JEE Mathematics 2025 Question with Solution

Given: $f(x+y)=f(x)f(y)$ for all $x,y\in\mathbb{R}$, $f'(0)=4a$, and $f''(x)-3a f'(x)-f(x)=0$ with $a>0$.

Find: The area of the region under $y=f(ax)$ from $x=0$ to $x=2$.

Differentiate the functional equation with respect to $y$ and then put $y=0$:

Since $f'(0)=4a$, we get

Using this, the differentiable solution has the exponential form

Now substitute in the differential equation:

Hence

so

that is,

Since $e^{4ax}\neq 0$ for all $x$,

As $a>0$,

Therefore,

and so

The required area is

Evaluating,

Therefore, the area of the region is $e^2-1$, so the correct option is A.

Given: $f(x+y)=f(x)f(y)$ and $f$ is twice differentiable.

Find: The area bounded by $y=f(ax)$, $x=0$, $x=2$, and the $x$-axis.

For a differentiable function satisfying

the standard form is

for some constant $k$. Also,

Here $f'(0)=4a$, so

and hence

Now use the given differential equation:

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

Substituting,

which gives

Thus,

and because $a>0$,

So,

since $a=\frac12$.

Therefore the area is

Hence the correct option is A.