Let f: R arrow R be a thrice differentiable odd function satisfying f^ (x) ≥ 0, f^ (x)=f(x), f(0)=0, f^ (0)=3 . Then 9 f ( _ c 3) is equal to _______.

Correct answer: 0 aligned & f^ (x)=f(x) & f^ (x) · f^ (x)=f^ (x) · f(x) & (f^ (x))^2 2 = (f(x))^2 2 +C & (f^ (x))^2=(f(x))^2+C^ aligned f(0)=0, f^ (0)=3 C^ =9 (f^ (x))^2=(f(x))^2+9 f^ (x)= (f(x))^2+9 f^ (x) ≥ 0 ∫ dy y^2+9 =∫ dx |y+ y^2+9 |=x+C f(0)=0 C= 3 y+ y^2+9 =3 e^ x at x= 3 ; y=4 9 f( 3)=36

JEE Main 20252 AprilMorning Shift

Exam: JEE Main
Year: 2025
Date: 2 April
Shift: Morning Shift
Subject: Mathematics
Chapter: Differential Equations

JEE Main 2025 · 2 April · Morning Shift · Mathematics · Differential Equations

Let f: R arrow R be a thrice differentiable odd function satisfying...

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 _______.

Correct answer

Step-by-step solution

\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}}

x=\ln 3 ; y=4

\therefore 9 \mathrm{f}(\ln 3)=36

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