JEE Mathematics 2025 Question with Solution

Given: $f(x) = \frac{1 + \sin^2 x}{\cos^2 x} + \frac{4 \sin 4x}{\sin^2 x \cos^2 x}$ for $x \in \mathbb{R}$.

Find: $M^4 - m^4$, where $M$ and $m$ are respectively the maximum and minimum values of $f(x)$.

From the solution working, the expression is evaluated by simplifying the corresponding determinant form and obtaining

Using $\sin^2 x + \cos^2 x = 1$,

Since $-1 \le \sin 4x \le 1$,

and

Now compute

Therefore, the value of $M^4 - m^4$ is $1280$. The correct option is D.

Given: After simplification in the solution, $f(x) = 2 + 4\sin 4x$.

Find: The required value $M^4 - m^4$.

The key observation is that the entire variation of $f(x)$ depends only on $\sin 4x$. Since the range of sine is

we get the range of $f(x)$ by substituting these extreme values.

At $\sin 4x = 1$,

So the maximum value is $M = 6$.

At $\sin 4x = -1$,

So the minimum value is $m = -2$.

Hence,

Therefore, the correct option is D.