JEE Mathematics 2025 Question with Solution

Given:

$$I=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{96x^2\cos^2 x}{1+e^x}\,dx = \pi(a\pi^2+\beta)$$

Find: $(a+\beta)^2$

Use the standard symmetry property

$$\int_{-m}^{m} f(x)\,dx = \int_{0}^{m} \left[f(x)+f(-x)\right]dx$$

with

$$f(x)=\frac{96x^2\cos^2 x}{1+e^x}.$$

Then

$$f(-x)=\frac{96x^2\cos^2 x}{1+e^{-x}}$$

and therefore

$$f(x)+f(-x)=96x^2\cos^2 x\left(\frac{1}{1+e^x}+\frac{1}{1+e^{-x}}\right)=96x^2\cos^2 x.$$

So the integral becomes

$$I=\int_0^{\frac{\pi}{2}}96x^2\cos^2 x\,dx.$$

Now use

$$\cos^2 x = \frac{1+\cos 2x}{2}.$$

Hence

$$I=48\int_0^{\frac{\pi}{2}}x^2\,dx+48\int_0^{\frac{\pi}{2}}x^2\cos 2x\,dx.$$

Evaluate the first term:

$$48\int_0^{\frac{\pi}{2}}x^2\,dx=48\left[\frac{x^3}{3}\right]_0^{\frac{\pi}{2}}=48\cdot \frac{\pi^3}{24}=2\pi^3.$$

For the second term, use integration by parts twice:

$$\int x^2\cos 2x\,dx=\frac{x^2\sin 2x}{2}-\int x\sin 2x\,dx$$

and

$$\int x\sin 2x\,dx=-\frac{x\cos 2x}{2}+\frac{\sin 2x}{4}.$$

So

$$\int x^2\cos 2x\,dx=\frac{x^2\sin 2x}{2}+\frac{x\cos 2x}{2}-\frac{\sin 2x}{4}.$$

Now substitute the limits $0$ to $\frac{\pi}{2}$:

$$\int_0^{\frac{\pi}{2}}x^2\cos 2x\,dx=\left[\frac{x^2\sin 2x}{2}+\frac{x\cos 2x}{2}-\frac{\sin 2x}{4}\right]_0^{\frac{\pi}{2}}=-\frac{\pi}{4}.$$

Thus

$$48\int_0^{\frac{\pi}{2}}x^2\cos 2x\,dx=48\left(-\frac{\pi}{4}\right)=-12\pi.$$

Therefore

$$I=2\pi^3-12\pi=\pi(2\pi^2-12).$$

Comparing with

$$I=\pi(a\pi^2+\beta),$$

we get

$$a=2,\qquad \beta=-12.$$

Hence

$$a+\beta=-10$$

and

$$(a+\beta)^2=100.$$

Therefore, the correct option is A.

The source solution text states the correct option is A and concludes $(a+\beta)^2=100$, but its intermediate reasoning is inconsistent because it incorrectly treats the original integrand as even. The correct derivation uses

$$\frac{1}{1+e^x}+\frac{1}{1+e^{-x}}=1$$

after pairing $f(x)$ and $f(-x)$ over symmetric limits. This yields the same final answer A.