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Provide solution for rd sharma maths class 12 chapter 9 Differentiability exercise Multiple choice question, question 12

Answers (1)

(b)

HINTS: put LHL =RHL at x=1 and LHD=RHL at x=1

GIVEN: $f(x)=\left\{\begin{array}{cl} \frac{1}{|x|} & |x| \geq 1 \\ a x^{2}+b & |x|<1 \end{array}\right.$

SOLUTION:

As f(x) continuous at x=1

$\begin{aligned} \lim _{x \rightarrow 1^{-}} f(x) &=\lim _{x \rightarrow 1} a x^{2}+b \\ &=a+b \\ \end{aligned}$

$\begin{aligned} f(1) &=\frac{1}{|1|}=1 \\ \end{aligned}$

$\begin{aligned} \lim _{x \rightarrow 1^{-}} f(x) &=f(1) \\ a+b &=1 \end{aligned}$

As f(x) is differentiable at $|x|=1$

$\lim _{|x| \rightarrow 1^{-}} \frac{f(x)-f(1)}{x-1}=\lim _{|x| \rightarrow 1} \frac{a x^{2}+b-1}{x-1}$

f(x) is differentiable at x=1

LHS=

$\begin{aligned} &\lim _{x \rightarrow 1^{-}} \frac{f(x)-f(1)}{x-1} \\ \end{aligned}$

$\begin{aligned}&=\lim _{h \rightarrow 0} \frac{f(1-h)-f(1)}{-h} \\ \end{aligned}$

$\begin{aligned} &=\lim _{x \rightarrow 0} \frac{a(1-h)^{2}+b-1}{-h} \\ \end{aligned}$

$\begin{aligned} &=\lim _{x \rightarrow 0} \frac{a+a h^{2}-2 a h+b-1}{-h} \\ \end{aligned}$

$\begin{aligned}&=\lim _{x \rightarrow 0} \frac{a h^{2}-2 a h}{-h} \\ &=\lim _{x \rightarrow 0}-a h+2 a \\ &=2 a \end{aligned}$

RHD=

$\begin{aligned} &\lim _{x \rightarrow 1^{+}} \frac{f(x)-f(1)}{x-1} \\ \end{aligned}$

$\begin{aligned} &=\lim _{h \rightarrow 0} \frac{f(1+h)-f(1)}{h} \\ \end{aligned}$

$\begin{aligned} &=\lim _{x \rightarrow 0} \frac{\frac{1}{1+h}}{h}-1 \\ \end{aligned}$

$\begin{aligned} &=\lim _{x \rightarrow 0} \frac{\frac{1-(1+h)}{1+h}}{h} \\ \end{aligned}$

$\begin{aligned} &=\lim _{x \rightarrow 0} \frac{-h}{h(1+h)}=-1 \end{aligned}$

LHD=RHD

$\begin{aligned} & 2 a=-1 \\ &\Rightarrow a =-\frac{1}{2} \\ &a+b=1 \\ &\frac{-1}{2}+b=1 \\ & b=1+\frac{-1}{2}=\frac{3}{2} \end{aligned}$

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