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  • If  <img alt="\mathrm{f(x)=\left\{\begin{array}{ll} {[\cos \pi x],} & x<1 \\ |x-2|, & 2>x \geq 1 \end{array}, \text { then } f(x)\right. \text { is }}" src="https://entrancecorne

If  \mathrm{f(x)=\left\{\begin{array}{ll} {[\cos \pi x],} & x<1 \\ |x-2|, & 2>x \geq 1 \end{array}, \text { then } f(x)\right. \text { is }}

Option: 1

discontinuous and non-differentiable at x=-1 and x=1


Option: 2

continuous and differentiable at x=0


Option: 3

discontinuous at x=1 / 2


Option: 4

continuous but not differentiable at x=2


Answers (1)

best_answer

We have \mathrm{ f(x)= \begin{cases}\mid \cos \pi x], & x<1 \\ |x-2|, & 1 \leq x<2\end{cases}}

\mathrm{ =\left\{\begin{aligned} 2-x, & 1 \leq x<2 \\ -1, & 1 / 2<x<1 \\ 0, & 0<x \leq 1 / 2 \\ 1, & x=0 \\ 0, & -1 / 2 \leq x<0 \\ -1, & -3 / 2<x<-1 / 2 \end{aligned}\right.}

It is evident from the definition that f(x) is discontinuous at x=1 / 2

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HARSH KANKARIA

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