JEE Mathematics 2025 Question with Solution

Given: $f(x) = \begin{cases} (1+ax)^{1/x} &, x < 0\\ 1+b &, x = 0\\ \frac{(x+4)^{1/2} - 2}{(x+c)^{1/3} - 2} &, x > 0 \end{cases}$

Find: The value of $e^a bc$ given that $f(x)$ is continuous at $x = 0$.

For continuity at $x = 0$,

From the left side,

So,

Also,

Hence,

For the right-hand limit,

Since both numerator and denominator approach zero, use L'Hospital's rule:

Now continuity gives

Because the denominator is $(x+c)^{1/3} - 2$ and the numerator becomes zero at $x=0$, for the right-hand limit to be of indeterminate form we must have

So,

Then

Hence,

and

Therefore,

The correct option is C.

Given: The function has different expressions for $x<0$, $x=0$, and $x>0$.

Find: Use continuity at $x=0$ to determine $e^a bc$.

First compute the left-hand limit:

Using the standard limit,

Next, the function value at the point is

Now evaluate the right-hand limit:

At $x=0$, the numerator becomes

So for L'Hospital's rule to apply as shown in the solution, the denominator must also vanish:

Substitute $c=8$ into the right-hand limit:

Applying L'Hospital's rule,

Now simplify:

So,

Hence continuity gives

Therefore,

Finally,

Therefore, the required value is $48$.