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

**Question 1. If f is defined by f(x) = x**^{2}, find f'(2).

^{2}, find f'(2).

**Solution:**

Hence, f'(2) = 4.

**Question 2. If f is defined by f(x) = x**^{2 }– 4x + 7, show that f'(5) = 2f'(7/2).

^{2 }– 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) = 2x**^{3} – 9x^{2} +12x + 9 at x = 1 and x = 2 are equal.

^{3}– 9x

^{2}+12x + 9 at x = 1 and x = 2 are equal.

**Solution:**

⇒ f'(1) = 0

Now,

⇒ f'(2) = 0

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

**Question 4. If for the function f(x) = ax**^{2} + 7x – 4, f'(5) = 97, find a.

^{2}+ 7x – 4, f'(5) = 97, find a.

**Solution:**

⇒ 97 = 10a +7

⇒ 10a = 90

⇒ a = 9

**Question 5. If f(x) = x**^{3 }+ 7x^{2} + 8x – 9, find f'(4).

^{3 }+ 7x

^{2}+ 8x – 9, find f'(4).

**Solution:**

⇒ f'(4) = 112

**Question 6. Find the derivative of f(x) = mx + c at x = 0.**

**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|}.

^{|x|}.

**Solution:**

For continuity:

(LHL at x = 0)

=

= e

^{0}= 1

(RHL at x = 0)

= e

^{0}= 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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