The equation of normal to the curve $y = x^{3} - 2 x^{2} + 4$ at the point $x = 2$ is-

x + 4 y = 0

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4 x - y = 0

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x + 4 y = 18

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4 x - y = 18

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Solution

The correct option is D $x + 4 y = 18$
At $x = 2$ , $y = 2^{3} - {2.2}^{2} + 4 = 4$
So the point is, $(2, 4)$
Now $y = x^{3} - 2 x^{2} + 4$
$\frac{d y}{d x} = 3 x^{2} - 4 x$
$∴ {(\frac{d y}{d x})}_{(2, 4)} = {3.2}^{2} - 4.2 = 4 = m$ (say)
$∴$ Slope of normal at the given point is $= - \frac{1}{m} = - \frac{1}{4}$
Therefore, equation of normal is, $(y - 4) = - \frac{1}{4} (x - 2)$
$\Rightarrow x + 4 y = 18$
Hence, option 'C' is correct.

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