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

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:
\(x=\frac { { sin }^{ 3 }t }{ \sqrt { cos2t } } \& y=\frac { { cos }^{ 3 }t }{ \sqrt { cos2t } } \)

Ex 5.6 Class 12 Maths Question 8.
\(x=a\left( cost+log\quad tan\frac { t }{ 2 } \right) ,y=a\quad sint \)
Solution:
\(x=a\left( cost+log\quad tan\frac { t }{ 2 } \right) ,y=a\quad sint \)

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
\(x=\sqrt { { a }^{ { sin }^{ -1 }t } } ,y=\sqrt { { a }^{ { cos }^{ -1 }t } } \)

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

NCERT Class 12 Maths Solutions

• Chapter 1 Relations and Functions
• Chapter 2 Inverse Trigonometric Functions
• Chapter 3 Matrices
• Chapter 4 Determinants
• Chapter 5 Continuity and Differentiability
• Chapter 6 Application of Derivatives
• Chapter 7 Integrals Ex 7.1
• Chapter 8 Application of Integrals
• Chapter 9 Differential Equations
• Chapter 10 Vector Algebra
• Chapter 11 Three Dimensional Geometry
• Chapter 12 Linear Programming
• Chapter 13 Probability Ex 13.1