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

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

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

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

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)

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