Suppose f- - 2-2- - R is defined by fx - - - 1for - 2 - x - 0- x - 1for0 - x - 2 -then the x - - 2-2-x - 0andf-x- - x -

The correct option is C { 12} The intersection point A will be x= 12 #160; on solving y=( 1 x) and #160; y=x #160; #160; #160; x= ...

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Suppose $f : [- 2, 2] \to R$ is defined by
$f (x) = {\begin{matrix} - 1 f o r - 2 \leq x \leq 0 x - 1 f o r 0 \leq x \leq 2 \end{matrix}$
then the ${x \in (- 2, 2) : x \leq 0 a n d f (| x |) = x} =$

{- 1}

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{0}

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{\frac{- 1}{2}}

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ϕ

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f (x) = ⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩ \begin{matrix} \frac{\sqrt{1 + c x} - \sqrt{1 - c x}}{x} & , & f o r & - 2 \leq x < 0 \frac{x + 3}{x + 1} & , & f o r & 0 \leq x \leq 2 \end{matrix}

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If $f : [- 2, 2] \to R$ is defined by $f (x) = {\begin{matrix} - 1, f o r - 2 \leq x \leq 0 t h e n x - 1, f o r 0 \leq x \leq 2 \end{matrix}}$

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