JEE 2025 — Mathematics PYQ

 f(x + y) = f(x) f(y) ∀ x, y ∈ R
From the functional equation f(x + y) = f(x) f(y):
Put y = 0, f(x) = f(x) = f(0)
This means either f’(0) = 1 or f(x) = 0 for all x
Since we have f’(0) = 4a > 0, f(x) cannot be
identiacially zero Therefore f(0) = 1
This implies that f is an exponential function.
f(x) = k^x for some constant ka

$f(x)=k^x\ln k$

$f^{\prime }(0)=\ln k=4a$

$k=e^{4a}$

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

$f^{\prime }(x)=4a{e}^{4ax}$

$f^{\prime \prime }(x)=16a^2{e}^{4ax}$

Now,

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

$16a^2{e}^{4ax}-12a{e}^{4ax}-e^{4ax}=0$

$(4a^2-12a-1)e^{4ax}=0$

$4a^2=1$

a = \frac{1}{2} \; (\because a > 0)

$y=e^{2x}$

$\therefore f(ax)=e^{4a(ax)}=e^x$

Area of region  
$R=\{(x,y):0\le y\le f(ax),0\le x\le 2\}$

$x=2$

$a={\int }_0^2{e}^{x}dx$

$={\left[e^x\right]}_0^2$

$=e^2-1$