NCERT Solutions for Class 12 Maths Exercise 5.6 Chapter 5 Continuity and Differentiability – FREE PDF Download
Free PDF download of NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.6 (Ex 5.6) and all chapter exercises at one place prepared by expert teacher as per NCERT (CBSE) books guidelines. Class 12 Maths Chapter 5 Continuity and Differentiability Exercise 5.6 Questions with Solutions to help you to revise complete Syllabus and Score More marks.
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Exercise 5.6 (Ex 5.6)
If 
and 
are connected parametrically by the equations given in Exercise 1 to 5, without eliminating the parameter, find 
1. 
and 
and 
= 
and 
= 
Now 
2. 
and 
and 
and 
and 
Now dydx=dy/dθdx/dθ=−bsinθ−asinθ=badydx=dy/dθdx/dθ=−bsinθ−asinθ=ba
3.
and 
and 
= 
Now 
4.
and 
= 
and 
= 
Now 
5.
and 
and dydθ=ddθsinθ−ddθsin2θdydθ=ddθsinθ−ddθsin2θ
and dydθ=cosθ−cos2θddθ2θdydθ=cosθ−cos2θddθ2θ
dxdθ=−sinθ+(sin2θ)2dxdθ=−sinθ+(sin2θ)2 and dydθ=cosθ−cos2θ×2dydθ=cosθ−cos2θ×2
and dydθ=cosθ−2cos2θdydθ=cosθ−2cos2θ
Now 
If and are connected parametrically by the equations given in Exercises 6 to 10, without eliminating the parameter, find
6. 
and 
and dydθ=addθ(1+cosθ)dydθ=addθ(1+cosθ)
and dydθ=a[ddθ(1)+ddθcosθ]dydθ=a[ddθ(1)+ddθcosθ]
and dydθ=a[0−sinθ]=−asinθdydθ=a[0−sinθ]=−asinθ
= 
= 
= 
= 
7.
and 
[By quotient rule]
= 
= 
= 
= 
And 
[By quotient rule]
= 
= 
= 
= 
= 
= 
= 
= −(4cos3t−3cost)3sint−4sin3t−(4cos3t−3cost)3sint−4sin3t
= 
8.
and 
=
=
=
=
=
= 
= 
And
= 
9.
and 
and dydθ=bsec2θdydθ=bsec2θ 
=
=
=
=
10.
and 
=
And
=
= 
11. If show that
show that 
= 
and y=acos−1t−−−−−√=(acos−1t)12y=acos−1t=(acos−1t)12 =
=
=
And
=a12cos−1tloga12.−11−t2√=a12cos−1tloga12.−11−t2
= 
= 
Hence proved.