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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:

f'(2) = lim_{h\to0}\frac{f(2+h)-f(2)}{h}

= lim_{h\to0}\frac{(2+h)^2-2^2}{h}

= lim_{h\to0}\frac{4+h^2+4h-4}{h}

= lim_{h\to0}(4+h)



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) = lim_{h\to0}\frac{f(5+h)-f(5)}{h}

lim_{h\to0}\frac{((5+h^2)-4(5+h)+7)-(25-20+7)}{h}

= lim_{h\to0}\frac{h^2+25+10h-20-4h+7-12}{h}

= lim_{h\to0}\frac{h^2+6h}{h}

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

f'(7/2) = lim_{h\to0}\frac{f(7/2+h)-f(7/2)}{h}



= lim_{h\to0}\frac{h^2+7h-14-4h+7+14-7}{h}

= lim_{h\to0}\frac{h^2+3h}{h}

= lim_{h\to0}(h+3)

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) = lim_{h\to0}\frac{f(1+h)-f(1)}{h}

= lim_{h\to0}\frac{[2(1+h)^2-9(1+h)^2+12(1+h)+9]-[2-9+12-9]}{h}



= lim_{h\to0}\frac{2+2h^3+6h^2+6h-9-9h^2-18h+12+12h+9-14}{h}

= lim_{h\to0}\frac{2h^3-3h^2}{h}

= lim_{h\to0}h(2h-3)

⇒ f'(1) = 0

Now, f'(2) = lim_{h\to0}\frac{f(2+h)-f(2)}{h}

= lim_{h\to0}\frac{[2(2+h)^3-9(2+h)^2+12(2+h)+9]-[16-36+24+9]}{h}

= lim_{h\to0}\frac{2h^3-3h^2}{h}

= lim_{h\to0}h(2h-3)

⇒ f'(2) = 0

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



Question 4. If for the function f(x) = ax2 + 7x – 4, f'(5) = 97, find a.

Solution:

f'(5) = lim_{h\to0}\frac{f(5+h)-f(5)}{h}

⇒ 97 = lim_{h\to0}\frac{a(25+h^2+10h)+35+7h-4-25a-35+4}{h}

= lim_{h\to0}\frac{25a+ah^2+10ah-25a+7h}{h}

= lim_{h\to0}\frac{ah^2+h(10a+7)}{h}

⇒ 97 = 10a +7

⇒ 10a = 90

⇒ a = 9

Question 5. If f(x) = x3 + 7x2 + 8x – 9, find f'(4).

Solution:

f'(4) = lim_{h\to0}\frac{f(4+h)-f(4)}{h}



= lim_{h\to0}\frac{[(4+h)^3+7(4+h)^2+8(4+h)-9]-[64+112+32-9]}{h}

= lim_{h\to0}\frac{64+h^3+48h+12h^2+112+7h^2+56h+32+8h-9]-[210-9]}{h}

= lim_{h\to0}\frac{h^3+19h^2+112h}{h}

= lim_{h\to0}\frac{h(h^2+19h+112)}{h}

⇒ f'(4) = 112

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

Solution:

f'(0) = lim_{h\to0}\frac{f(0+h)-f(0)}{h}

= lim_{h\to0}\frac{f(h)-h(0)}{h}

= lim_{h\to0}\frac{(mh+c)-(c)}{h}

= lim_{h\to0}\frac{mh+c-c}{h}



= lim_{h\to0}\frac{mh}{h}

⇒ f'(0) = m.

Question 7. Examine the differentiability of f(x) = \begin{cases}2x+3\ \ \ \ \ \ \ ,-3\lex<-2\\x+1\ \ \ \ \ \ \ \ \ ,-2\lex<0\\x+2\ \ \ \ \ \ \ \ \ ,0\lex1\end{cases}.

Solution:

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

Differentiability at x = –2

(LHD at x = -2) = lim_{h\to-2^-}\frac{f(x)-f(-2)}{x-(-2)}

= lim_{h\to-2^-}\frac{2x+3+1}{x+2}

= lim_{h\to-2^-}\frac{2(x+2)}{x+2}

= 2

(RHD at x = -2) = lim_{h\to-2^+}\frac{f(x)-f(-2)}{x-(-2)}



= lim_{h\to-2^-}\frac{x+1+1}{x+2}

= lim_{h\to-2^-}\frac{x+2}{x+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) = lim_{h\to0}\frac{f(x)-f(0)}{x-0}

= lim_{h\to0}\frac{x+1-2}{x}

= lim_{h\to0}\frac{x-1}{x}

= ∞

(RHD at x = 0) = lim_{h\to0}\frac{f(x)-f(0)}{x-0}

= lim_{h\to0}\frac{x+2-2}{x}

= lim_{h\to0}\frac{x}{x}

= 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:

f(x) = e|x| = \begin{cases}e^{-x} \ \ \ \ \ \  \ \ ,x<0\\e^x\ \ \ \ \ \ \ \ \ \ ,x\ge 0\end{cases}

For continuity:

(LHL at x = 0) = lim_{x\to0^-}f(x)

lim_{h\to0}f(0-h)

= lim_{h\to0}e^{0-h}

= lim_{h\to0}e^{-h}

= e0

= 1

(RHL at x = 0) = lim_{x\to0^+}f(x)

= lim_{h\to0}f(0+h)

= lim_{h\to0}e^{0+h}

= lim_{h\to0}e^h

= e0

= 1

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

For Differentiability:



(LHD at x = 0) = lim_{x\to0^-}\frac{f(x)-f(0)}{x-0}

= lim_{h\to0}\frac{f(x)-f(0)}{x-0}

= lim_{h\to0}\frac{e^h-1}{-h}

= –1

(RHD at x = 0) = lim_{x\to0^+}\frac{f(x)-f(0)}{x-0}

= lim_{h\to0}\frac{e^h-1}{h}

= 1

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

Question 11. Discuss the differentiability of f(x) = \begin{cases}(x-c)cos\frac{1}{x-c}\ \ \ \ \ \ ,x≠c\\0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ,x=c\end{cases}.

Solution:

(LHD at x = c) = lim_{x\to c^-}\frac{f(c-h)-f(c)}{-h}



= lim_{h\to0}\frac{cos(-1/h)}{h}

= lim_{h\to0}\frac{cos(1/h)}{h}

= k

(RHD at x = c) = lim_{x\to c^+}\frac{f(c+h)-f(c)}{h}

= lim_{h\to0}\frac{cos(1/h)}{h}

= 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:

f(x) = |sinx| = \begin{cases}-sinx \ \ \ \ \ \ \  \ , x<nπ\\sinx \ \ \ \ \ \ \ \ \ \ \ ,x>nπ\end{cases}

 (LHD at x = nπ) = lim_{x\to nπ^-}\frac{f(x)-f(nπ)}{x-nπ}

= lim_{h\to0}\frac{-sin(nπ-h)-sinnπ}{nπ-h-nπ}

= lim_{h\to0}\frac{sinh}{-h}

= –1

 (RHD at x = nπ) = lim_{h\to0}\frac{sin(nπ+h)-sinnπ}{h}

= lim_{h\to0}\frac{sinh}{-h}

= 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.

Attention reader! Don’t stop learning now. Participate in the Scholorship Test for First-Step-to-DSA Course for Class 9 to 12 students.




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