NCERT Solutions - Continuity & Differentiability, Exercise 5.2 JEE Notes | EduRev

Mathematics (Maths) Class 12

JEE : NCERT Solutions - Continuity & Differentiability, Exercise 5.2 JEE Notes | EduRev

The document NCERT Solutions - Continuity & Differentiability, Exercise 5.2 JEE Notes | EduRev is a part of the JEE Course Mathematics (Maths) Class 12.
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Continuity & Differentiability

 Question 1: Differentiate the functions with respect to x.

NCERT Solutions - Continuity & Differentiability, Exercise 5.2 JEE Notes | EduRev

Answer

NCERT Solutions - Continuity & Differentiability, Exercise 5.2 JEE Notes | EduRev

NCERT Solutions - Continuity & Differentiability, Exercise 5.2 JEE Notes | EduRev

NCERT Solutions - Continuity & Differentiability, Exercise 5.2 JEE Notes | EduRev


Question 2: Differentiate the functions with respect to x. cos(sinx)

Answer

NCERT Solutions - Continuity & Differentiability, Exercise 5.2 JEE Notes | EduRev

Thus, f is a composite function of two functions. 

Put t = u (x) = sin x  

NCERT Solutions - Continuity & Differentiability, Exercise 5.2 JEE Notes | EduRev

NCERT Solutions - Continuity & Differentiability, Exercise 5.2 JEE Notes | EduRev

By chain rule, 
Alternate method  

NCERT Solutions - Continuity & Differentiability, Exercise 5.2 JEE Notes | EduRev

NCERT Solutions - Continuity & Differentiability, Exercise 5.2 JEE Notes | EduRev


Question 3: Differentiate the functions with respect to x.

sin(ax + b)

Answer
NCERT Solutions - Continuity & Differentiability, Exercise 5.2 JEE Notes | EduRev
Alternate method
NCERT Solutions - Continuity & Differentiability, Exercise 5.2 JEE Notes | EduRev


Question 4: Differentiate the functions with respect to x.

NCERT Solutions - Continuity & Differentiability, Exercise 5.2 JEE Notes | EduRev

Answer
NCERT Solutions - Continuity & Differentiability, Exercise 5.2 JEE Notes | EduRev
Hence, by chain rule, we obtain

NCERT Solutions - Continuity & Differentiability, Exercise 5.2 JEE Notes | EduRev


Question 5: Differentiate the functions with respect to x.

NCERT Solutions - Continuity & Differentiability, Exercise 5.2 JEE Notes | EduRev

Answer

The given function is
NCERT Solutions - Continuity & Differentiability, Exercise 5.2 JEE Notes | EduRev
Put y = p (x) = cx + d
NCERT Solutions - Continuity & Differentiability, Exercise 5.2 JEE Notes | EduRev


Question 6: Differentiate the functions with respect to x. NCERT Solutions - Continuity & Differentiability, Exercise 5.2 JEE Notes | EduRev

Answer

NCERT Solutions - Continuity & Differentiability, Exercise 5.2 JEE Notes | EduRev


Question 7:  Differentiate the functions with respect to x.

NCERT Solutions - Continuity & Differentiability, Exercise 5.2 JEE Notes | EduRev

Answer

NCERT Solutions - Continuity & Differentiability, Exercise 5.2 JEE Notes | EduRev

Question 8: Differentiate the functions with respect to x. NCERT Solutions - Continuity & Differentiability, Exercise 5.2 JEE Notes | EduRev

Answer

NCERT Solutions - Continuity & Differentiability, Exercise 5.2 JEE Notes | EduRev
Clearly, f is a composite function of two functions, u and v, such that

NCERT Solutions - Continuity & Differentiability, Exercise 5.2 JEE Notes | EduRev
Alternate method
NCERT Solutions - Continuity & Differentiability, Exercise 5.2 JEE Notes | EduRev


Question 9:
 Prove that the function f given by NCERT Solutions - Continuity & Differentiability, Exercise 5.2 JEE Notes | EduRev is not differentiable at x = 1.

Answer
The given function is NCERT Solutions - Continuity & Differentiability, Exercise 5.2 JEE Notes | EduRev

It is known that a function f is differentiable at a point x = c in its domain if both

NCERT Solutions - Continuity & Differentiability, Exercise 5.2 JEE Notes | EduRev are finite and equal.

To check the differentiability of the given function at x = 1,  

consider the left hand limit of f at x = 1  

NCERT Solutions - Continuity & Differentiability, Exercise 5.2 JEE Notes | EduRev

NCERT Solutions - Continuity & Differentiability, Exercise 5.2 JEE Notes | EduRev
Since the left and right hand limits of f at x = 1 are not equal, f is not differentiable at x = 1


Question 10:
Prove that the greatest integer function  NCERT Solutions - Continuity & Differentiability, Exercise 5.2 JEE Notes | EduRev defined by is not differentiable at x = 1 and x = 2.


Answer
The given function f is  NCERT Solutions - Continuity & Differentiability, Exercise 5.2 JEE Notes | EduRev
It is known that a function f is differentiable at a point x = c in its domain if both

NCERT Solutions - Continuity & Differentiability, Exercise 5.2 JEE Notes | EduRev are finite and equal. 

To check the differentiability of the given function at x = 1, consider the left hand limit  of f at x = 1  

NCERT Solutions - Continuity & Differentiability, Exercise 5.2 JEE Notes | EduRev

Since the left and right hand limits of f at x = 1 are not equal, f is not differentiable at x = 1

To check the differentiability of the given function at x = 2, consider the left hand limit of f at x = 2

NCERT Solutions - Continuity & Differentiability, Exercise 5.2 JEE Notes | EduRev
Since the left and right hand limits of f at x = 2 are not equal, f is not differentiable at x = 2

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