# Class 12 RD Sharma Solutions – Chapter 10 Differentiability – Exercise 10.2

Last Updated : 16 May, 2021

### 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 â€“ 4x + 7, show that f'(5) = 2f'(7/2).

Solution:

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

f'(7/2) = 3

â‡’ 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 â€“ 9x2 +12x + 9 at x = 1 and x = 2 are equal.

Solution:

â‡’ f'(1) = 0

Now,

â‡’ f'(2) = 0

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

Solution:

â‡’ 97 = 10a +7

â‡’ 10a = 90

â‡’ a = 9

Solution:

â‡’ f'(4) = 112

Solution:

â‡’ 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 = â€“2

= 2

= 1

Since, LHD at x = â€“2  â‰  RHD at x = â€“2

Hence f(x) is not differentiable at x = â€“2.

Now, Differentiability at x = 0

(LHD at x = 0)

= âˆž

(RHD at x = 0)

= 1

Since, LHD at x = â€“2  â‰  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 â€“ 1| + |x â€“ 2| + |x â€“ 3| + |x â€“ 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) =

= â€“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 = nÏ€)

= â€“1

(RHD at x = nÏ€)

= 1

Since, LHD at x = nÏ€  â‰  RHD at x = nÏ€

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

Now, f(x) = cos|x|

Since, cos(â€“x) = cosx

Thus, f(x) = cos x

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

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