KC Sinha Mathematics Solution Class 12 Chapter 10 अवकलनीयता (Differentiability) Exercise 10.1 (Q1-Q3)

Exercise 10.1

Question 1 

यदि (If) $f(x)=\frac{x-2}{x^{2}-3 x+2}$ , when x≠2
=1 , when x=2
तो (then find) $f^{\prime}(2)$
Sol :

When  x≠2 , $f(x)=\frac{x-2}{x^{2}-3 x+2}$

$=\frac{x -2}{(x-1)(x-2)}=\frac{1}{x-1}$

$f(2)=\left\{\begin{array}{cl}\frac{1}{x-1} & \text { when } x\neq 2 \\ 1 & , \text{when }x=2\end{array}\right.$

$f^{\prime}(2)=\lim _{h \rightarrow 0} \frac{f(2+h)-f(2)}{h}$

$=\lim _{h \rightarrow 0} \frac{\frac{1}{2+h-1}-1}{h}$

$=\lim _{h \rightarrow 0} \frac{\frac{1}{1+h}-1}{h}$

$=\lim _{h \rightarrow 0} \frac{\frac{1-(1+h)}{1+h}}{h}$

$=\lim_{h \rightarrow 0} \frac{1-1-h}{h(1+h)}$

$=\lim_{h \rightarrow 0} \frac{-h}{h(1+h)}$

$=\frac{-1}{1+0}=-1$

Question 2

यदि (If) $f(x)=x+x^{3} \sin \frac{1}{x}$ , when x≠0
=0 , when x=0
 तो x=0 पर f(x) का अवकलन निकालें (then find the derivative of f(x) at x=0)

Sol :

$f^{\prime}(0)=\lim _{h \rightarrow 0} \frac{f(0+h)-f(0)}{h}$

$=\lim _{h \rightarrow 0} \frac{f(h)-f(0)}{h}$

$=\lim _{h \rightarrow 0} \frac{h+h^{3} \cdot \sin \frac{1}{h}-0}{h}$

$=\lim _{h \rightarrow 0} \frac{h\left(1+h^{2} \sin \frac{1}{h}\right)}{h}$

$=\lim _{h \rightarrow 0}\left[1+h \cdot \frac{\sin \frac{1}{h}}{\frac{1}{h}}\right]$

$=\lim_{h \rightarrow 0}(1)+\lim_{h \rightarrow 0} h . \lim_{h \rightarrow 0} \frac{\sin \frac{1}{h}}{\frac{1}{h}}$

=1+0×1=1

Question 3

क्या (Is) |x+1| पर अवकलनी differentiable at x=-1 ?
Sol :
A+x=-1

L.H.D

$f^{\prime}\left(-1^{-}\right)=\lim_{h\rightarrow 0^-} \frac{f(-1-h)-f(-1)}{-h}$

$=\lim _{h \rightarrow 0} \frac{|-1-h+1|-|-1+1|}{-h}$

$=\lim_{h\rightarrow 0} \frac{|-h|-0}{-h}$

$=\lim _{h \rightarrow 0} \frac{h}{-h}=-1$


R.H.D

$f^{\prime}\left(-1^{+}\right)=\lim _{h \rightarrow 0^{+}} \frac{f(-1+4)-f(-1)}{h}$

$=\lim_{h \rightarrow 0} \frac{|-1+h+1|-|-1+1|}{h}$

$=\lim_{h \rightarrow 0} \frac{|h|-0}{h}$

$=\lim _{h \rightarrow 0} \frac{h}{h}=1$

$f^{\prime}\left(-1^{-}\right) \neq f\left(-1^{+}\right)$

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