JEE MAIN - Mathematics (2025 - 22nd January Morning Shift - No. 19)
Let $f: \mathbf{R} \rightarrow \mathbf{R}$ be a twice differentiable function such that $f(x+y)=f(x) f(y)$ for all $x, y \in \mathbf{R}$. If $f^{\prime}(0)=4 \mathrm{a}$ and $f$ satisfies $f^{\prime \prime}(x)-3 \mathrm{a} f^{\prime}(x)-f(x)=0, \mathrm{a}>0$, then the area of the region $\mathrm{R}=\{(x, y) \mid 0 \leq y \leq f(a x), 0 \leq x \leq 2\}$ is :
$\mathrm{e}^2-1$
$e^4+1$
$\mathrm{e}^2+1$
$e^4-1$
Explanation
$$ \begin{aligned} & \mathrm{f}(\mathrm{x}+\mathrm{y})=\mathrm{f}(\mathrm{x}) \mathrm{f}(\mathrm{y}) \\ & \Rightarrow \mathrm{f}(\mathrm{x})=\mathrm{e}^{\lambda \mathrm{x}} \mathrm{f}^{\prime}(0)=4 \mathrm{a} \\ & \Rightarrow \mathrm{f}^{\prime}(\mathrm{x})=\lambda \mathrm{e}^{\lambda \mathrm{x}} \Rightarrow \lambda=4 \mathrm{a} \end{aligned} $$
So, $\mathrm{f}(\mathrm{x})=\mathrm{e}^{4 \mathrm{ax}}$
$$ \begin{aligned} & \mathrm{f}^{\prime \prime}(\mathrm{x})-3 \mathrm{af}^{\prime}(\mathrm{x})-\mathrm{f}(\mathrm{x})=0 \\ & \Rightarrow \lambda^2-3 \mathrm{a} \lambda-1=0 \\ & \Rightarrow 16 \mathrm{a}^2-12 \mathrm{a}^2-1=0 \Rightarrow 4 \mathrm{a}^2=1 \Rightarrow \mathrm{a}=\frac{1}{2} \end{aligned} $$
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$\begin{aligned} & \mathrm{F}(\mathrm{x})=\mathrm{e}^{2 \mathrm{x}} \\ & \text { Area }=\int_0^2 \mathrm{e}^{\mathrm{x}} \mathrm{dx}=\mathrm{e}^2-1\end{aligned}$Comments (0)
