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Explain solution rd sharma class 12 chapter 9 Differentiability exercise Multiple choice question, question 25

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HINTS: understand the definition of differentiability.

GIVEN: f(x)=|x-3| \cos x


f(x)= \begin{cases}-(x-3) \cos x & x<3 \\ (x-3) \cos x & x>3\end{cases}

At x=3

\begin{aligned} &=\lim _{x \rightarrow-1^{-}} \frac{f(x)-f(3)}{x-3} \\ &=\lim _{h \rightarrow 0} \frac{f(3-h)-0}{-h} \\ \end{aligned}

\begin{aligned} &=\lim _{h \rightarrow 0} \frac{-(3-h-3) \cos (3-h)}{-h} \\ &=\lim _{h \rightarrow 0} \cos (3-h)=-\cos 3 \end{aligned}


\begin{aligned} &=\lim _{x \rightarrow-3^{-}} \frac{f(x)-f(3)}{x-3} \\ &=\lim _{h \rightarrow 0} \frac{f(3+h)-f(3)}{x-3} \\ \end{aligned}

\begin{aligned} &=\lim _{h \rightarrow 0} \frac{(3+h-3) \cos (3+h)-0}{h} \\ &=\lim _{h \rightarrow 0} \cos (3+h)=\cos 3 \end{aligned}

So f(x) is not differentiable at x=-3.At other point f(x) is a product of two continuous and differential function (x-3 and cos x).So f(x) is differentiable  at R-3

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