JEE Main 2025 — 22 January, Morning Shift — Question 2

$f(x+y)=f(x) . f(y)$

$\Rightarrow \mathrm{f}(\mathrm{x})=\mathrm{e}^{\lambda \mathrm{x}} \mathrm{f}^{\prime}(0)=4 \mathrm{a}$

$\Rightarrow \mathrm{f}^{\prime}(\mathrm{x})=\lambda \mathrm{e}^{\lambda \mathrm{x}} \Rightarrow \lambda=4 \mathrm{a}$

So, $f(x)=e^{4ax}$

$f^{\prime \prime}(x)-3 a f^{\prime}(x)-f(x)=0$

$\Rightarrow \lambda^{2}-3 \mathrm{a} \lambda-1=0$

$\Rightarrow 16 \mathrm{a}^{2}-12 \mathrm{a}^{2}-1=0 \Rightarrow 4 \mathrm{a}^{2}=1 \Rightarrow \mathrm{a}=\frac{1}{2}$

$F(x)=e^{2 x}$

Area $=\int_{0}^{2} \mathrm{e}^{\mathrm{x}} \mathrm{dx}=\mathrm{e}^{2}-1$