 Class 12 Maths NCERT Solutions of exercise 5.2 Chapter- Continuity and Differentiability - Future Study Point # Class 12 Maths NCERT Solutions of exercise 5.2 Chapter- Continuity and Differentiability # Class 12 Maths NCERT Solutions of exercise 5.2 Chapter- Continuity and Differentiability

Class 12 maths NCERT  solution of exercise 5.2 -Continuity and differentiability will help all students of 12 class who are studying mathematics. From NCERT  Solutions of exercise 5.2 -Continuity and differentiability class 12 maths textbook you can understand the differentiation of the complex functions like trigonometric functions, rational functions etc.and the derivation of the greatest integer function. On these NCERT solutions of exercise 5.2- Continuity and differentiability, you will get complete knowledge of the chain rule of differentiation. The NCERT Solutions of exercise 5.2 is very easy to understand for the student so students can get the advantage of this by studying exercise 5.2 – Continuity and differentiability. Exercise 5.1-Continuity and Differentiability

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## Class 12 Maths NCERT Solutions of exercise 5.2 Chapter- Continuity and Differentiability

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### Exercise 5.2

Differentiate the functions with respect to x in Questions 1 to 8.

Q1.sin( x² + 5)

Let  y = sin( x² + 5)  and t = x² + 5

y = sin t

t = x² + 5

= 2x.cos(x² + 5)

Q2. cos(sin x)

Let y = cos(sinx) and t = sin x

y = cos t

t = sin x

Substituting t = sinx

Q3. sin(ax + b)

Let y = sin(ax + b) and t = ax + b

y = sint

t = ax + b

Substituting t = ax + b

Q4.sec(tan√x)

Ans. Let y = sec(tan√x) and t = tan √x

y = sec t

t = tan √x

According to the chain rule

Now, we have

Substituting the value of t

Now evaluating, by chain rule

We have

Differentiating the function with respect to x both sides

Let t= x5

t= x5

Similarly, we can evaluate

Therefore

Ans.

Q8. cot (√x)

Let y =cot (√x)

Let √x = t

Substituting value of t

Q9. Prove that the function f  given by

, x ∈ R is not differentiable at x = 1

Ans.

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

LHS limit = RHS limit

LHS limit

x = 1

LHS limit

LHL  ≠ RHL

Therefore f(x) is not differentiable at  x= 1

Q10.Prove that the greatest integer function defined by is not differentiable at x= 1 and  x= 2.

Ans. Given:

LHL

=

The greatest value of the function at  1-h   im the domain 0 < x < 3  is = 0

= ∞

RHL

The greatest value at  1+ h of the function when h tends to 0 is =1

RHL ≠  LHL

Therefore the function is not differentiable at x =1

Let’s check at x = 2

LHL

The value of greatest iteger fuctio [2-h ] = 1

= ∞

RHL

Simce x = 2

The value of greatest fuctio at [ 2 + h] = 2

Therefore f(x) is ot differetiale at x = 2

Himt:  1) If there exist the greatest integer fuctiom [x]   the value of  [1.099] = 1 ,[0.99] = 0 , [2.099] =2, [1.999]=1

[1 +h] = 1 and  [ 1- h] = 0, [2 + h] = 2 and [2-h] where h →0 .

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