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  • Need solution for rd sharma maths class 12 chapter 9 Differentiability exercise Multiple choice question, question 21

Need solution for rd sharma maths class 12 chapter 9 Differentiability exercise Multiple choice question, question 21

Answers (1)

best_answer

(b)

Hint: LHD = RHD at x = 0

Given : f(x)=a+b|x|+c|x|^{4}

Explanation: LHD at x = 0

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

\begin{aligned} &=\lim _{h \rightarrow 0} \frac{a+b(-(-h))+c(-h)^{4}-a}{-h} \\ &=\lim _{h \rightarrow 0} \frac{b h+c h^{4}}{-h}=-b\end{aligned}

RHD at x = 0

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

\begin{aligned} &=\lim _{h \rightarrow 0} \frac{a+b h+c h^{4}-a}{h} \\ &=\lim _{h \rightarrow 0} \frac{b h+c h^{4}}{h}=b \end{aligned}

Thus f(x) is differentiable at x = 0

LHD = RHD

\begin{aligned} &-b=b \\ &2 b=0 \\ &b=0 \end{aligned}

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