Let $f: \mathbf{R} \rightarrow \mathbf{R}$ be a thrice differentiable odd function satisfying $f^{\prime}(\mathrm{x}) \geq 0, f^{\prime}(\mathrm{x})=f(\mathrm{x}), f(0)=0, f^{\prime}(0)=3$. Then $9 f\left(\log _{\mathrm{c}} 3\right)$ is equal to _______.

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Aicharya · Accepted Answer

$\begin{aligned} & \mathrm{f}^{\prime \prime}(\mathrm{x})=\mathrm{f}(\mathrm{x}) \\ & \Rightarrow \mathrm{f}^{\prime}(\mathrm{x}) \cdot \mathrm{f}^{\prime \prime}(\mathrm{x})=\mathrm{f}^{\prime}(\mathrm{x}) \cdot \mathrm{f}(\mathrm{x}) \\ & \Rightarrow \frac{\left(\mathrm{f}^{\prime}(\mathrm{x})\right)^2}{2}=\frac{(\mathrm{f}(\mathrm{x}))^2}{2}+\mathrm{C} \\ & \Rightarrow\left(\mathrm{f}^{\prime}(\mathrm{x})\right)^2=(\mathrm{f}(\mathrm{x}))^2+\mathrm{C}^{\prime}\end{aligned}$
$\mathrm{f}(0)=0, \mathrm{f}^{\prime}(0)=3 \quad \Rightarrow \mathrm{C}^{\prime}=9$
$\therefore\left(\mathrm{f}^{\prime}(\mathrm{x})\right)^2=(\mathrm{f}(\mathrm{x}))^2+9$
$\mathrm{f}^{\prime}(\mathrm{x})=\sqrt{(\mathrm{f}(\mathrm{x}))^2+9} \quad \because \mathrm{f}^{\prime}(\mathrm{x}) \geq 0$
$\int \frac{\mathrm{dy}}{\sqrt{\mathrm{y}^2+9}}=\int \mathrm{dx} \Rightarrow \ln \left|\mathrm{y}+\sqrt{\mathrm{y}^2+9}\right|=\mathrm{x}+\mathrm{C}$
$\Rightarrow \mathrm{f}(0)=0 \Rightarrow \mathrm{C}=\ln 3$
$\Rightarrow \mathrm{y}+\sqrt{\mathrm{y}^2+9}=3 \mathrm{e}^{\mathrm{x}}$
at $x=\ln 3 ; y=4$
$\therefore 9 \mathrm{f}(\ln 3)=36$

Let $f: \mathbf{R} \rightarrow \mathbf{R}$ be a thrice differentiable odd function satisfying…

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