Let be a thrice differentiable odd function satisfying . Then is equal to _________. [2025]

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Solution

Q.
Let $f : R \to R$ be a thrice differentiable odd function satisfying $f' (x) \geq 0, f'' (x) = f (x), f (0) = 0, f' (0) = 3$ . Then $9 f (\log_{e} 3)$ is equal to _________. [2025]

Ans.
(36)

Since, f(x) is a thrice differentiable odd function,

$f'' (x) = f (x) \Rightarrow f' (x) \cdot f'' (x) = f' (x) \cdot f (x)$

$\Rightarrow \frac{(f' {(x))}^{2}}{2} = \frac{(f {(x))}^{2}}{2} + C \Rightarrow (f' {(x))}^{2} = (f {(x))}^{2} + C'$

$∵ (f' {(x))}^{2} = (f {(x))}^{2} + 9$ [ $∵ f' (0) = 3, f (0) = 0$ ]

$y = f (x) \Rightarrow \frac{d y}{d x} = f' (x) = \sqrt{(f {(x))}^{2} + 9}$

$= \int d x \Rightarrow \log | y + \sqrt{y^{2} + 9} | = x + C$

$\Rightarrow f (0) = 0 \Rightarrow C = \log 3 \Rightarrow y + \sqrt{y^{2} + 9} = 3 e^{x}$

At x = log 3; y = 4

$∴ 9 f (\log_{e} 3) = 36$

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