations 7

Ex 9.2 Class 12 Maths Question 8.
y-cosy=x:(ysiny+cosy+x){ y }^{ I }=y
Solution:
y-cosy=x:(ysiny+cosy+x){ y }^{ I }=y
vedantu class 12 maths Chapter 9 Differential Equations 8

Ex 9.2 Class 12 Maths Question 9.
x+y={ ta }n^{ -1 }y;{ y }^{ 2 }{ y }^{ I }+{ y }^{ 2 }+1=0
Solution:
x+y={ ta }n^{ -1 }y;{ y }^{ 2 }{ y }^{ I }+{ y }^{ 2 }+1=0
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations 9
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations 9.1

Ex 9.2 Class 12 Maths Question 10.
y=\sqrt { { a }^{ 2 }-{ x }^{ 2 } } x\in (-a,a);x+y\frac { dy }{ dx } =0,(y\neq 0)
Solution:
y=\sqrt { { a }^{ 2 }-{ x }^{ 2 } } x\in (-a,a);x+y\frac { dy }{ dx } =0,(y\neq 0)
vedantu class 12 maths Chapter 9 Differential Equations 10

Ex 9.2 Class 12 Maths Question 11.
The number of arbitrary constants in the general solution of a differential equation of fourth order are:
(a) 0
(b) 2
(c) 3
(d) 4
Solution:
(b) The general solution of a differential equation of fourth order has 4 arbitrary constants.
Because it contains the same number of arbitrary constants as the order of differential equation.

Ex 9.2 Class 12 Maths Question 12.
The number of arbitrary constants in the particular solution of a differential equation of third order are:
(a) 3
(b) 2
(c) 1
(d) 0
Solution:
(d) Number of arbitrary constants = 0
Because particular solution is free from arbitrary constants.

NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations Hindi Medium Ex 9.2

NCERT Solutions for Class 12 Maths Exercise 9.2 of Differential Equations
NCERT Solutions for Class 12 Maths Exercise 9.2
12 Maths ex 9.2
NCERT Solutions for Class 12 Maths Exercise 9.2 in Hindi Medium PDF
9.2 of 12 Maths
12 Maths Exercise 9.2 solutions

HC Verma Concepts of Physics NCERT Solutions Homepage RD Sharma Solutions

 

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NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations Ex 9.3

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NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations Ex 9.3

Get Free NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations Ex 9.3 PDF in Hindi and English Medium. Sets Class 12 Maths NCERT Solutions are extremely helpful while doing your homework. Differential Equations Exercise 9.3 Class 12 Maths NCERT Solutions were prepared by Experienced LearnCBSE.in Teachers. Detailed answers of all the questions in Chapter 9 Class 12 Differential Equations Ex 9.3 provided in NCERT Textbook.

Free download NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations Ex 9.3 PDF in Hindi Medium as well as in English Medium for CBSE, Uttarakhand, Bihar, MP Board, Gujarat Board, BIE, Intermediate and UP Board students, who are using NCERT Books based on updated CBSE Syllabus for the session 2019-20.

NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations Ex 9.3

In each of the following, Q. 1 to 5 form a differential equation representing the given family of curves by eliminating arbitrary constants a and b.

Ex 9.3 Class 12 Maths Question 1.
\frac { x }{ a } +\frac { y }{ b } =1
Solution:
Given that \frac { x }{ a } +\frac { y }{ b } =1 …(i)
differentiating (i) w.r.t x, we get
\frac { 1 }{ a } +\frac { 1 }{ b } { y }^{ I }=0 …(ii)
again differentiating w.r.t x, we get
\frac { 1 }{ b } { y }^{ II }=0\Rightarrow { y }^{ II }=0
which is the required differential equation

Ex 9.3 Class 12 Maths Question 2.
y² = a(b² – x²)
Solution:
given that
y² = a(b² – x²)…(i)
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations 2
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations 2.1

Ex 9.3 Class 12 Maths Question 3.
y = ae3x+be-2x
Solution:
Given that
y = ae3x+be-2x …(i)
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations 3

Ex 9.3 Class 12 Maths Question 4.
y = e2x (a+bx)
Solution:
y = e2x (a+bx)
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations 4

Ex 9.3 Class 12 Maths Question 5.
y = ex(a cosx+b sinx)
Solution:
The curve y = ex(a cosx+b sinx) …(i)
differentiating w.r.t x
byjus class 12 maths Chapter 9 Differential Equations 5

Ex 9.3 Class 12 Maths Question 6.
Form the differential equation of the family of circles touching the y axis at origin
Solution:
The equation of the circle with centre (a, 0) and radius a, which touches y- axis at origin
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations 6

Ex 9.3 Class 12 Maths Question 7.
Form the differential equation of the family of parabolas having vertex at origin and axis along positive y-axis.
Solution:
The equation of parabola having vertex at the origin and axis along positive y-axis is
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations 7

Ex 9.3 Class 12 Maths Question 8.
Form the differential equation of family of ellipses having foci on y-axis and centre at origin.
Solution:
The equation of family ellipses having foci at y- axis is
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations 8

Ex 9.3 Class 12 Maths Question 9.
Form the differential equation of the family of hyperbolas having foci on x-axis and centre at the origin.
Solution:
Equation of the hyperbola is \frac { { x }^{ 2 } }{ { a }^{ 2 } } -\frac { { y }^{ 2 } }{ { b }^{ 2 } } =1
Differentiating both sides w.r.t x
byjus class 12 maths Chapter 9 Differential Equations 9
which is the req. differential eq. of the hyperbola.

Ex 9.3 Class 12 Maths Question 10.
Form the differential equation of the family of circles having centre on y-axis and radius 3 units
Solution:
Let centre be (0, a) and r = 3
Equation of circle is
x² + (y – a)² = 9 …(i)
Differentiating both sides, we get
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations 10
which is required equation

Ex 9.3 Class 12 Maths Question 11.
Which of the following differential equation has y={ c }_{ 1 }{ e }^{ x }+{ c }_{ 2 }{ e }^{ -x } as the general solution ?
(a) \frac { { d }^{ 2 }y }{ { dx }^{ 2 } } +y=0
(b) \frac { { d }^{ 2 }y }{ { dx }^{ 2 } } -y=0
(c) \frac { { d }^{ 2 }y }{ { dx }^{ 2 } } +1=0
(d) \frac { { d }^{ 2 }y }{ { dx }^{ 2 } } -1=0
Solution:
(b) y={ c }_{ 1 }{ e }^{ x }+{ c }_{ 2 }{ e }^{ -x }\Rightarrow \frac { dy }{ dx } ={ c }_{ 1 }{ e }^{ x }-{ c }_{ 2 }{ e }^{ -x }
\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } ={ c }_{ 1 }{ e }^{ x }+{ c }_{ 2 }{ e }^{ -x }\Rightarrow \frac { { d }^{ 2 }y }{ { dx }^{ 2 } } -y=0

Ex 9.3 Class 12 Maths Question 12.
Which of the following differential equations has y = x as one of its particular solution ?
(a) \frac { { d }^{ 2 }y }{ { dx }^{ 2 } } -{ x }^{ 2 }\frac { dy }{ dx } +xy=x
(b) \frac { { d }^{ 2 }y }{ { dx }^{ 2 } } +{ x }\frac { dy }{ dx } +xy=x
(c) \frac { { d }^{ 2 }y }{ { dx }^{ 2 } } -{ x }^{ 2 }\frac { dy }{ dx } +xy=0
(d) \frac { { d }^{ 2 }y }{ { dx }^{ 2 } } +{ x }\frac { dy }{ dx } +xy=0
Solution:
(c) y = x
\frac { dy }{ dx } =1,\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } =0
\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } -{ x }^{ 2 }\frac { dy }{ dx } +xy=0

NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations Hindi Medium Ex 9.3

NCERT Solutions for Class 12 Maths Exercise 9.3 of Differential Equations
NCERT Solutions for Class 12 Maths Exercise 9.3
12 Maths Exercise 9.3
12 Maths Exercise 9.3 solutions
12 Maths Exercise 9.3 all answers
12 Maths Exercise 9.3 in English Medium
NCERT Solutions for Class 12 Maths Exercise 9.3 for up board
NCERT Solutions for Class 12 Maths Exercise 9.3 in Hindi medium

HC Verma Concepts of Physics NCERT Solutions Homepage RD Sharma Solutions

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NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations Ex 9.4

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NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations Ex 9.4

Get Free NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations Ex 9.4 PDF in Hindi and English Medium. Sets Class 12 Maths NCERT Solutions are extremely helpful while doing your homework. Differential Equations Exercise 9.4 Class 12 Maths NCERT Solutions were prepared by Experienced LearnCBSE.in Teachers. Detailed answers of all the questions in Chapter 9 Class 12 Differential Equations Ex 9.4 provided in NCERT Textbook.

Free download NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations Ex 9.4 PDF in Hindi Medium as well as in English Medium for CBSE, Uttarakhand, Bihar, MP Board, Gujarat Board, BIE, Intermediate and UP Board students, who are using NCERT Books based on updated CBSE Syllabus for the session 2019-20.

NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations Ex 9.4

For each of the following D.E in Q. 1 to 10 find the general solution:

Ex 9.4 Class 12 Maths Question 1.
\frac { dy }{ dx } =\frac { 1-cosx }{ 1+cosx }
Solution:
\frac { dy }{ dx } =\frac { 1-cosx }{ 1+cosx }
\frac { dy }{ dx } =\frac { 1-cosx }{ 1+cosx } =\frac { { 2sin }^{ 2 }\left( \frac { x }{ 2 } \right) }{ { 2cos }^{ 2 }\left( \frac { x }{ 2 } \right) } ={ tan }^{ 2 }\left( \frac { x }{ 2 } \right)
integrating both sides, we get
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations 1

Ex 9.4 Class 12 Maths Question 2.
\frac { dy }{ dx } =\sqrt { 4-{ y }^{ 2 } } (-2<y<2)
Solution:
\frac { dy }{ dx } =\sqrt { 4-{ y }^{ 2 } } \Rightarrow \int { \frac { dy }{ \sqrt { { 4-y }^{ 2 } } } } =\int { dx }
\Rightarrow { sin }^{ -1 }\frac { y }{ 2 } =x+C
\Rightarrow y=2sin(x+C)

Ex 9.4 Class 12 Maths Question 3.
\frac { dy }{ dx } +y=1(y\neq 1)
Solution:
\frac { dy }{ dx } +y=1\Rightarrow \int { \frac { dy }{ y-1 } } =-\int { dx }
\Rightarrow log(y-1)=-x+c\Rightarrow y=1+{ e }^{ -x }.{ e }^{ c }
Hence\quad y=1+{ Ae }^{ -x }
which is required solution

Ex 9.4 Class 12 Maths Question 4.
sec² x tany dx+sec² y tanx dy = 0
Solution:
we have
sec² x tany dx+sec² y tanx dy = 0
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations 4

Ex 9.4 Class 12 Maths Question 5.
\left( { e }^{ x }+{ e }^{ -x } \right) dy-\left( { e }^{ x }-{ e }^{ -x } \right) dx=0
Solution:
we have
\left( { e }^{ x }+{ e }^{ -x } \right) dy-\left( { e }^{ x }-{ e }^{ -x } \right) dx=0
Integrating on both sides
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations 5

Ex 9.4 Class 12 Maths Question 6.
\frac { dy }{ dx } =\left( { 1+x }^{ 2 } \right) \left( { 1+y }^{ 2 } \right)
Solution:
\frac { dy }{ { 1+y }^{ 2 } } =\left( { 1+x }^{ 2 } \right) dx
integrating on both side we get
{ tan }^{ -1 }y={ x+\frac { 1 }{ 3 } }x^{ 3 }+c
which is required solution

Ex 9.4 Class 12 Maths Question 7.
y logy dx – x dy = 0
Solution:
\because \quad y\quad logy\quad dx=x\quad dy\Rightarrow \frac { dy }{ y\quad logy } =\frac { dx }{ x }
integrating we get
tiwari academy class 12 maths Chapter 9 Differential Equations 7

Ex 9.4 Class 12 Maths Question 8.
{ x }^{ 5 }\frac { dy }{ dx } =-{ y }^{ 5 }
Solution:
{ x }^{ 5 }\frac { dy }{ dx } =-{ y }^{ 5 }\Rightarrow \int { { y }^{ -5 }dy } =-\int { { x }^{ -5 }dx }
\Rightarrow -\frac { 1 }{ { y }^{ 4 } } =\frac { 1 }{ { x }^{ 4 } } +4c\Rightarrow { x }^{ -4 }+{ y }^{ -4 }=k

Ex 9.4 Class 12 Maths Question 9.
solve the following
\frac { dy }{ dx } ={ sin }^{ -1 }x
Solution:
\frac { dy }{ dx } ={ sin }^{ -1 }x\Rightarrow \int { dy } =\int { { sin }^{ -1 }xdx }
integrating both sides we get
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations 9

Ex 9.4 Class 12 Maths Question 10.
{ e }^{ x }tany\quad dx+{ (1-e }^{ x }){ sec }^{ 2 }dy=0
Solution:
{ e }^{ x }tany\quad dx+{ (1-e }^{ x }){ sec }^{ 2 }dy=0
we can write in another form
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations 10

Find a particular solution satisfying the given condition for the following differential equation in Q.11 to 14.

Ex 9.4 Class 12 Maths Question 11.
\left( { x }^{ 3 }+{ x }^{ 2 }+x+1 \right) \frac { dy }{ dx } ={ 2x }^{ 2 }+x;y=1,when\quad x=0
Solution:
here
dy=\frac { { 2x }^{ 2 }+x }{ \left( { x }^{ 3 }+{ x }^{ 2 }+x+1 \right) } dx
integrating we get
tiwari academy class 12 maths Chapter 9 Differential Equations 11

Ex 9.4 Class 12 Maths Question 12.
x\left( { x }^{ 2 }-1 \right) \frac { dy }{ dx } =1,y=0\quad when\quad x=2
Solution:
x\left( { x }^{ 2 }-1 \right) \frac { dy }{ dx } =1,y=0\quad when\quad x=2
\Rightarrow \int { dy } =\int { \frac { dy }{ x(x+1)(x-1) } }
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations 12
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations 12.1

Ex 9.4 Class 12 Maths Question 13.
cos\left( \frac { dy }{ dx } \right) =a,(a\epsilon R),y=1\quad when\quad x=0
Solution:
cos\left( \frac { dy }{ dx } \right) =a\quad \therefore \frac { dy }{ dx } ={ cos }^{ -1 }a
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations 13

Ex 9.4 Class 12 Maths Question 14.
\frac { dy }{ dx } =ytanx,y=1\quad when\quad x=0
Solution:
\frac { dy }{ dx } =ytanx\Rightarrow \int { \frac { dy }{ y } } =\int { tanx\quad dx }
=> logy = logsecx + C
When x = 0, y = 1
=> log1 = log sec0 + C => 0 = log1 + C
=> C = 0
∴ logy = log sec x
=> y = sec x.

Ex 9.4 Class 12 Maths Question 15.
Find the equation of the curve passing through the point (0,0) and whose differential equation { y }^{ I }={ e }^{ x }sinx
Solution:
{ y }^{ I }={ e }^{ x }sinx
\Rightarrow dy={ e }^{ x }sinx\quad dx
tiwari academy class 12 maths Chapter 9 Differential Equations 15

Ex 9.4 Class 12 Maths Question 16.
For the differential equation xy\frac { dy }{ dx } =(x+2)(y+2) find the solution curve passing through the point (1,-1)
Solution:
The differential equation isxy\frac { dy }{ dx } =(x+2)(y+2)
or xydy=(x + 2)(y+2)dx
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations 16

Ex 9.4 Class 12 Maths Question 17.
Find the equation of a curve passing through the point (0, -2) given that at any point (pc, y) on the curve the product of the slope of its tangent and y-coordinate of the point is equal to the x-coordinate of the point
Solution:
According to the question y\frac { dy }{ dx } =x
\Rightarrow \int { ydy } =\int { xdx } \Rightarrow \frac { { y }^{ 2 } }{ 2 } =\frac { { x }^{ 2 } }{ 2 } +c
0, – 2) lies on it.c = 2
∴ Equation of the curve is : x² – y² + 4 = 0.

Ex 9.4 Class 12 Maths Question 18.
At any point (x, y) of a curve the slope of the tangent is twice the slope of the line segment joining the point of contact to the point (-4,-3) find the equation of the curve given that it passes through (- 2,1).
Solution:
Slope of the tangent to the curve = \frac { dy }{ dx }
slope of the line joining (x, y) and (- 4, – 3)
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations 18
tiwari academy class 12 maths Chapter 9 Differential Equations 18.1

Ex 9.4 Class 12 Maths Question 19.
The volume of a spherical balloon being inflated changes at a constant rate. If initially its radius is 3 units and offer 3 seconds it is 6 units. Find the radius of balloon after t seconds.
Solution:
Let v be volume of the balloon.
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations 19

Ex 9.4 Class 12 Maths Question 20.
In a bank principal increases at the rate of r% per year. Find the value of r if Rs 100 double itself in 10 years
Solution:
Let P be the principal at any time t.
According to the problem
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations 20

Ex 9.4 Class 12 Maths Question 21.
In a bank principal increases at the rate of 5% per year. An amount of Rs 1000 is deposited with this bank, how much will it worth after 10 years
Solution:
Let p be the principal Rate of interest is 5%
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations 21

Ex 9.4 Class 12 Maths Question 22.
In a culture the bacteria count is 1,00,000. The number is increased by 10% in 2 hours. In how many hours will the count reach 2,00,000 if the rate of growth of bacteria is proportional to the number present
Solution:
Let y denote the number of bacteria at any instant t • then according to the question
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations 22

Ex 9.4 Class 12 Maths Question 23.
The general solution of a differential equation \frac { dy }{ dx } ={ e }^{ x+y } is
(a) { e }^{ x }+{ e }^{ -y }=c
(b) { e }^{ x }+{ e }^{ y }=c
(c) { e }^{ -x }+{ e }^{ y }=c
(d) { e }^{ -x }+{ e }^{ -y }=c
Solution:
(a) \frac { dy }{ dx } ={ e }^{ x }.{ e }^{ y }\Rightarrow \int { { e }^{ -y }dy } =\int { { e }^{ x }dx }
\Rightarrow { e }^{ -y }={ e }^{ x }+k\Rightarrow { e }^{ x }+{ e }^{ -y }=c

NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations Hindi Medium Ex 9.4

NCERT Solutions for Class 12 Maths Exercise 9.4 of Differential Equations
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NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations Ex 9.5

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NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations Ex 9.5

Get Free NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations Ex 9.5 PDF in Hindi and English Medium. Sets Class 12 Maths NCERT Solutions are extremely helpful while doing your homework. Differential Equations Exercise 9.5 Class 12 Maths NCERT Solutions were prepared by Experienced LearnCBSE.in Teachers. Detailed answers of all the questions in Chapter 9 Class 12 Differential Equations Ex 9.5 provided in NCERT Textbook.

Free download NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations Ex 9.5 PDF in Hindi Medium as well as in English Medium for CBSE, Uttarakhand, Bihar, MP Board, Gujarat Board, BIE, Intermediate and UP Board students, who are using NCERT Books based on updated CBSE Syllabus for the session 2019-20.

NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations Ex 9.5

Show that the given differential equation is homogeneous and solve each of them in Questions 1 to 10

Ex 9.5 Class 12 Maths Question 1.
(x²+xy)dy = (x²+y²)dx
Solution:
(x²+xy)dy = (x²+y²)dx
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations 1
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations 1.1

Ex 9.5 Class 12 Maths Question 2.
{ y }^{ I }=\frac { x+y }{ x }
Solution:
{ y }^{ I }=\frac { x+y }{ x }
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations 2

Ex 9.5 Class 12 Maths Question 3.
(x-y)dy-(x+y)dx=0
Solution:
\frac { dy }{ dx } =\frac { x+y }{ x-y } =\frac { 1+\frac { y }{ x } }{ 1-\frac { y }{ x } }
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations 3

Ex 9.5 Class 12 Maths Question 4.
(x²-y²)dx+2xy dy=0
Solution:
\frac { dy }{ dx } =\frac { { y }^{ 2 }-{ x }^{ 2 } }{ 2xy }
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations 4

Ex 9.5 Class 12 Maths Question 5.
{ x }^{ 2 }\frac { dy }{ dx } ={ x }^{ 2 }-{ 2y }^{ 2 }+xy
Solution:
\frac { dy }{ dx } =1-2{ \left( \frac { y }{ x } \right) }^{ 2 }+\frac { y }{ x }
vedantu class 12 maths Chapter 9 Differential Equations 5

Ex 9.5 Class 12 Maths Question 6.
xdy-ydx=\sqrt { { x }^{ 2 }+{ y }^{ 2 } } dx
Solution:
xdy-ydx=\sqrt { { x }^{ 2 }+{ y }^{ 2 } } dx
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations 6

Ex 9.5 Class 12 Maths Question 7.
\left\{ xcos\left( \frac { y }{ x } \right) +ysin\left( \frac { y }{ x } \right) \right\} ydx=\left\{ ysin\left( \frac { y }{ x } \right) -xcos\left( \frac { y }{ x } \right) \right\} xdy
Solution:
\left\{ xcos\left( \frac { y }{ x } \right) +ysin\left( \frac { y }{ x } \right) \right\} ydx=\left\{ ysin\left( \frac { y }{ x } \right) -xcos\left( \frac { y }{ x } \right) \right\} xdy
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations 7
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations 7.1
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations 7.2

Ex 9.5 Class 12 Maths Question 8.
x\frac { dy }{ dx } -y+xsin\left( \frac { y }{ x } \right) =0
Solution:
x\frac { dy }{ dx } -y+xsin\left( \frac { y }{ x } \right) =0\Rightarrow \frac { dy }{ dx } =\frac { y }{ x } -sin\frac { y }{ x }
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations 8

Ex 9.5 Class 12 Maths Question 9.
ydx+xlog\left( \frac { y }{ x } \right) dy-2xdy=0
Solution:
\frac { dy }{ dx } =\frac { y }{ 2x-xlog\frac { y }{ x } } =\frac { \frac { y }{ x } }{ 2-log\frac { y }{ x } }
vedantu class 12 maths Chapter 9 Differential Equations 9

Ex 9.5 Class 12 Maths Question 10.
\left( { 1+e }^{ \frac { x }{ y } } \right) dx+{ e }^{ \frac { x }{ y } }\left( 1-\frac { x }{ y } \right) dy=0
Solution:
\frac { dx }{ dy } =-\frac { { e }^{ \frac { x }{ y } }\left( 1-\frac { x }{ y } \right) }{ { 1+e }^{ \frac { x }{ y } } } =\frac { \left( \frac { x }{ y } -1 \right) { e }^{ \frac { x }{ y } } }{ { 1+e }^{ \frac { x }{ y } } } =f(x,y)
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations 10

For each of the following differential equation in Q 11 to 15 find the particular solution satisfying the given condition:

Ex 9.5 Class 12 Maths Question 11.
(x + y) dy+(x – y)dx = 0,y = 1 when x = 1
Solution:
given
(x + y) dy+(x – y)dx = 0
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations 11
vedantu class 12 maths Chapter 9 Differential Equations 11.1

Ex 9.5 Class 12 Maths Question 12.
x²dy+(xy+y²)dx=0, y=1 when x=1
Solution:
\frac { dy }{ dx } =\frac { xy+{ y }^{ 2 } }{ { x }^{ 2 } } =f(x,y)
f(x,y) is homogeneous
∴ put y = vx
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations 12

Ex 9.5 Class 12 Maths Question 13.
\left( x{ sin }^{ 2 }\frac { y }{ x } -y \right) dx+xdy=0,y=\frac { \pi }{ 4 } ,when\quad x=1
Solution:
\left( x{ sin }^{ 2 }\frac { y }{ x } -y \right) dx+xdy=0
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations 13

Ex 9.5 Class 12 Maths Question 14.
\frac { dy }{ dx } -\frac { y }{ x } +cosec\left( \frac { y }{ x } \right) =0,y=0\quad when\quad x=1
Solution:
\frac { dy }{ dx } -\frac { y }{ x } +cosec\left( \frac { y }{ x } \right) =0
which is a homogeneous differential equation
vedantu class 12 maths Chapter 9 Differential Equations 14

Ex 9.5 Class 12 Maths Question 15.
2xy-{ y }^{ 2 }-{ 2x }^{ 2 }\frac { dy }{ dx } =0,y=2,when\quad x=1
Solution:
\frac { dy }{ dx } =\frac { y }{ x } +\frac { 1 }{ 2 } { \left( \frac { y }{ x } \right) }^{ 2 } …(i)
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations 15

Ex 9.5 Class 12 Maths Question 16.
A homogeneous equation of the form \frac { dx }{ dy } =h\left( \frac { x }{ y } \right) can be solved by making the substitution,
(a) y=vx
(b) v=yx
(c) x=vy
(d) x=v
Solution:
(c) option x = vy

Ex 9.5 Class 12 Maths Question 17.
Which of the following is a homogeneous differential equation?
(a) (a) (4x + 6y + 5)dy-(3y + 2x + 4)dx = 0
(b) (xy)dx-({ x }^{ 3 }+{ y }^{ 3 })dy
(c) ({ x }^{ 3 }+{ 2y }^{ 2 })dx+2xydy=0
(d) { y }^{ 2 }dx+{ (x }^{ 2 }-xy-{ y }^{ 2 })dy=0
Solution:
(d)

NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations Hindi Medium Ex 9.5

NCERT Solutions for Class 12 Maths Exercise 9.5 of Differential Equations
NCERT Solutions for Class 12 Maths Exercise 9.5
NCERT Solutions for Class 12 Maths Exercise 9.5 in PDF
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NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations Ex 9.6

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NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations Ex 9.6

Get Free NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations Ex 9.6 PDF in Hindi and English Medium. Sets Class 12 Maths NCERT Solutions are extremely helpful while doing your homework. Differential Equations Exercise 9.6 Class 12 Maths NCERT Solutions were prepared by Experienced LearnCBSE.in Teachers. Detailed answers of all the questions in Chapter 9 Class 12 Differential Equations Ex 9.6 provided in NCERT Textbook.

Free download NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations Ex 9.6 PDF in Hindi Medium as well as in English Medium for CBSE, Uttarakhand, Bihar, MP Board, Gujarat Board, BIE, Intermediate and UP Board students, who are using NCERT Books based on updated CBSE Syllabus for the session 2019-20.

NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations Ex 9.6

Find the general solution of the following differential equations in Q.1 to 12

Ex 9.6 Class 12 Maths Question 1.
\frac { dy }{ dx } +2y=sinx
Solution:
Given equation is a linear differential equation of the form \frac { dy }{ dx } +Py=Q;
Here, P = 2, Q = sin x
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations 1

Ex 9.6 Class 12 Maths Question 2.
\frac { dy }{ dx } +3y={ e }^{ -2x }
Solution:
\frac { dy }{ dx } +3y={ e }^{ -2x }
Here P = 3, IF={ e }^{ \int { p.dx } }={ e }^{ 3x }
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations 2
which is required equation

Ex 9.6 Class 12 Maths Question 3.
\frac { dy }{ dx } +\frac { y }{ x } ={ x }^{ 2 }
Solution:
\frac { dy }{ dx } +\frac { y }{ x } ={ x }^{ 2 }
IF={ e }^{ \int { \frac { 1 }{ x } dx } }={ e }^{ logx }=x
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations 3

Ex 9.6 Class 12 Maths Question 4.
\frac { dy }{ dx } +(secx)y=tanx\left( 0\le x<\frac { \pi }{ 2 } \right)
Solution:
Here, P = secx, Q = tanx; IF={ e }^{ \int { p.dx } }={ e }^{ \int { secx.dx } }
={ e }^{ log|secx+tanx| }
= sec x + tan x
i.e., The solu. is y.× I.F. = ∫Q × I.F. dx + c
or y × (secx+tanx) = ∫tanx(secx+tanx)dx+c
Reqd. sol. is
∴ y(secx + tanx) = (secx + tanx)-x + c

Ex 9.6 Class 12 Maths Question 5.
{ cos }^{ 2 }x\frac { dy }{ dx } +y=tanx\left( 0\le x\le \frac { \pi }{ 2 } \right)
Solution:
\frac { dy }{ dx } +{ y\quad sec }^{ 2 }x={ sec }^{ 2 }x\quad tanx
⇒ integrating factor = { e }^{ \int { { sec }^{ 2 }xdx } }={ e }^{ tanx }
byjus class 12 maths Chapter 9 Differential Equations 5

Ex 9.6 Class 12 Maths Question 6.
x\frac { dy }{ dx } +2y={ x }^{ 2 }logx
Solution:
\frac { dy }{ dx } +\frac { 2 }{ x } y\quad =\quad x\quad logx
Here P = \frac { 2 }{ x } and Q = x logx
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations 6

Ex 9.6 Class 12 Maths Question 7.
xlogx\frac { dy }{ dx } +y=\frac { 2 }{ x } logx
Solution:
\frac { dy }{ dx } +\frac { 1 }{ xlogx } y=\frac { 2 }{ { x }^{ 2 } }
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations 7

Ex 9.6 Class 12 Maths Question 8.
(1+x²)dy+2xy dx = cotx dx(x≠0)
Solution:
(1+x²)dy+2xy dx = cotx dx
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations 8

Ex 9.6 Class 12 Maths Question 9.
x\frac { dy }{ dx } +y-x+xy\quad cotx=0(x\neq 0)
Solution:
x\frac { dy }{ dx } +y-x+xy\quad cotx=0
x\frac { dy }{ dx } +(1+xcot x)y=x
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations 9
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations 9.1

Ex 9.6 Class 12 Maths Question 10.
(x+y)\frac { dy }{ dx } =1
Solution:
(x+y)\frac { dy }{ dx } =1
\frac { 1 }{ (x+y) } \frac { dx }{ dy } =1\Rightarrow \frac { dx }{ dy } =x+y
byjus class 12 maths Chapter 9 Differential Equations 10

Ex 9.6 Class 12 Maths Question 11.
ydx+(x-{ y }^{ 2 })dy=0
Solution:
ydx+(x-{ y }^{ 2 })dy=0
\Rightarrow y\frac { dx }{ dy } +x-{ y }^{ 2 }=0
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations 11

Ex 9.6 Class 12 Maths Question 12.
\left( { x+3y }^{ 2 } \right) \frac { dy }{ dx } =y(y>0)
Solution:
y\frac { dx }{ dy } =x+{ 3y }^{ 2 }\quad or\quad \frac { dx }{ dy } -\frac { x }{ y } =3y
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations 12
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations 12.1

For each of the following Questions 13 to is find a particular solution, satisfying the given condition:

Ex 9.6 Class 12 Maths Question 13.
\frac { dy }{ dx } +2ytanx=sinx,y=0\quad when\quad x=\frac { \pi }{ 3 }
Solution:
\frac { dy }{ dx } +(2tanx)y=sinx,P=2tanx
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations 13

Ex 9.6 Class 12 Maths Question 14.
\left( 1+{ x }^{ 2 } \right) \frac { dy }{ dx } +2xy=\frac { 1 }{ 1+{ x }^{ 2 } } ,y=0\quad when\quad x=1
Solution:
\frac { dy }{ dx } +\frac { 2x }{ 1+{ x }^{ 2 } } y=\frac { 1 }{ { \left( { 1+x }^{ 2 } \right) }^{ 2 } }
byjus class 12 maths Chapter 9 Differential Equations 14

Ex 9.6 Class 12 Maths Question 15.
\frac { dy }{ dx } -3ycotx=sin2x,y=2\quad when\quad x=\frac { \pi }{ 2 }
Solution:
Here P = -3cot x
Q = sin 2x
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations 15

Ex 9.6 Class 12 Maths Question 16.
Find the equation of the curve passing through the origin given that the slope of the tangent to the curve at any point (x,y) is equal to the sum of the coordinates of the point
Solution:
\frac { dy }{ dx } =x+y\Rightarrow \frac { dy }{ dx } -y=x\Rightarrow P=-1
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations 16

Ex 9.6 Class 12 Maths Question 17.
Find the equation of the curve passing through the point (0, 2) given that the sum of the coordinates of any point on the curve exceeds the magnitude of the slope of the tangent to the curve at that point by 5
Solution:
By the given condition
x+y-\left| \frac { dy }{ dx } \right|=5
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations 17
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations 17.1

Ex 9.6 Class 12 Maths Question 18.
The integrating factor of the differential equation x\frac { dy }{ dx } -y={ 2x }^{ 2 }
(a) { e }^{ -x }
(b) { e }^{ -y }
(c) \frac { 1 }{ x }
(d) x
Solution:
(c) P=\frac { -1 }{ x } \therefore IF={ e }^{ -\int { \frac { 1 }{ x } dx } }={ e }^{ -logx }=\frac { 1 }{ x }

Ex 9.6 Class 12 Maths Question 19.
The integrating factor of the differential equation \left( { 1-y }^{ 2 } \right) \frac { dx }{ dy } +yx=ay(-1<y<1) is
(a) \frac { 1 }{ { y }^{ 2 }-1 }
(b) \frac { 1 }{ \sqrt { { y }^{ 2 }-1 } }
(c) \frac { 1 }{ 1-{ y }^{ 2 } }
(d) \frac { 1 }{ \sqrt { { 1-y }^{ 2 } } }
Solution:
(d) \left( { 1-y }^{ 2 } \right) \frac { dx }{ dy } +yx=ay
byjus class 12 maths Chapter 9 Differential Equations 19

NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations Hindi Medium Ex 9.6

NCERT Solutions for Class 12 Maths Exercise 9.6 of Differential Equations
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NIOS Class 12 Date Sheet 2019 | Download 12th Class Date Sheet @ nios.ac.in

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NIOS Class 12 Date Sheet 2019: NIOS Class 12 Date Sheet has been released. The National Institute of Open Schooling has released the Class 12 date sheet for Block 2. NIOS administers the Class 12 exams twice a year. The class 12 exam is conducted once in the month of April-May (Block 1) and then in the month of October-November (Block 2). Students enrolled under NIOS regular and vocational exams can check their 12th Date Sheet October 2019 released by NIOS through the official website link provided below.

On September 4th, National Institute of Open Schooling (NIOS) has published the time table for 12th October session on its website – nios.ac.in. As per the calendar, the exams will be conducted from October 03, 2019, to November 1st, 2019. The practical exams will be conducted from September 16th to September 30th, 2019. The timings of the exams will be 2.30 PM to 5.30 PM.

The NIOS is a government-approved institute and is one of the biggest open schooling systems in the world. It conducts exams for both senior secondary and secondary students belonging to NIOS. Read the complete article to get the NIOS Class 12 Date Sheet.

Check Out NIOS Class 10 Date Sheet

NIOS Class 12 Date Sheet 2019

The date sheet for NIOS class 12th examination has been designed in a tabular form for all the students so that they can prepare for the exam accordingly. Check the table below

DayExam DatesSubject Name
ThursdayOctober 3, 2019Sanskrit
FridayOctober 4, 2019Employability Skills and Entrepreneurship
WednesdayOctober 9, 2019Psychology
ThursdayOctober 10, 2019Urdu
FridayOctober 11, 2019Hindi
MondayOctober 14, 2019Geography
TuesdayOctober 15, 2019Bengali, Tamil, Odiya, Gujarati, Punjabi, Arabic, Persian
WednesdayOctober 16, 2019English
FridayOctober 18, 2019Home Science
SaturdayOctober 19, 2019Physics, History, Library, and Inform. Science, Sanskrit Vyakaran
MondayOctober 21, 2019Economics
TuesdayOctober 22, 2019Chemistry, Political Science, Mass Communication, Sanskrit Sahitya
WednesdayOctober 23, 2019Mathematics
ThursdayOctober 24, 2019Data Entry Operations
FridayOctober 25, 2019Painting Theory
WednesdayOctober 30, 2019Computer Science, Environmental Science, Sociology, Bharatiya Darshan
ThursdayOctober 31, 2019Biology, Accountancy, Introduction to Law, Veda Adhyayan
FridayNovember 1, 2019Business Studies

NIOS Date Sheet for Practical Exam

DatesSubjects
September 16 to 20, 2019Home Science (321), Biology (314), Geography (316), Painting (332)
September 21 to 25, 2019Chemistry (313)
Physics (312)
Environmental Science (333)
September 26 to September 30, 2019Computer science (330)
Data Entry Operations (336)
Mass Communication (335)
Library and Information Science (339)

How To Download NIOS 12th Date Sheet?

To download the date sheet for NIOS 12th or senior secondary class exam, students have to follow the below-given steps.

  • Visit the official website of National Institute of Open Schooling, nios.ac.in.
  • Click on the link given for Examination/Result at the home page.
  • Check the notification declared for date sheet and click on the relevant link.
  • The PDF format of the 12th date sheet will get opened on your screen.
  • Download the pdf and keep for the reference.

Download NIOS Date Sheet 2019 (For All India Exam Centres)

Download NIOS Date Sheet 2019 (For Overseas Exam Centres)

Instructions To be Followed

  • Students can download the Intimation-cum-Hall Ticket from the NIOS official website.
  • Practical exams will be conducted according to the schedule mentioned in the above table, at the respective Accredited Institute (AIs) of NIOS.
  • Students will be divided into batches. The number of candidates in a batch will be decided based on the capacity of the lab for the practical exam. Therefore, students are requested to contact the center superintendent or coordinator.
  • The results will be declared after 6 weeks of the conduct of the exam, regarding which the notification will be published on the official website.
  • Marksheets, provisional certificate, and Migration-cum-transfer certificate will be issued by the respective Accredited Institute (AIs) of NIOS.
  • There will be no change in the dates of the exam.

About NIOS

NIOS is “Open School” to provide to the requirements of a heterogeneous group of students up to pre-degree level. It was commenced as a project with in-built extensibility by the Central Board of Secondary Education (CBSE) in 1979. In 1986, the National Policy on Education proposed establishing of Open School System for enlarging open learning facilities in a phased method at secondary level all over the country as an autonomous system with its own curriculum and examination driving to certification.

The post NIOS Class 12 Date Sheet 2019 | Download 12th Class Date Sheet @ nios.ac.in appeared first on Learn CBSE.

NIOS Class 10 Date Sheet 2019 (Released) | Download 10th Class Date Sheet @ nios.ac.in

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NIOS Class 10 Date Sheet 2019: National Institute of open schooling also known as NIOS conducts the NIOS exam for class 10 twice in a year. The block 1 exam is conducted in the month of April/May while the exams for block 2 are conducted in October/November. Recently, NIOS has released recently the NIOS class 10 date sheet for block 1. According to this date sheet, NIOS class, the 10-time table is given the article. Also, NIOS officials, later on, revealed the revised time table for practical exams. NIOS class block 1 practical exam will be conducted from March 16, 2019, to March 30, 2019. While the theory exams will be conducted from April 3, 2019, to May 4, 2019. The official NIOS class 10 date sheet released by the NIOS gives you a detailed exam sheet for class 10.

NIOS class block 2 practical exam will be conducted from September 16, 2019 and ends on September 30th 2019. While the theory exams will be conducted from October 4th, 2019, to November 5th, 2019. The official NIOS class 10 date sheet released by the NIOS gives you a detailed exam sheet for class 10.

Download NIOS Class 10 Date Sheet 2019 for All India Exam Centres

NIOS Class 10 Date Sheet Overview

Name of the ExamNIOS 10th Block 2 Exam
Conducting BodyNational Institute of Open Schooling (NIOS)
Exam ModeOffline
Exam Start DatePractical – September 16th, 2019
Theory – October 4th, 2019
Exam End DatePractical – 30th September 2019
Theory – November 1st, 2019
Official Websitenios.ac.in

Check Out NIOS Class 12 Date Sheet

NIOS Class 10 Date Sheet 2019 for Theory Exam

Below are the dates for all the exams conducted by the NIOS for class 10

DayExam DatesSubject Name
FridayOctober 4, 2019Employability Skills
WednesdayOctober 9, 2019Bengali, Marathi, Telugu, Gujarati, Kannada, Punjabi, Assamese, Nepali, Malayalam, Arabic, Persian, Tamil
ThursdayOctober 10, 2019Sanskrit
FridayOctober 11, 2019Urdu
MondayOctober 14, 2019English
TuesdayOctober 15, 2019Home science
WednesdayOctober 16, 2019Indian Culture and Heritage
FridayOctober 18, 2019Social Science
SaturdayOctober 19, 2019Hindi
MondayOctober 21, 2019Science & Technology
TuesdayOctober 22, 2019Psychology, Sanskrit Sahitya
WednesdayOctober 23, 2019Data entry operations
ThursdayOctober 24, 2019Panting Theory, Sanskrit Vyakaran
FridayOctober 25, 2019Mathematics
WednesdayOctober 30, 2019Business Studies, Bhartiya Darshan
ThursdayOctober 31, 2019Economics, Veda Adhyayan
FridayNovember 1, 2019Accountancy

NIOS Class 10 Date Sheet for Overseas

NIOS Class 10 Date Sheet for Practical Exams

Below is the practical exam date sheet for NIOS class 10

DatesExams
September 16 to 20, 2019Science and Technology
September 21 to 25, 2019Painting, Maths
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How To Download NIOS Class 10 Date From Official NIOS Website?

  • Go to the official website of NIOS which is nios.ac.in.
  • On the main page, scroll down and you will find a link that reads “Examination results”. Click on this link.
  • Now, for class 10 date sheet, click on the “Date sheet for secondary class (10th class)” link.
  • Your class 10 date sheet will be displayed on the screen. Download the NIOS class 10 date sheet and keep one of its copies handy with you.

Details Mentioned On NIOS Class 10 Date Sheet

The following details will be mentioned on the NIOS class 10 date sheet

  • Day of the exam
  • Date of the exam
  • Timings of the exam
  • Subject code
  • Important instructions
  • List of all the subjects

Important Points for NIOS Class 10 Date Sheet

  1. There will be no changes made in the exam dates. The dates mentioned in the date sheet will be final.
  2. Students are advised to carry their admit card or hall ticket to the exam center. No one will allow entering the examination center without a valid admit card.
  3. Candidates can download their intimidation cum admit card which will be available on NIOS official website.
  4. Practical exams for the candidates will be conducted on the given AIs where the candidates can enroll their admission.
  5. For practical exams, students at the particular AI will be separated in various batches depending upon the capacity of the lab and various other factors.
  6. The practical exams will be done in small batches, so the candidate must contact the coordinator of AI or center superintendent in well advance to know the given dates along with the batches allotted to them.

The post NIOS Class 10 Date Sheet 2019 (Released) | Download 10th Class Date Sheet @ nios.ac.in appeared first on Learn CBSE.


NCERT Solutions for Class 11 Maths Chapter 1 Sets Ex 1.2

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NCERT Solutions for Class 11 Maths Chapter 1 Sets Ex 1.2

Get Free NCERT Solutions for Class 11 Maths Chapter 1 Sets Ex 1.2 PDF in Hindi and English Medium. Sets Class 11 Maths NCERT Solutions are extremely helpful while doing your homework. Sets Exercise 1.2 Class 11 Maths NCERT Solutions were prepared by Experienced LearnCBSE.in Teachers. Detailed answers of all the questions in Chapter 1 Class 11 Sets Ex 1.2 provided in NCERT Textbook.

Free download NCERT Solutions for Class 11 Maths Chapter 1 Sets Ex 1.2 PDF in Hindi Medium as well as in English Medium for CBSE, Uttarakhand, Bihar, MP Board, Gujarat Board, BIE, Intermediate and UP Board students, who are using NCERT Books based on updated CBSE Syllabus for the session 2019-20.

Class 11 Maths NCERT Solutions Chapter 1 Sets Ex 1.2

Q.1: Which of the following given below is null set?
(i).  Set of odd natural numbers which is divisible by 2.
(ii).  Set of even numbers which are prime
(iii).  {x: x is a natural number, x<5 and x>7}
(iv).  {y: y is a point common to any two parallel lines}

Solution:
sets class 11

Q.2:  State whether the following sets are infinite or finite:
(i).  A set of months of a year.
(ii).  {1, 2, 3 ….}
(iii).  {1, 2, 3…99, 100}
(iv).  The set of positive integers which are greater than 100.
(v).  The set of prime numbers which are less than 99

Solution:
class 11 maths ncert solutions chapter 1

Q.3: State whether the following sets are infinite or finite:
(i).  The set of lines parallel to the x – axis.
(ii).  The set of letters in the vowels.
(iii).  The set of numbers multiple of 10.
(iv).  The set of humans living on Earth.
(v).  The set of circles passing through the origin (0, 0).

Solution:
ncert solutions for class 11 maths chapter 1

Q.4: In the following set given below, state whether A = B or not:
(i).  A = {w, x, y, z}
B = {z, y, x, w}
(ii).  A = {5, 9, 13, 17}
B = {9, 5, 17, 19}
(iii).  A = {4, 2, 6, 10, 8}
B = {x: x is positive even integer and x10 }
(iv).   A = {x: x is a multiple of 10}
B = {10, 15, 20, 25, 30 …}

Solution:
ncert solutions for class 11 maths chapter 1 pdf

Q.5 In the following set given below, is the pair of sets equal?
(i). A = {3, 4}
B = {y: y is solution of +5y+6=0}
(ii).  A = {a: a is a letter in the word FOLLOW}
B = {b: b is a letter in the word WOLF}

Solution:
+1 maths NCERT Solutions Ex 1.2 Q 5

Q.6: From the following sets, select equal sets:
A = {2, 4, 8, 12}
B = {1, 2, 3, 4}
C = {4, 8, 12, 14}
D = {3, 1, 4, 2}
E = {–1, 1}
F = {0, a}
G = {1, –1}
H = {0, 1}

Solution:
class 11 maths ch 1
class 11 maths chapter 1 Ex 1.2

NCERT Solutions for Class 11 Maths Chapter 1 Sets Ex 1.2 in Hindi

Class 11 Maths Ex 1.2 Sets in Hindi
(v) 99 से छोटे अभाज्य पूर्णांकों का समुच्चय {2, 3, 5, 7, …… 97} है जिसमें अवयवों की संख्या निश्चित है।
अत: यह एक परिमित समुच्चय है।

प्रश्न 3.
निम्नलिखित समुच्चयों में से प्रत्येक के लिए बताइए कि कौन परिमित है और कौन अपरिमित है?
(i) x-अक्ष के समांतर रेखाओं का समुच्चय।
(ii) अंग्रेजी वर्णमाला के अक्षरों का समुच्चय।
(iii) उन संख्याओं का समुच्चय जो 5 के गुणज हैं।
(iv) पृथ्वी पर रहने वाले जानवरों का समुच्चय
(v) मूल बिन्दु (0, 0) से होकर जाने वाले वृत्तों का समुच्चय।
हल:
(i) x-अक्ष के समांतर अनंत रेखाएँ खींची जा सकती हैं। अत: यह एक अपरिमित समुच्चय है।
(ii) अंग्रेजी वर्णमाला में कुल 26 अक्षर होते हैं। इन अक्षरों से बनने वाला समुच्चय परिमित होगा।
(iii) 5 से विभाजित होने वाली संख्याओं का समुच्चय {5, 10, 15, 20, ….} है, जिसमें अनंत अवयव हैं। अतः यह एक अपरिमित समुच्चय है।
(iv) पृथ्वी पर रहने वाले जानवरों का समुच्चय परिमित होगा।
(v) मूल बिन्दु को केन्द्र मानकर अनन्त वृत्त चे जा सकते हैं। अत: यह अपरिमित होगा।

प्रश्न 4.
निम्नलिखित में बताइए कि A = B है अथवा नहीं है।
(i) A = {a, b, c, a}, B = {a, c, b, a}
(ii) A = {4, 8, 12, 16}, B = {8, 4, 16, 18}
(iii) A = {2, 4, 6, 8, 10}, B = {x : x सम धन पूर्णाक है और x ≤ 10}
(iv) A = {x : x संख्या 10 का एक गुणज है}, B = {10, 15,20, 25, 30, …}
हल:
(i) A और B दोनों समुच्चयों के अवयव a, b, c, d हैं अतः A = B.
(ii) A में अवयव 12 है परन्तु B में नहीं है अतः A ≠ B.
(iii) A और B दोनों समुच्चयों में अवयव 2, 4, 6, 8 और 10 हैं। अतः A = B.
(iv) A = {10, 20, 30, 40, …..}, B = {10, 15, 25, 30, ….}
10 के गुणजों में 5, 15, 25 नहीं आता है। अतः A ≠ B.

प्रश्न 5.
क्या निम्नलिखित समुच्चय युग्म समान हैं ? कारण सहित बताइए।
(i) A = {2, 3}
B = {x : x समीकरण x² + 5x + 6 = 0 का एक हल है।}
(ii) A = {k : x शब्द ‘FOLLOW’ का एक अक्षर है।}
B = {y : y शब्द ‘WOLF का एक अक्षर है।}
हल:
(i) A = {2, 3}, B = x : x समीकरण x² + 5x + 6 = 0} = {-2, -3}
स्पष्ट है कि समुच्चय A और B के अवयव भिन्न हैं।
अत: A ≠ B.
(ii) A = {F, O, L, W}, B = {W, O, L, F}
समुच्च्य A और B के अवयव समान हैं। अत: A = B.

प्रश्न 6.
नीचे दिए गए समुच्चयों में से समान समुच्चयों का चयन कीजिए:
A = {2, 4, 8, 12}
B = {1, 2, 3, 4}
C = {4, 8, 12, 14}
D = {3, 1, 4, 2}
E = {- 1, 1}
F = {0, a}
G = {1, -1}
H = {0, 1}
हल:
यहाँ समुच्चय B और D के अवयव 1, 2, 3, 4, हैं।
B = D
तथा समुच्चय E और G में -1, 1 अवयव समान हैं।
E = G

The post NCERT Solutions for Class 11 Maths Chapter 1 Sets Ex 1.2 appeared first on Learn CBSE.

DTE Scholarship 2019 Maharashtra | Dates, Eligibility, Awards, Application Process, Documents 

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DTE Maharashtra Scholarship 2019: Directorate of Technical Education (DTE), Maharashtra released the notification for DTE Scholarship 2019. DTE Scholarship scheme has initiated for the economically weak, and state minority communities students. DTE, Maharashtra offers Scholarship to the candidates studying in higher education or Post HSC courses. This scholarship provides financial help to economically weak and minority communities students for pursuing professional and technical courses. Candidates who are domiciled in Maharashtra State can only apply for this scholarship scheme.

Application for DTE scholarship will be available on the official website of DTE, Maharashtra. Candidates can fill the application form and apply for this scholarship on or before 28th February 2020. The article below provides the information that is the date eligibility, awards, and so forth about DTE Scholarship 2019.

DTE Scholarship 2019

DTE Scholarship 2019 is one of the initiatives of DTE, Maharashtra. It encourages lower-income students for pursuing professional and technical courses. This scholarship helps economically backward students to complete higher education without financial hassle. This scholarship was initiated with the help of the mahaDBT mahait team by the Minority Department of DTE, Maharashtra. This scholarship is also available for different religion students. Those who study diploma, degree and engineering courses every year.

Different religion students including Muslim, Christians, Buddhists, and so forth avail this scholarship. DTE, Maharashtra is an exceptionally flexible and universally competitive higher education institute. It is responsive to the institutional, and socio developmental needs of the Maharashtra people. It is working towards economic growth and technical manpower development to come across the needs of the industry. DTE Scholarship mainly consists of the following two scholarships.

Detailed List of DTE Scholarship in Maharashtra

Scholarship nameProviderApplication Timeline
Rajarshi Chhatrapati Shahu Maharaj Shikshan Shulkh Shishyavrutti YojnaDirectorate of Technical Education (DTE), Maharashtra StateSeptember – January
Dr. Panjabrao Deshmukh Vasatigruh Nirvah Bhatta YojnaDirectorate of Technical Education (DTE), Maharashtra StateSeptember – January

Scholarships for Students

DTE Scholarship Overview

ParticularsDetails
Conducting BodyDirectorate of Technical Education (DTE), Maharashtra
Scholarship NameDTE Scholarship
Applicable StateMaharashtra
Websitewww.dtemaharashtra.gov.in

DTE Scholarship 2019 – Important Dates

EventsImportant Dates
Start of DTE Scholarship ApplicationSeptember 4, 2019
Deadline to Submit DTE Scholarship ApplicationFebruary 28, 2020

DTE Scholarship 2019 – Eligibility Criteria

Candidates must fulfill the following eligibility criteria to apply for DTE Scholarship:

  • Candidate must be an Indian national.
  • Candidate should be the domicile or permanent resident of Maharashtra state.
  • Candidates from outside the state are not eligible.
  • Candidate should be “Bonafide Student of Institute”. Also, admitted for the professional and technical courses.
  • Candidates from the deemed university are not eligible for this scholarship.
  • Candidates must be admitted through the Centralized Admission Process (CAP).
  • Candidates cannot avail any other scholarship while applying for this scholarship.
  • Only 2 children from a family are allowed for the benefit of the scheme for the current academic year.
  • Candidates annual family income should not be more than 8 Lakhs.
  • The minimum 50% attendance in the previous semester is required for this scholarship.
  • Candidates should not have a gap of two or more years during the course duration.

DTE Scholarship 2019 – Selection Process

The Selections for DTE scholarship are made on the basis of Merit. The final selection considers academic merit prescribed for each selected courses. The financial needs of the candidates also take into account for the selection. State Govt. is responsible for the selection of candidates after verifying the application. Candidates are required to link their Aadhar number to the Bank account to avail the benefits of this scholarship. The scholarship amount will be transferred directly to the selected candidate’s bank account. The AapleSarkar DBT portal will be used to transfer the scholarship amount.

DTE Scholarship 2019 – Awards

There are variable awards agreed to the permanent resident students of Maharashtra under DTE scholarship. These awards can be accessed through the mahaDBT portal. Refer to the below award details that a candidate can avail under these 2 DTE scholarships.

Rajarshi Chhatrapati Shahu Maharaj Shikshan Shulkh Shishyavrutti Yojna

  • Candidates of this scholarship will receive 50% of the tuition and exam fees.

Dr. Panjabrao Deshmukh Vasatigruh Nirvah Bhatta Yojna

  • The child of registered labor/Alpabhudharak (marginal landholder) will be given the following awards:
  • Candidates who belong to the institutes in MMRDA/ PMRDA/ Aurangabad City/ Nagpur City will be awarded Rs. 30,000/- for 10 months.
  • The scholarship amount of Rs. 20,000/- is awarded to candidates in other areas for 10 months.
  • Candidates whose family income is up to 8 Lakhs per annum will be given the following awards:
  • Candidates who belong to the institutes in MMRDA/ PMRDA/ Aurangabad City/ Nagpur City will be awarded Rs. 10,000/- for 10 months.
  • The scholarship amount of Rs. 8,000/- is awarded to candidates in other areas for 10 months.

DTE Scholarship 2019 – Application Process

The application for DTE Scholarship will be available online on September 4, 2019. Candidates can fill and submit the application on or before 28th February 2020. Candidates can apply for this scholarship through Aaple Sarkar DBT portal. It is a unique platform run by the Govt. of Maharashtra to help its citizens for availing benefits of different schemes. From the academic session 2018-19, it is mandatory to link an Aadhaar number. So, candidates should have an Aadhar number to apply for any scholarship through the portal. Candidates must read the instructions carefully and keep ready all the documents prior to fill-up the form. Refer to the below points to apply for DTE Scholarship 2019.

  • Visit the official website www.dtemaharashtra.gov.in.
  • Click on “Click here for Online Application System” available under Scholarships section.
  • It will go to the homepage of the Aaple Sarkar DBT portal. Click on the Directorate of Technical Education (DTE) link to see the scholarships. Click on the scholarship to register yourself.
  • Click on the “New Applicant Registration” link to register for the DTE scholarship. Then, enter the information of Aadhar number and proceed.
  • Candidates have to authenticate their Aadhar number in 2 ways. Either by choosing OTP or Biometric authentication.
  • If the candidate’s mobile number is registered with Aadhar then they can select OTP authentication. Otherwise, they have to select Biometric authentication.
  • Fill all the necessary information and create a user Id and password to access the portal.
  • Login to mahaDBT website by clicking on “Applicant Login”.
  • Fill all the necessary details of the profile correctly.
  • Recheck the filled application to avoid rejection by the authorities during verification.
  • Upload the required documents along with the application.
  • Finally, submit the filled application along with the documents.

DTE Scholarship 2019 – Checklist of Documents

The below-mentioned documents must be uploaded along with the application to apply for DTE Scholarship.

  • Candidates need to upload the mark sheets of class 10th and onwards.
  • Candidates need to provide domicile certificate of Maharashtra State.
  • Candidates should provide caste certificate
  • Annual family income certificate and affidavit of belonging to the minority community
  • Candidates need to provide School leaving / transfer certificate
  • Candidates have to provide an undertaking “In the current year not more than 2 beneficiaries from family”
  • Candidates need to upload a CAP related document
  • Candidates also need to upload a proof of Biometric attendance through the interface UIDAI
  • Residential proof (PAN card, voter Id, aadhar card, and so forth.)
  • Register labor certificate or Alpabhudarak (marginal landholder) certificate. For those who are applying for Dr. Panjabrao Deshmukh Vasatigruh Nirvah Bhatta Yojna scholarship.
  • Candidates who want to opt for hostel need to provide hosteller documents.
  • In the case of the private hostels or paying guest, agreement with the owner should be uploaded

DTE Scholarship 2019 – Contact Details

In case of any queries related to DTE Scholarship or any assistance contact on the below-mentioned details.

Phone Number022-22641150
AddressDirectorate of Technical Education (DTE), Mahapalika Marg, Mumbai 3, Opp.Metro Cinema, Post Box No. 1967, Mumbai – 400001, Maharashtra State

FAQ’s on DTE Scholarship 2019

Question 1.
What is the use of Aaple Sarkar DBT in  DTE Scholarship?

Answer:
Govt. of Maharashtra has launched Aaple Sarkar DBT portal. This portal uses to transfer directly the benefits and subsidies of the schemes into the bank account of the beneficiary. These schemes include pension, disaster, e-scholarships, and so forth.

Question 2.
Can the applicant edit the application form after submission?

Answer:
Yes, if the institute sends back the application to the applicant for modification then the applicant can make the necessary changes.

Question 3.
How the Aadhar based DBT is helpful for the beneficiaries?

Answer:
The Aadhar based DBT confirms that nobody can claim a share of the benefits by imitating the candidate.

Question 4.
What are the DTE Scholarship application starting and closing date?

Answer:
The DTE Scholarship application starting date is 4th September 2019. Whereas, the application closing date is 28th February 2020.

Question 5.
How is the DTE Scholarship helpful to Maharashtra state students?

Answer:
DTE Scholarship offers financial help to economically weak and minority communities students for pursuing professional and technical courses.

Hope this article will help you to get more information about DTE Scholarship 2019. If you have queries related to DTE Scholarship, then leave it in the comment box to get in touch with us.

The post DTE Scholarship 2019 Maharashtra | Dates, Eligibility, Awards, Application Process, Documents  appeared first on Learn CBSE.

NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra Ex 10.2

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NCERT Solutions for Class 12th Maths Chapter 10 Vector Algebra Ex 10.2

Get Free NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra Ex 10.2 PDF in Hindi and English Medium. Sets Class 12 Maths NCERT Solutions are extremely helpful while doing your homework. Vector Algebra Exercise 10.2 Class 12 Maths NCERT Solutions were prepared by Experienced LearnCBSE.in Teachers. Detailed answers of all the questions in Chapter 10 Class 12 Vector Algebra Ex 10.2 provided in NCERT Textbook.

Free download NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra Ex 10.2 PDF in Hindi Medium as well as in English Medium for CBSE, Uttarakhand, Bihar, MP Board, Gujarat Board, BIE, Intermediate and UP Board students, who are using NCERT Books based on updated CBSE Syllabus for the session 2019-20.

NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra Ex 10.2

Ex 10.2 Class 12 Maths Question 1.
Compute the magnitude of the following vectors:
\overrightarrow { a } =\hat { i } +\hat { j } +\hat { k } ,\overrightarrow { b } =\hat { 2i } -\hat { 7j } -\hat { 3k }
\overrightarrow { c } =\frac { 1 }{ \sqrt { 3 } } \hat { i } +\frac { 1 }{ \sqrt { 3 } } \hat { j } -\frac { 1 }{ \sqrt { 3 } } \hat { k }
Solution:
\overrightarrow { a } =\hat { i } +\hat { j } +\hat { k }
\left| \overrightarrow { a } \right| =\sqrt { { 1 }^{ 2 }+{ 1 }^{ 2 }+{ 1 }^{ 2 } }
NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra 1
NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra 1.1

Ex 10.2 Class 12 Maths Question 2.
Write two different vectors having same magnitude.
Solution:
\overrightarrow { a } =\hat { i } +\hat { 2j } +\hat { 3k } ,\overrightarrow { b } =\hat { 3i } +\hat { 2j } +\hat { k }
NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra 2
Such possible answers are infinite

Ex 10.2 Class 12 Maths Question 3.
Write two different vectors having same direction.
Solution:
Let the two vectors be
\overrightarrow { a } =\hat { i } +\hat { j } +\hat { k } ,\overrightarrow { b } =\hat { 3i } +\hat { 3j } +\hat { 3k }
tiwari academy class 12 maths Chapter 10 Vector Algebra 3
Hence vectors \overrightarrow { a } ,\overrightarrow { b } have the same direction but different magnitude

Ex 10.2 Class 12 Maths Question 4.
Find the values of x and y so that the vectors \overrightarrow { 2i } +\overrightarrow { 3j } \quad and\quad \hat { xi } +\hat { yj } are equal.
Solution:
We are given \overrightarrow { 2i } +\overrightarrow { 3j } \quad and\quad \hat { xi } +\hat { yj }
If vectors are equal, then their respective components are equal. Hence x = 2, y = 3.

Ex 10.2 Class 12 Maths Question 5.
Find the scalar and vector components of the vector with initial point (2,1) and terminal point (-5,7).
Solution:
LetA(2, 1) be the initial point and B(-5,7) be the terminal point \overrightarrow { AB } =\left( { x }_{ 2 }-{ x }_{ 1 } \right) \hat { i } +\left( { y }_{ 2 }-{ y }_{ 1 } \right) \hat { j } =-\hat { 7i } +\hat { 6j }
∴The vector components are -\hat { 7i } and\hat { 6j } and scalar components are – 7 and 6.

Ex 10.2 Class 12 Maths Question 6.
Find the sum of three vectors:
\overrightarrow { a } =\hat { i } -\hat { 2j } +\hat { k } ,\overrightarrow { b } =-2\hat { i } +\hat { 4j } +5\hat { k } \quad and\quad \overrightarrow { c } =\hat { i } -\hat { 6j } -\hat { 7k } ,
Solution:
\overrightarrow { a } =\hat { i } -\hat { 2j } +\hat { k } ,\overrightarrow { b } =-2\hat { i } +\hat { 4j } +5\hat { k } \quad and\quad \overrightarrow { c } =\hat { i } -\hat { 6j } -\hat { 7k } ,
\overrightarrow { a } +\overrightarrow { b } +\overrightarrow { c } =\hat { 0i } -\hat { 4j } -\hat { k } =-4\hat { i } -\hat { k }

Ex 10.2 Class 12 Maths Question 7.
Find the unit vector in the direction of the vector
\overrightarrow { a } =\hat { i } +\hat { j } +\hat { 2k }
Solution:
\overrightarrow { a } =\hat { i } +\hat { j } +\hat { 2k }
NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra 7

Ex 10.2 Class 12 Maths Question 8.
Find the unit vector in the direction of vector \overrightarrow { PQ } , where P and Q are the points (1,2,3) and (4,5,6) respectively.
Solution:
The points P and Q are (1, 2, 3) and (4, 5, 6) respectively
\overrightarrow { PQ } =(4-1)\hat { i } +(5-2)\hat { j } +(6-3)\hat { k }
tiwari academy class 12 maths Chapter 10 Vector Algebra 8

Ex 10.2 Class 12 Maths Question 9.
For given vectors \overrightarrow { a } =2\hat { i } -\hat { j } +2\hat { k } \quad and\quad \overrightarrow { b } =-\hat { i } +\hat { j } -\hat { k } find the unit vector in the direction of the vector \overrightarrow { a } +\overrightarrow { b }
Solution:
\overrightarrow { a } =2\hat { i } -\hat { j } +2\hat { k } \quad and\quad \overrightarrow { b } =-\hat { i } +\hat { j } -\hat { k }
NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra 9

Ex 10.2 Class 12 Maths Question 10.
Find a vector in the direction of 5\hat { i } -\hat { j } +2\hat { k } which has magnitude 8 units.
Solution:
The given vector is \overrightarrow { a } =5\hat { i } -\hat { j } +2\hat { k }
NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra 10

Ex 10.2 Class 12 Maths Question 11.
Show that the vector 2\hat { i } -3\hat { j } +4\hat { k } \quad and\quad -4\hat { i } +6\hat { j } -8\hat { k } are collinear.
Solution:
\overrightarrow { a } =2\hat { i } -3\hat { j } +4\hat { k } \quad and\quad \overrightarrow { b } =-4\hat { i } +6\hat { j } -8\hat { k }
=-2(2\hat { i } -3\hat { j } +4\hat { k } )
vector \overrightarrow { a } \quad and\quad \overrightarrow { b } have the same direction they are collinear.

Ex 10.2 Class 12 Maths Question 12.
Find the direction cosines of the vector \hat { i } +2\hat { j } +3\hat { k }
Solution:
let \overrightarrow { p } =\hat { i } +2\hat { j } +3\hat { k }
Now a = 1,b = 2,c = 3
tiwari academy class 12 maths Chapter 10 Vector Algebra 12

Ex 10.2 Class 12 Maths Question 13.
Find the direction cosines of the vector joining the points A (1,2, -3) and B(-1, -2,1), directed fromAtoB.
Solution:
Vector joining the points A and B is
({ x }_{ 2 }-{ x }_{ 1 })\hat { i } +({ y }_{ 2 }-{ y }_{ 1 })\hat { j } +({ z }_{ 2 }-{ z }_{ 1 })\hat { k }
NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra 13

Ex 10.2 Class 12 Maths Question 14.
Show that the vector \hat { i } +\hat { j } +\hat { k } are equally inclined to the axes OX, OY, OZ.
Solution:
Let \hat { i } +\hat { j } +\hat { k } =\overrightarrow { a } , Direction cosines of vector x\hat { i } +y\hat { j } +z\hat { k } are
NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra 14
which shows that the vector a is equally inclined to the axes OX, OY, OZ.

Ex 10.2 Class 12 Maths Question 15.
Find the position vector of a point R which divides the line joining the points whose positive vector are P(\hat { i } +2\hat { j } -\hat { k } )\quad and\quad Q(-\hat { i } +\hat { j } +\hat { k } ) in the ratio 2:1
(i) internally
(ii) externally.
Solution:
(i) The point R which divides the line joining the point P(\overrightarrow { a } )\quad and\quad Q(\overrightarrow { b } ) in the ratio m : n
NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra 15
tiwari academy class 12 maths Chapter 10 Vector Algebra 15.1

Ex 10.2 Class 12 Maths Question 16.
Find position vector of the mid point of the vector joining the points P (2,3,4) and Q (4,1, -2).
Solution:
Let \overrightarrow { OP } =2\hat { i } +3\hat { j } +4\hat { k } \quad and\quad \overrightarrow { OQ } =4\hat { i } +\hat { j } -2\hat { k }
NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra 16

Ex 10.2 Class 12 Maths Question 17.
Show that the points A, B and C with position vector \overrightarrow { a } =3\hat { i } -4\hat { j } -4\hat { k } ,\overrightarrow { b } =2\hat { i } -\hat { j } +\hat { k } and\quad \overrightarrow { c } =\hat { i } -3\hat { j } -5\hat { k } respectively form the vertices of a right angled triangle.
Solution:
\overrightarrow { AB } =\overrightarrow { b } -\overrightarrow { a } =-\hat { i } +3\hat { j } +5\hat { k }
NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra 17
NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra 17.1

Ex 10.2 Class 12 Maths Question 18.
In triangle ABC (fig.), which of the following is not
NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra 18
(a) \overrightarrow { AB } +\overrightarrow { BC } +\overrightarrow { CA } =\overrightarrow { 0 }
(b) \overrightarrow { AB } +\overrightarrow { BC } -\overrightarrow { AC } =\overrightarrow { 0 }
(c) \overrightarrow { AB } +\overrightarrow { BC } -\overrightarrow { CA } =\overrightarrow { 0 }
(d) \overrightarrow { AB } -\overrightarrow { CB } +\overrightarrow { CA } =\overrightarrow { 0 }
Solution:
We know that
\overrightarrow { AB } +\overrightarrow { BC } +\overrightarrow { CA } =\overrightarrow { 0 }
\overrightarrow { AB } +\overrightarrow { BC } -\overrightarrow { AC } =\overrightarrow { 0 }
Hence option (c) is not correct

Ex 10.2 Class 12 Maths Question 19.
If \overrightarrow { a } ,\overrightarrow { b } are two collinear vectors then which of the following are incorrect:
(a) \overrightarrow { b } =\lambda \overrightarrow { a } , for some scalar λ.
(b) \overrightarrow { a } =\pm \overrightarrow { b }
(c) the respective components of \overrightarrow { a } ,\overrightarrow { b } are proportional.
(d) both the vectors \overrightarrow { a } ,\overrightarrow { b } have same direction, but different magnitudes.
Solution:
Options (d) is incorrect since both the vectors \overrightarrow { a } ,\overrightarrow { b } , being collinear, are not necessarily in the same direction. They may have opposite directions. Their magnitudes may be different.

NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra Hindi Medium Ex 10.2

NCERT Solutions for Class 12 Maths Exercise 10.2 of Vector Algebra
NCERT Solutions for Class 12 Maths Exercise 10.2
12 Maths ex 9.2
10.2 of 12 Maths
NCERT Solutions for Class 12 Maths Exercise 10.2 in Hindi Medium PDF
NCERT Solutions for Class 12 Maths Exercise 10.2 for up board

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NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra Ex 10.3

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NCERT Solutions for Class 12th Maths Chapter 10 Vector Algebra Ex 10.3

Get Free NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra Ex 10.3 PDF in Hindi and English Medium. Sets Class 12 Maths NCERT Solutions are extremely helpful while doing your homework. Vector Algebra Exercise 10.3 Class 12 Maths NCERT Solutions were prepared by Experienced LearnCBSE.in Teachers. Detailed answers of all the questions in Chapter 10 Class 12 Vector Algebra Ex 10.3 provided in NCERT Textbook.

Free download NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra Ex 10.3 PDF in Hindi Medium as well as in English Medium for CBSE, Uttarakhand, Bihar, MP Board, Gujarat Board, BIE, Intermediate and UP Board students, who are using NCERT Books based on updated CBSE Syllabus for the session 2019-20.

NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra Ex 10.3

Ex 10.3 Class 12 Maths Question 1.
Find the angle between two vectors \overrightarrow { a } ,\overrightarrow { b } with magnitudes √3 and 2 respectively, and such that \overrightarrow { a } \cdot \overrightarrow { b } =\sqrt { 6 }
Solution:
Angle θ between two vectors \overrightarrow { a } ,\overrightarrow { b } ,
NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra 1

Ex 10.3 Class 12 Maths Question 2.
Find the angle between the vectors \hat { i } -2\hat { j } +3\hat { k } \quad and\quad 3\hat { i } -2\hat { j } +\hat { k }
Solution:
Let \overrightarrow { a } =\hat { i } -2\hat { j } +3\hat { k } \quad and\quad \overrightarrow { b } =3\hat { i } -2\hat { j } +\hat { k }
Let θ be the angle between \overrightarrow { a } ,\overrightarrow { b } ,
NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra 2

Ex 10.3 Class 12 Maths Question 3.
Find the projection of the vector \overrightarrow { i } -\overrightarrow { j } , on the line represented by the vector \overrightarrow { i } +\overrightarrow { j } ,
Solution:
let \overrightarrow { a } =\hat { i } -\hat { j } \quad and\quad \overrightarrow { b } =\hat { i } +\hat { j }
NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra 3

Ex 10.3 Class 12 Maths Question 4.
Find the projection of the vector \hat { i } +3\hat { j } +7\hat { k } on the vector 7\hat { i } -\hat { j } +8\hat { k }
Solution:
let \overrightarrow { a } =\hat { i } +3\hat { j } +7\hat { k } \quad and\quad \overrightarrow { b } =7\hat { i } -\hat { j } +8\hat { k } then
tiwari academy class 12 maths Chapter 10 Vector Algebra 4

Ex 10.3 Class 12 Maths Question 5.
Show that each of the given three vectors is a unit vector \frac { 1 }{ 7 } \left( 2\hat { i } +3\hat { j } +6\hat { k } \right) ,\frac { 1 }{ 7 } \left( 3\hat { i } -6\hat { j } +2\hat { k } \right) ,\frac { 1 }{ 7 } \left( 6\hat { i } +2\hat { j } -3\hat { k } \right) Also show that they are mutually perpendicular to each other.
Solution:
Let\quad \overrightarrow { a } =\frac { 1 }{ 7 } \left( 2\hat { i } +3\hat { j } +6\hat { k } \right) ,\overrightarrow { b } =\frac { 1 }{ 7 } \left( 3\hat { i } -6\hat { j } +2\hat { k } \right) ,\overrightarrow { c } =\frac { 1 }{ 7 } \left( 6\hat { i } +2\hat { j } -3\hat { k } \right)
NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra 5

Ex 10.3 Class 12 Maths Question 6.
Find\left| \overrightarrow { a } \right| and\left| \overrightarrow { b } \right| if\left( \overrightarrow { a } +\overrightarrow { b } \right) \cdot \left( \overrightarrow { a } -\overrightarrow { b } \right) =8\quad and\left| \overrightarrow { a } \right| =8\left| \overrightarrow { b } \right|
Solution:
Given \left( \overrightarrow { a } +\overrightarrow { b } \right) \cdot \left( \overrightarrow { a } -\overrightarrow { b } \right) =8
tiwari academy class 12 maths Chapter 10 Vector Algebra 6

Ex 10.3 Class 12 Maths Question 7.
Evaluate the product :
\left( 3\overrightarrow { a } -5\overrightarrow { b } \right) \cdot \left( 2\overrightarrow { a } +7\overrightarrow { b } \right)
Solution:
\left( 3\overrightarrow { a } -5\overrightarrow { b } \right) \cdot \left( 2\overrightarrow { a } +7\overrightarrow { b } \right)
=6\overrightarrow { a } .\overrightarrow { a } -10\overrightarrow { b } \overrightarrow { a } +21\overrightarrow { a } .\overrightarrow { b } -35\overrightarrow { b } .\overrightarrow { b }
=6{ \left| \overrightarrow { a } \right| }^{ 2 }-11\overrightarrow { a } \overrightarrow { b } -35{ \left| \overrightarrow { b } \right| }^{ 2 }

Ex 10.3 Class 12 Maths Question 8.
Find the magnitude of two vectors \overrightarrow { a } ,\overrightarrow { b } having the same magnitude and such that the angle between them is 60° and their scalar product is \frac { 1 }{ 2 }
Solution:
We know that \overrightarrow { a } .\overrightarrow { b } =\left| \overrightarrow { a } \right| \left| \overrightarrow { b } \right| cos\theta
NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra 8

Ex 10.3 Class 12 Maths Question 9.
Find \left| \overrightarrow { x } \right| , if for a unit vector \overrightarrow { a } ,(\overrightarrow { x } -\overrightarrow { a } )\cdot (\overrightarrow { x } +\overrightarrow { a } )=12
Solution:
Given
\overrightarrow { a } ,(\overrightarrow { x } -\overrightarrow { a } )\cdot (\overrightarrow { x } +\overrightarrow { a } )=12
NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra 9

Ex 10.3 Class 12 Maths Question 10.
If \overrightarrow { a } =2\hat { i } +2\hat { j } +3\hat { k } ,\overrightarrow { b } =-\hat { i } +2\hat { j } +\hat { k } and\overrightarrow { c } =3\hat { i } +\hat { j } such that \overrightarrow { a } +\lambda \overrightarrow { b } \bot \overrightarrow { c } , then find the value of λ.
Solution:
Given
\overrightarrow { a } =2\hat { i } +2\hat { j } +3\hat { k } ,\overrightarrow { b } =-\hat { i } +2\hat { j } +\hat { k } and\overrightarrow { c } =3\hat { i } +\hat { j }
tiwari academy class 12 maths Chapter 10 Vector Algebra 10

Ex 10.3 Class 12 Maths Question 11.
Show that \left| \overrightarrow { a } \right| \overrightarrow { b } +\left| \overrightarrow { b } \right| a\quad \bot \quad \left| \overrightarrow { a } \right| \cdot \overrightarrow { b } -\left| \overrightarrow { b } \right| a for any two non-zero vectors \overrightarrow { a } ,\overrightarrow { b }
Solution:
\overrightarrow { a } ,\overrightarrow { b } are any two non zero vectors
NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra 11

Ex 10.3 Class 12 Maths Question 12.
If \overrightarrow { a } \cdot \overrightarrow { a } =0\quad and\quad \overrightarrow { a } \cdot \overrightarrow { b } =0, then what can be concluded about the vector \overrightarrow { b } ?
Solution:
\overrightarrow { a } \overrightarrow { a } =0\quad and\quad \overrightarrow { a } .\overrightarrow { b } =0 ,
=> \overrightarrow { b } = 0
Hence b is any vector.

Ex 10.3 Class 12 Maths Question 13.
If \overrightarrow { a } ,\overrightarrow { b } ,\overrightarrow { c } are the unit vector such that \overrightarrow { a } +\overrightarrow { b } +\overrightarrow { c } =0 , then find the value of \overrightarrow { a } .\overrightarrow { b } +\overrightarrow { b } .\overrightarrow { c } +\overrightarrow { c } .\overrightarrow { a }
Solution:
We have
\overrightarrow { a } +\overrightarrow { b } +\overrightarrow { c } =0
tiwari academy class 12 maths Chapter 10 Vector Algebra 13

Ex 10.3 Class 12 Maths Question 14.
If either vector \overrightarrow { a } =0\quad or\quad \overrightarrow { b } =0 then \overrightarrow { a } .\overrightarrow { b } =0. But the converse need not be true. Justify your answer with an example.
Solution:
Given: \overrightarrow { a } =0\quad or\quad \overrightarrow { b } =0
To prove: \overrightarrow { a } .\overrightarrow { b } =0
NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra 14

Ex 10.3 Class 12 Maths Question 15.
If the vertices A,B,C of a triangle ABC are (1,2,3) (-1,0,0), (0,1,2) respectively, then find ∠ABC.
Solution:
Let O be the origin then.
\frac { 1 }{ 2 }
NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra 15
NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra 15.1

Ex 10.3 Class 12 Maths Question 16.
Show that the points A (1,2,7), B (2,6,3) and C (3,10, -1) are collinear.
Solution:
The position vectors of points A, B, C are
NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra 16

Ex 10.3 Class 12 Maths Question 17.
Show that the vectors 2\hat { i } -\hat { j } +\hat { k } ,\hat { i } -3\hat { j } -5\hat { k } and \left( 3\hat { i } -4\hat { j } -4\hat { k } \right) from the vertices of a right angled triangle.
Solution:
The position vectors of the points A, B and C are
2\hat { i } -\hat { j } +\hat { k } ,\hat { i } -3\hat { j } -5\hat { k } and \left( 3\hat { i } -4\hat { j } -4\hat { k } \right)
NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra 17
NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra 17.1

Ex 10.3 Class 12 Maths Question 18.
If \overrightarrow { a } is a non-zero vector of magnitude ‘a’ and λ is a non- zero scalar, then λ \overrightarrow { a } is unit vector if
(a) λ = 1
(b) λ = – 1
(c) a = |λ|
(d) a = \frac { 1 }{ \left| \lambda \right| }
Solution:
\left| \overrightarrow { a } \right| =a
Given : \lambda \overrightarrow { a } is a unit vectors
NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra 18

NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra Hindi Medium Ex 10.3

NCERT Solutions for Class 12 Maths Exercise 10.3 of Vector Algebra
NCERT Solutions for Class 12 Maths Exercise 10.3
12 Maths Exercise 10.3
12 Maths Exercise 10.3 solutions
12 Maths Exercise 10.3 all answers
12 Maths Exercise 10.3 in English Medium
12 Maths Exercise 10.3 in Hindi Medium
Class 12 Maths Exercise 10.3 in Hindi medium
Class 12 Maths Exercise 10.3 for 2019-20
Class 12 Maths Exercise 10.3 updated

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NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra Ex 10.4

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NCERT Solutions for Class 12th Maths Chapter 10 Vector Algebra Ex 10.4

Get Free NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra Ex 10.4 PDF in Hindi and English Medium. Sets Class 12 Maths NCERT Solutions are extremely helpful while doing your homework. Vector Algebra Exercise 10.4 Class 12 Maths NCERT Solutions were prepared by Experienced LearnCBSE.in Teachers. Detailed answers of all the questions in Chapter 10 Class 12 Vector Algebra Ex 10.4 provided in NCERT Textbook.

Free download NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra Ex 10.4 PDF in Hindi Medium as well as in English Medium for CBSE, Uttarakhand, Bihar, MP Board, Gujarat Board, BIE, Intermediate and UP Board students, who are using NCERT Books based on updated CBSE Syllabus for the session 2019-20.

NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra Ex 10.4

Ex 10.4 Class 12 Maths Question 1.
Find \left| \overrightarrow { a } \times \overrightarrow { b } \right| ,if\quad \overrightarrow { a } =\hat { i } -7\hat { j } +7\hat { k } \quad and\quad \overrightarrow { b } =3\hat { i } -2\hat { j } +2\hat { k }
Solution:
Given
\overrightarrow { a } =\hat { i } -7\hat { j } +7\hat { k } \quad and\quad \overrightarrow { b } =3\hat { i } -2\hat { j } +2\hat { k }
NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra 1

NCERT Maths Class 12 Chapter 10

Ex 10.4 Class 12 Maths Question 2.
Find a unit vector perpendicular to each of the vector \overrightarrow { a } +\overrightarrow { b } \quad and\quad \overrightarrow { a } -\overrightarrow { b } , where \overrightarrow { a } =3\hat { i } +2\hat { j } +2\hat { k } \quad and\quad \overrightarrow { b } =\hat { i } +2\hat { j } -2\hat { k }
Solution:
we have
\overrightarrow { a } =3\hat { i } +2\hat { j } +2\hat { k } \quad and\quad \overrightarrow { b } =\hat { i } +2\hat { j } -2\hat { k }
NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra 2
NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra 2.1

Ex 10.4 Class 12 Maths Question 3.
If a unit vector \overrightarrow { a } makes angle \frac { \pi }{ 3 } with\quad \hat { i } ,\frac { \pi }{ 4 } with\quad \hat { j } and an acute angle θ with \overrightarrow { k } ,then find θ and hence the components of \overrightarrow { a } .
Solution:
Let\quad \overrightarrow { a } ={ a }_{ 1 }\hat { i } +{ a }_{ 2 }\hat { j } +{ a }_{ 3 }\hat { k } such\quad that\quad \left| \overrightarrow { a } \right| =1
NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra 3

Ex 10.4 Class 12 Maths Question 4.
Show that \left( \overrightarrow { a } -\overrightarrow { b } \right) \times \left( \overrightarrow { a } +\overrightarrow { b } \right) =2\left( \overrightarrow { a } \times \overrightarrow { b } \right)
Solution:
LHS = \left( \overrightarrow { a } -\overrightarrow { b } \right) \times \left( \overrightarrow { a } +\overrightarrow { b } \right)
vedantu class 12 maths Chapter 10 Vector Algebra 4

Ex 10.4 Class 12 Maths Question 5.
Find λ and μ if
\left( 2\hat { i } +6\hat { j } +27\hat { k } \right) \times \left( \hat { i } +\lambda \hat { j } +\mu \hat { k } \right) =0
Solution:
\left( 2\hat { i } +6\hat { j } +27\hat { k } \right) \times \left( \hat { i } +\lambda \hat { j } +\mu \hat { k } \right) =0
NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra 5

Ex 10.4 Class 12 Maths Question 6.
Given that \overrightarrow { a } .\overrightarrow { b } =0\quad and\quad \overrightarrow { a } \times \overrightarrow { b } =0. What can you conclude about the vectors \overrightarrow { a } ,\overrightarrow { b } ?
Solution:
\overrightarrow { a } .\overrightarrow { b } =0\quad and\quad \overrightarrow { a } \times \overrightarrow { b } =0
NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra 6

Ex 10.4 Class 12 Maths Question 7.
Let the vectors \overrightarrow { a } ,\overrightarrow { b } ,\overrightarrow { c } are given { a }_{ 1 }\hat { i } +{ a }_{ 2 }\hat { j } +{ a }_{ 3 }\hat { k } ,{ b }_{ 1 }\hat { i } +{ b }_{ 2 }\hat { j } +{ b }_{ 3 }\hat { k } ,{ c }_{ 1 }\hat { i } +{ c }_{ 2 }\hat { j } +{ c }_{ 3 }\hat { k } . Then show that \overrightarrow { a } \times \left( \overrightarrow { b } +\overrightarrow { c } \right) =\overrightarrow { a } \times \overrightarrow { b } +\overrightarrow { a } \times \overrightarrow { c }
Solution:
Given
\overrightarrow { a } ,\overrightarrow { b } ,\overrightarrow { c } are given { a }_{ 1 }\hat { i } +{ a }_{ 2 }\hat { j } +{ a }_{ 3 }\hat { k } ,{ b }_{ 1 }\hat { i } +{ b }_{ 2 }\hat { j } +{ b }_{ 3 }\hat { k } ,{ c }_{ 1 }\hat { i } +{ c }_{ 2 }\hat { j } +{ c }_{ 3 }\hat { k }
vedantu class 12 maths Chapter 10 Vector Algebra 7
NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra 7.1

Ex 10.4 Class 12 Maths Question 8.
If either \overrightarrow { a } =0\quad or\quad \overrightarrow { b } =0\quad then\quad \hat { a } \times \hat { b } =0.Is the
converse true? Justify your answer with an example.
Solution:
\overrightarrow { a } =0\Rightarrow \left| \overrightarrow { a } \right| =0
NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra 8

Ex 10.4 Class 12 Maths Question 9.
Find the area of the triangle with vertices A (1,1,2), B (2,3,5) and C (1,5,5).
Solution:
A (1,1,2), B (2,3,5) and C (1,5,5).
NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra 9
NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra 9.1

Ex 10.4 Class 12 Maths Question 10.
Find the area of the parallelogram whose adjacent sides are determined by the vectors \overrightarrow { a } =\hat { i } -\hat { j } +3\hat { k } ,\overrightarrow { b } =2\hat { i } -7\hat { j } +\hat { k }
Solution:
We have \overrightarrow { a } =\hat { i } -\hat { j } +3\hat { k } ,\overrightarrow { b } =2\hat { i } -7\hat { j } +\hat { k }
vedantu class 12 maths Chapter 10 Vector Algebra 10

Ex 10.4 Class 12 Maths Question 11.
Let the vectors\overrightarrow { a } ,\overrightarrow { b } such that \left| \overrightarrow { a } \right| =3,\left| \overrightarrow { b } \right| =\frac { \sqrt { 2 } }{ 3 } then \overrightarrow { a } \times \overrightarrow { b } is a unit vector if the angle between \overrightarrow { a } ,\overrightarrow { b } is
(a) \frac { \pi }{ 6 }
(b) \frac { \pi }{ 4 }
(c) \frac { \pi }{ 3 }
(d) \frac { \pi }{ 2 }
Solution:
Given
\left| \overrightarrow { a } \times \overrightarrow { b } \right| =1
\left| \overrightarrow { a } \right| =3,\left| \overrightarrow { b } \right| =\frac { \sqrt { 2 } }{ 3 }
NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra 11

Ex 10.4 Class 12 Maths Question 12.
Area of a rectangles having vertices
A\left( -\hat { i } +\frac { 1 }{ 2 } \hat { j } +4\hat { k } \right) ,B\left( \hat { i } +\frac { 1 }{ 2 } \hat { j } +4\hat { k } \right) ,
C\left( \hat { i } -\frac { 1 }{ 2 } \hat { j } +4\hat { k } \right) ,D\left( -\hat { i } -\frac { 1 }{ 2 } \hat { j } +4\hat { k } \right) ,
(a) \frac { 1 }{ 2 } sq units
(b) 1sq.units
(c) 2sq.units
(d) 4sq.units
Solution:
\overrightarrow { OA } =\left( -\hat { i } +\frac { 1 }{ 2 } \hat { j } +4\hat { k } \right)
\overrightarrow { OB } =\left( \hat { i } +\frac { 1 }{ 2 } \hat { j } +4\hat { k } \right)
NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra 12

NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra Hindi Medium Ex 10.4

NCERT Solutions for Class 12 Maths Exercise 10.4 of Vector Algebra
NCERT Solutions for Class 12 Maths Exercise 10.4 of Vector Algebra in PDF
NCERT Solutions for Class 12 Maths Exercise 10.4
NCERT Solutions for Class 12 Maths Exercise 10.4 in PDF
NCERT Solutions for Class 12 Maths Exercise 10.4 for up board
NCERT Solutions for Class 12 Maths Exercise 10.4 in Hindi medium
NCERT Solutions for Class 12 Maths Exercise 10.4 for 2019-20

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