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

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HINTS: Understand the  definition of differentiability

GIVEN: f(x)=a|\sin x|+b e^{|x|}+e|x|^{3}

SOLUTION:The f(x) is differntaible

If x=0


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

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

\begin{aligned} &\lim _{h \rightarrow 0} \frac{a\left(\sin (-4)+b e^{-(-h)}+c\left(-(-h)^{3}\right)-b\right.}{-h} \\ \end{aligned}

\begin{aligned} &=\lim _{h \rightarrow 0} \frac{a \sinh +b e^{h}+c h^{3}-b}{-h}=f^{\prime}(0) \end{aligned}


\begin{aligned} &\left | \sin x \right |= \begin{cases}-\sin x & x<0 \\ \sin x & x>0\end{cases} \\ &|x|= \begin{cases}x & x>0 \\ -x & x<0\end{cases} \\ &|x|^{3}= \begin{cases}x^{3} & x>0 \\ -x^{3} & x<0\end{cases} \end{aligned}

\begin{aligned} &\lim _{h \rightarrow 0} \frac{a \sinh +b\left(e^{h}-1\right)+c h^{3}-b}{-h}=\lim _{h \rightarrow 0} \frac{a \sinh +b\left(e^{h}-1\right)+c h^{3}}{-h} \\ \end{aligned}

\begin{aligned} &=a \lim _{h \rightarrow 0} \frac{\sinh }{-h}-b \lim _{h \rightarrow 0} \frac{\left(e^{h}-1\right)}{h}+c \lim _{h \rightarrow 0} \frac{+h^{3}}{-h} \\ \end{aligned}

\begin{aligned} &=a \lim _{h \rightarrow 0} \frac{\sinh }{-h}-b \lim _{h \rightarrow 0} \frac{\left(e^{h}-1\right)}{h}+c \lim _{h \rightarrow 0} \frac{+h^{3}}{-h}-\frac{h^{3}}{h} \\ \end{aligned}\begin{aligned} &-a \times 1-b \times 1+0=a \times 1+b \times 1 \quad\left[\text { as } \lim _{h \rightarrow 0} \frac{\sinh }{-h}=1 \text { and } \lim _{h \rightarrow 0} \frac{\left(e^{h}-1\right)}{h}=1\right] \\\end{aligned}\begin{aligned} &-a=a \quad \&-b=b \\ &\quad a=0 \quad \& \quad b=0 \end{aligned}


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