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  • Let f(x) = |sinx|. Then A. f is everywhere differentiable B. f is everywhere continuous but not differentiable C. f is everywhere continuous but not differentiable at x = (2n + 1) D. None of these

Let f(x) = |sinx|. Then
A. f is everywhere differentiable
B. f is everywhere continuous but not differentiable x = n\pi,n\in Z
C. f is everywhere continuous but not differentiable at  x = (2n+1)\frac{\pi}{2},n\in Z
D. None of these

Answers (1)

B)

Given that,  f(x) = |sinx|

Let g(x) = sinx and h(x) = |x|

Then, f(x) = hog(x)

We know that, modulus function and sine function are continuous everywhere.

Since, composition of two continuous functions is a continuous function.

Hence, f(x) = hog(x) is continuous everywhere.

Now, v(x)=|x| is not differentiable at x=0.

\begin{aligned} &LHD=\lim _{0^{-}} \frac{\mathrm{v}(\mathrm{x})-\mathrm{v}(0)}{\mathrm{x}-0}\\ &\lim _{h \rightarrow 0} \frac{v(0-h)-v(0)}{(0-h)-0}\\ &\lim _{=h \rightarrow 0} \frac{|0-h|-|0|}{-h} \quad(\because v(x)=|x|)\\ &\lim _{h \rightarrow 0} \frac{|-h|}{-h}\\ &=\lim _{h \rightarrow 0} \frac{h}{-h}\\ &=\lim _{h \rightarrow 0}-1=-1\\ \end{aligned}

\\RHD = \lim _{h \rightarrow 0} \frac{v(0+h)-v(0)}{(0+h)-0} \\=\lim _{h \rightarrow 0} \frac{|0+h|-|0|}{h} \quad(\because v(x)=|x|) \\\lim _{h \rightarrow 0} \frac{|h|}{h} \\=\lim _{h \rightarrow 0} \frac{h}{h} \\=\lim _{h \rightarrow 0} 1=1 \\\Rightarrow \mathrm{LY}^{\prime}(0) \neq \mathrm{R} \mathrm{C}^{\prime}(0) \\\Rightarrow|\mathrm{x}| \text{is not differentiable at} \ \mathrm{x}=0 . \\\Rightarrow \mathrm{h}(\mathrm{x}) \text{is not differentiable at} \ \mathrm{x}=0 .
So, f(x) is not differentiable where \sin x=0
We know that \sin x=0 at x=n \pi, n \in Z
Hence,  f(x)  is everywhere continuous but not differentiable x=\mathrm{n} \pi, \mathrm{n} \in \mathrm{Z}.

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