The equation of the normal to the curve $y = | x^{2} - | x | |$ at $x = - 2$ is

x = 3 y + 8

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y = 3 x + 8

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3 y = x + 8

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3 x = y + 8

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Solution

The correct option is C $3 y = x + 8$
In the neighbourhood of $x = - 2$ , $y = x^{2} + x$
Hence, the point on the curve is $(- 2, 2)$
$\frac{d y}{d x} = 2 x + 1 \Rightarrow {(\frac{d y}{d x})}_{x = - 2} = - 3$
So the slope of the normal at $(- 2, 2)$ is $\frac{1}{3}$
Hence the equation of the normal is
$y - 2 = \frac{1}{3} (x + 2)$
$\Rightarrow 3 y = x + 8$

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List-I	List-II
a) Equation of tangent to the curve $y = b e^{- x / a}$ at $x = 0$	1) $x - 2 y = 2$
b) Equation of tangent to the curve $y = x^{2} + 1$ at $(1, 2)$	2) $y = 2 x$
c) Equation of normal to the curve $y = 2 x - x^{2}$ at $(2, 0)$	3) $x - y = π$
d) Equation of normal to the curve $y = sin x$ at $x = π$	4) $\frac{x}{a} + \frac{y}{b} = 1$

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