Let a function $y = y (x)$ be defined parametrically by $x = 2 t - | t |, y = t^{2} + t | t |$ . Then for $x > 0$ , $y^{'} (x) =$

0

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4 x

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2 x

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N o t d i f f e r e n t i a b l e

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Solution

The correct option is B $4 x$
$x = 2 t - | t | = {\begin{matrix} t, t \geq 0 3 t, t < 0 \end{matrix} \Rightarrow t = {\begin{matrix} x, x \geq 0 x / 3, x < 0 \end{matrix}$
$y = t^{2} + t | t | = {\begin{matrix} 2 t^{2}, t \geq 0 0, t < 0 \end{matrix} \Rightarrow y = {\begin{matrix} 2 x^{2}, x \geq 0 0, x < 0 \end{matrix} \Rightarrow \frac{d y}{d x} = {\begin{matrix} 4 x, x \geq 0 0, x < 0 \end{matrix}$

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