JEE Main · 2025 · 22 Jan 2025 · Shift 1 · Differential Equations

Since $f(x+y)=f(x)f(y)$ for all real $x,y$, setting $x=y=0$ gives $f(0)=f(0{)}^2$, so $f(0)=0$ or $f(0)=1$. If $f(0)=0$ then $f(x)=f(x)f(0)=0$ for every $x$, which would force $f^{\prime }(0)=0\ne 4a$ since $a>0$; so $f(0)=1$.

A differentiable solution of this functional equation is exponential: $f(x)=e^{\lambda x}$ for some constant $\lambda$. Then $f^{\prime }(x)=\lambda e^{\lambda x}$, and $f^{\prime }(0)=\lambda =4a$, so


Substitute into $f^{\prime \prime }(x)-3a{f}^{\prime }(x)-f(x)=0$. Since $f^{\prime \prime }(x)={\lambda }^2{e}^{\lambda x}$,

As $a>0$, $a=\frac{1}{2}$.

So $f(x)=e^{4\cdot \frac{1}{2}x}=e^{2x}$, and $f(ax)=f\left(\frac{x}{2}\right)=e^{2\cdot x/2}=e^x$.

The region is $R=\{(x,y):0\le y\le e^x,0\le x\le 2\}$, whose area is