# RD Sharma Class 12 Ex 10.2 Solutions Chapter 10 Differentiability

Here we provide RD Sharma Class 12 Ex 10.2 Solutions Chapter 10 Differentiability for English medium students, Which will very helpful for every student in their exams. Students can download the RD Sharma Class 12 Ex 10.2 Solutions Chapter 10 Differentiability book pdf download. Now you will get step-by-step solutions to each question.

## RD Sharma Class 12 Ex 10.2 Solutions Chapter 10 Differentiability

### Question 1. If f is defined by f(x) = x2, find f'(2).

Solution:

Hence, f'(2) = 4.

### Question 2.  If f is defined by f(x) = x2 – 4x + 7, show that f'(5) = 2f'(7/2).

Solution: f'(5) = 6                                      …….(1)

f'(7/2) = 3

⇒ 2f'(7/2) = 6                           ……(2)

From (1) and (2)

f'(5) = 2f'(7/2).

### Question 3. Show that the derivative of the function f given by f(x) = 2x3 – 9x2 +12x + 9 at x = 1 and x = 2 are equal.

Solution:

⇒ f'(1) = 0

Now, ⇒ f'(2) = 0

Hence f'(1) = f'(2) = 0.

Solution:

⇒ 97 = 10a +7

⇒ 10a = 90

⇒ a = 9

Solution:

⇒ f'(4) = 112

Solution:

⇒ f'(0) = m.

### Question 7. Examine the differentiability of Solution:

Since f(x) is a polynomial function, it is continuous and differentiable everywhere.

Differentiability at x = –2

= 2

= 1

Since, LHD at x = –2  ≠ RHD at x = –2

Hence f(x) is not differentiable at x = –2.

Now, Differentiability at x = 0

(LHD at x = 0) = ∞

(RHD at x = 0) = 1

Since, LHD at x = –2  ≠ RHD at x = 0

Hence f(x) is not differentiable at x = 0.

### Question 8. Write an example of a function which is everywhere continuous but fails to be differentiable at exactly five points.

Solution:

We know the modulus function f(x) = |x| is continuous but not differentiable at x = 0.

Hence, f(x) = |x| + |x – 1| + |x – 2| + |x – 3| + |x – 4| is continuous but not fails to be differentiable at x = 0,1,2,3,4.

### Question 9. Discuss the continuity and differentiability of f(x) = |log|x||.

Solution:

Graph of f(x) = |log|x||:

From the graph above, it is clear that f(x) is continuous everywhere, but not differentiable at 1 and -1.

### Question 10.Discuss the continuity and differentiability of f(x) = e|x|.

Solution:

For continuity:

(LHL at x = 0)  = e0

= 1

(RHL at x = 0) = e0

= 1

Hence f(x) is continuous at x = 0.

For Differentiability:

(LHD at x = 0) = = –1

(RHD at x = 0) = 1

Thus, f(x) is not differentiable at x = 0.

### Question 11. Discuss the differentiability of Solution:

(LHD at x = c) = k

(RHD at x = c) = = k

Clearly (LHD at x = c) = (RHD at x = c)

f(x) is differentiable at x = c.

### Question 12. Is |sinx| differentiable? What about cos|x|?

Solution:

(LHD at x = nπ) = –1

(RHD at x = nπ) = 1

Since, LHD at x = nπ  ≠ RHD at x = nπ

Hence f(x) = |sinx|  is not differentiable at x = nπ.

Now, f(x) = cos|x|

Since, cos(–x) = cosx

Thus, f(x) = cos x

Hence f(x) = cos|x| is differentiable everywhere.

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