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 =
Now
2.
and
and
and
Now dydx=dy/dθdx/dθ=−bsinθ−asinθ=badydx=dy/dθdx/dθ=−bsinθ−asinθ=ba
3.
and =
Now
4.
=
and =
Now
5.
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 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.
[By quotient rule]
=
=
=
=
And [By quotient rule]
=
=
=
=
=
=
=
= −(4cos3t−3cost)3sint−4sin3t−(4cos3t−3cost)3sint−4sin3t
=
8.
=
=
=
=
= = =
And
=
9.
and dydθ=bsec2θdydθ=bsec2θ
=
=
=
=
10.
=
And
= =
11. If 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.