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  • please solve rd sharma class 12 chapter 9 Differentiability exercise Multiple choice question, question 5 maths textbook solution

please solve rd sharma class 12 chapter 9 Differentiability exercise Multiple choice question, question 5 maths textbook solution

Answers (1)

(a) ,(c)

HINTS: Understand the  definition of continuity and differentiability

GIVEN: f(x)=(x+|x|)(|x|)

SOLUTION:\begin{aligned} f(x) &=(x+|x|)(|x|) \\ f(x) &= \begin{cases}(x+x) x & x>0 \\( x-x)(-x) & x<0\end{cases} \\ &=\left\{\begin{array}{cc} 2 x^{2} & x>0 \\ 0 & x<0 \end{array}\right. \end{aligned}

Check the continuity of  f(x) at x<0

\begin{aligned} &\lim _{x \rightarrow 0^{+}} f(x)=\lim _{h \rightarrow 0} f(0+h)=\lim _{x \rightarrow 0^{+}} 2 h^{2}=0 \\ \end{aligned}

As

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

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

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} \\ &=\lim _{h \rightarrow 0} \frac{0-0}{-h} \\ &=0 \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(0+h)-f(0)}{h} \\ &=\lim _{h \rightarrow 0} \frac{2 h^{2}-0}{h} \\ &=\lim _{h \rightarrow 0} 2 h \\ &=0 \end{aligned}

As LHD at x=0 and RHD at x=0

Hence, f(x) is  differentiable.

f^{\prime}(x)= \begin{cases}4 x & x>0 \\ 0 & x<0\end{cases}

At x=0

\begin{aligned} \lim _{x \rightarrow 0^{+}} f(x) &=\lim _{h \rightarrow 0} f(0+h) \\ &=\lim _{h \rightarrow 0} 4 h \\ &=0 \\ \end{aligned}

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

Hence f'(x) is continuous

Now,

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

RHD=

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

Hence f'(x) is not differentiable

f'(x) does not exist

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