Consider a function $y = f (x)$ satisfies differential equation $y'' x^{2} + x {(y')}^{2} = 2 x y'$ where $f (0) = 0 a n d f'' (1) = 1$ . Let A be area bounded by $y = f (x), y - a x i s$ and line $y = 3 - \frac{π}{2}$ then

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f(x) is odd function

f(x) is strictly increasing function

$A = 2 - l n 2$

f(x) is symmetric with respect to y -axis

answer is A, B, C.

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Detailed Solution

$\frac{2 x y' - x^{2} y''}{{(y')}^{2}} = x \Rightarrow \int d (\frac{x^{2}}{y'}) = \int x d x$

Given,

$f'' (1) = 1 \Rightarrow f' (1) = 1, f' (x) = \frac{2 x^{2}}{x^{2} + 1} \Rightarrow y = \int \frac{2 x^{2}}{x^{2} + 1} d x = 2 x - 2 \tan^{- 1} x + c$

$c = 0 ∵ f (1) = 0 \Rightarrow f (x)$ is strictly increasing

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