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  • <img alt="\mathrm{\text { Let } f(x)=\left\{\begin{array}{ll} -1, & -2 \leq x \leq 0 \\ x-1, & 0<x \leq 2 \end{array} \text {. Then } g(x)=f(|x|)+|f(x)|\right. \text { is not }}" src="ht

\mathrm{\text { Let } f(x)=\left\{\begin{array}{ll} -1, & -2 \leq x \leq 0 \\ x-1, & 0<x \leq 2 \end{array} \text {. Then } g(x)=f(|x|)+|f(x)|\right. \text { is not }} 

 differentiable at 

Option: 1

only x=0


Option: 2

only x=1


Option: 3

Both  x=0and x=1 


Option: 4

differentiable at all points 


Answers (1)

best_answer

\mathrm{f(|x|)=|x|-1=\left\{\begin{array}{rr} -x-1, & -2 \leq x \leq 0 \\ x-1, & 0<x \leq 2 \end{array}\right.}

\mathrm{|f(x)|=\left\{\begin{array}{lr} 1, & -2 \leq x \leq 0 \\ 1-x, & 0<x \leq 1 \\ x-1, & 1<x \leq 2 \end{array} \quad \therefore g(x)=\left\{\begin{array}{lr} -x, & -2 \leq x \leq 0 \\ 0, & 0<x \leq 1 \\ 2(x-1), & 1<x \leq 2 \end{array}\right.\right.}

g(x) is continuous and not differentiable at x= 0, 1.

 

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Divya Prakash Singh

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