NCERT Solutions class 12 Maths Exercise 5.6 (Ex 5.6) Chapter 5 Continuity and Differentiability


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. 

Ans. Given:  and  

  and 

  =  and  = 

Now 


2. 

 

Ans. Given:      and  

   and      

  and 

  and 

Now dydx=dy/dθdx/dθ=bsinθasinθ=badydx=dy/dθdx/dθ=−bsin⁡θ−asin⁡θ=ba


3. 

 

Ans. Given:  and  

  and 

Now 


4. 

 

Ans. Given:  and  

  = 

and 

 

Now 


5. 

 

Ans. Given:  and  

  and dydθ=ddθsinθddθsin2θdydθ=ddθsin⁡θ−ddθsin⁡2θ

  and dydθ=cosθcos2θddθ2θdydθ=cos⁡θ−cos⁡2θddθ2θ

 dxdθ=sinθ+(sin2θ)2dxdθ=−sin⁡θ+(sin⁡2θ)2 and dydθ=cosθcos2θ×2dydθ=cos⁡θ−cos⁡2θ×2

  and dydθ=cosθ2cos2θdydθ=cos⁡θ−2cos⁡2θ

Now 


If  and  are connected parametrically by the equations given in Exercises 6 to 10, without eliminating the parameter, find 

 

6. 

Ans. Given:  and  

  and dydθ=addθ(1+cosθ)dydθ=addθ(1+cos⁡θ)

  and dydθ=a[ddθ(1)+ddθcosθ]dydθ=a[ddθ(1)+ddθcos⁡θ]

   and dydθ=a[0sinθ]=asinθdydθ=a[0−sin⁡θ]=−asin⁡θ

  = 


7. 

 

Ans. Given:  and  

   [By quotient rule]

And  [By quotient rule]

  

(4cos3t3cost)3sint4sin3t−(4cos3t−3cos⁡t)3sin⁡t−4sin3t


8. 

 

Ans. Given:  and 
 




 =  = 
And 
  =  


9. 

 

Ans. Given:  and 
  and dydθ=bsec2θdydθ=bsec2θ
 



 


10. 

 

Ans. Given:  and 
  

And 
 = 
  


11. If  show that 

 

Ans. Given: 
and y=acos1t−−−−−√=(acos1t)12y=acos−1t=(acos−1t)12 =

 

And      

=a12cos1tloga12.11t2=a12cos−1tlog⁡a12.−11−t2

 
 = 
Hence proved.