Let a function y=fx be defined parametrically by x=2t−t,y=t2+tt then

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Let a function $y = f (x)$ be defined parametrically by $x = 2 t - |t|$ , $y = t^{2} + t |t|$ then

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$y^{1} (x) = 4 x$ when x>0

$y^{1} (x)$ does not exist at x=0

$y^{1} (x) = 0$ When x<0

$y^{1} (x) = 2$ for all x

answer is A, C.

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Detailed Solution

$x = 2 t - |t| = \{\begin{cases} t t \geq, 0 \\ 3 t t < 0 \end{cases}$

$∴ t = \{\begin{cases} x x \geq 0 \\ \frac{x}{3} x < 0 \end{cases}$

$y = t^{2} + t |t| = \{\begin{cases} 2 t^{2} x \geq 0 \\ 0 x < 0 \end{cases}$

$= \{\begin{cases} 2 x^{2} x \geq 0 \\ 0 x < 0 \end{cases}$

Hence $y^{1} (x) = \{\begin{cases} 4 x x \geq 0 \\ 0 x < 0 \end{cases}$

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