JEE Mathematics 2026 Question with Solution

Given: $f(\theta) = 4\left(\sin^4\left(\frac{7\pi}{2} - \theta\right) + \sin^4(11\pi + \theta)\right) - 2\left(\sin^6\left(\frac{3\pi}{2} - \theta\right) + \sin^6(9\pi - \theta)\right)$

Find: The value of $\alpha + 2\beta$, where $\alpha$ is the maximum value and $\beta$ is the minimum value of $f(\theta)$.

Using allied angle identities,

$$\sin\left(\frac{7\pi}{2} - \theta\right) = \cos\theta, \qquad \sin(11\pi + \theta) = -\sin\theta$$

and

$$\sin\left(\frac{3\pi}{2} - \theta\right) = -\cos\theta, \qquad \sin(9\pi - \theta) = \sin\theta$$

Therefore,

$$f(\theta) = 4(\cos^4\theta + \sin^4\theta) - 2(\cos^6\theta + \sin^6\theta)$$

Now use the identities

$$\sin^4\theta + \cos^4\theta = 1 - 2\sin^2\theta\cos^2\theta$$ $$\sin^6\theta + \cos^6\theta = 1 - 3\sin^2\theta\cos^2\theta$$

Substituting,

$$\begin{aligned} f(\theta) &= 4(1 - 2\sin^2\theta\cos^2\theta) - 2(1 - 3\sin^2\theta\cos^2\theta) \\ &= 4 - 8\sin^2\theta\cos^2\theta - 2 + 6\sin^2\theta\cos^2\theta \\ &= 2 - 2\sin^2\theta\cos^2\theta \\ &= 2 - \frac{1}{2}\sin^2(2\theta) \end{aligned}$$

Since $0 \leq \sin^2(2\theta) \leq 1$,

• maximum value occurs when $\sin^2(2\theta) = 0$, so $\alpha = 2$
• minimum value occurs when $\sin^2(2\theta) = 1$, so $\beta = 2 - \frac{1}{2} = \frac{3}{2}$

Hence,

$$\alpha + 2\beta = 2 + 2\left(\frac{3}{2}\right) = 5$$

Therefore, the correct option is C.

Given: The expression contains terms of the form $\sin^4 x + \cos^4 x$ and $\sin^6 x + \cos^6 x$ after allied-angle reduction.

Find: Maximum and minimum values of the resulting function.

First reduce the angles to get

$$f(\theta) = 4(\cos^4\theta + \sin^4\theta) - 2(\cos^6\theta + \sin^6\theta)$$

This is a standard pattern. Write everything in terms of $\sin^2\theta\cos^2\theta$:

$$\sin^4\theta + \cos^4\theta = 1 - 2\sin^2\theta\cos^2\theta$$ $$\sin^6\theta + \cos^6\theta = 1 - 3\sin^2\theta\cos^2\theta$$

So,

$$f(\theta) = 2 - 2\sin^2\theta\cos^2\theta = 2 - \frac{1}{2}\sin^2(2\theta)$$

Now the range follows immediately because $\sin^2(2\theta) \in [0,1]$. Hence $f(\theta) \in \left[\frac{3}{2}, 2\right]$. Therefore,

$$\alpha = 2, \qquad \beta = \frac{3}{2}$$

and

$$\alpha + 2\beta = 5$$

Therefore, the correct option is C.