NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.6

Get Free NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.6 PDF in Hindi and English Medium. Sets Class 12 Maths NCERT Solutions are extremely helpful while doing your homework. Continuity and Differentiability Exercise 5.6 Class 12 Maths NCERT Solutions were prepared by Experienced LearnCBSE.in Teachers. Detailed answers of all the questions in Chapter 5 Class 12 Continuity and Differentiability Ex 5.6 provided in NCERT Textbook.

Free download NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.6 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.

The topics and sub-topics included in the Continuity and Differentiability chapter are the following:

Continuity and Differentiability
Introduction
Algebra of continuous functions
Differentiability
Derivatives of composite functions
Derivatives of implicit functions
Derivatives of inverse trigonometric functions
Exponential and Logarithmic Functions
Logarithmic Differentiation
Derivatives of Functions in Parametric Forms
Second Order Derivative
Mean Value Theorem
Summary

There are total eight exercises and one misc exercise(144 Questions fully solved) in the class 12th maths chapter 5 Continuity and Differentiability.

NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.6

If x and y are connected parametrically by the equations given in Questions 1 to 10, without eliminating the parameter. Find $\\ \frac { dy }{ dx }$ .

Ex 5.6 Class 12 Maths Question 1.
x = 2at², y = at⁴
Solution:
$\frac { dx }{ dt } =4at,\frac { dy }{ dt } ={ 4at }^{ 3 }\quad \therefore \frac { dy }{ dx } =\frac { \frac { dy }{ dt } }{ \frac { dx }{ dt } } =\frac { { 4at }^{ 3 } }{ { 4at } } ={ t }^{ 2 }$

Ex 5.6 Class 12 Maths Question 2.
x = a cosθ,y = b cosθ
Solution:
$\frac { dx }{ d\theta } =-asin\theta ,\frac { dy }{ d\theta } =-sinb\quad sin\theta =>\frac { dy }{ dx } =\frac { b }{ a }$

Ex 5.6 Class 12 Maths Question 3.
x = sin t, y = cos 2t
Solution:
$\therefore \frac { dx }{ dt } =cos\quad t\quad and\frac { dy }{ dt } =-sin2t.2=-2sin2t$
$\frac { dy }{ dx } =\frac { -2sin2t }{ cost } =\frac { -2.2sintcost }{ cost } =-4sint$

Ex 5.6 Class 12 Maths Question 4.
$x=4t,y=\frac { 4 }{ t }$
Solution:
$\frac { dx }{ dt } =4;\frac { dy }{ dt } =\frac { -4 }{ { t }^{ 2 } } =>\frac { dy }{ dx } =\frac { -4 }{ { t }^{ 2 } } \times \frac { 1 }{ 4 } =\frac { -1 }{ { t }^{ 2 } }$

Ex 5.6 Class 12 Maths Question 5.
x = cos θ – cos 2θ, y = sin θ – sin 2θ
Solution:
$\frac { dx }{ d\theta } =-sin\theta -(-sin2\theta ).2=2sin2\theta -sin\theta$
$\frac { dy }{ d\theta } =cos\theta -2cos2\theta \quad \therefore \frac { dy }{ dx } =\frac { cos\theta -2cos2\theta }{ 2sin2\theta -sin\theta }$

Ex 5.6 Class 12 Maths Question 6.
x = a(θ – sinθ), y = a(1 + cosθ)
Solution:
$\frac { dx }{ d\theta } =a\left[ 1-cos\theta \right] \& \frac { dy }{ d\theta } =-asin\theta$
$\frac { dy }{ dx } =\frac { -asin\theta }{ a(1-cos\theta ) } =\frac { -2sin\frac { \theta }{ 2 } .cos\frac { \theta }{ 2 } }{ 2{ sin }^{ 2 }\frac { \theta }{ 2 } } =-cot\frac { \theta }{ 2 }$

Ex 5.6 Class 12 Maths Question 7.
$x=\frac { { sin }^{ 3 }t }{ \sqrt { cos2t } } \& y=\frac { { cos }^{ 3 }t }{ \sqrt { cos2t } }$
Solution:
$x=\frac { { sin }^{ 3 }t }{ \sqrt { cos2t } } \& y=\frac { { cos }^{ 3 }t }{ \sqrt { cos2t } }$
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability 7

Ex 5.6 Class 12 Maths Question 8.
$x=a\left( cost+log\quad tan\frac { t }{ 2 } \right) ,y=a\quad sint$
Solution:
$x=a\left( cost+log\quad tan\frac { t }{ 2 } \right) ,y=a\quad sint$
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability 8

Ex 5.6 Class 12 Maths Question 9.
x = a sec θ,y = b tan θ
Solution:
x = a sec θ,y = b tan θ
$\frac { dx }{ d\theta } =a\quad sec\theta \quad tan\theta \quad and\frac { dy }{ d\theta } =b{ sec }^{ 2 }\theta$
$\frac { dy }{ dx } =\frac { { bsec }^{ 2 }\theta }{ asec\theta tan\theta } \frac { b }{ a } cosec\theta$

Ex 5.6 Class 12 Maths Question 10.
x = a(cosθ+θsinθ), y = a(sinθ-θcosθ)
Solution:
x = a(cosθ+θsinθ), y = a(sinθ-θcosθ)
$\frac { dx }{ d\theta } =a\left[ -sin\theta +\theta .cos\theta +sin\theta \right] =a\theta cos\theta$
$\frac { dy }{ d\theta } =a\theta sin\theta =>\frac { dy }{ dx } =\frac { a\theta sin\theta }{ a\theta cos\theta } =tan\theta$

Ex 5.6 Class 12 Maths Question 11.
If $x=\sqrt { { a }^{ { sin }^{ -1 }t } } ,y=\sqrt { { a }^{ { cos }^{ -1 }t } }$ show that $\frac { dy }{ dx } =-\frac { y }{ x }$
Solution:
Given that
$x=\sqrt { { a }^{ { sin }^{ -1 }t } } ,y=\sqrt { { a }^{ { cos }^{ -1 }t } }$
byjus class 12 maths Chapter 5 Continuity and Differentiability 11

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