JEE Mathematics 2025 Question with Solution

Given: $\omega_1 = (8 + i) \sin \theta + (7 + 4i) \cos \theta$ and $\omega_2 = (1 + 8i) \sin \theta + (4 + 7i) \cos \theta$.

Find: The value of $p + q$, where $p$ and $q$ are the maximum and minimum values of $\alpha + \beta$ if $\omega_1\omega_2 = \alpha + i\beta$.

Write

$$\omega_1 = (8\sin\theta + 7\cos\theta) + i(\sin\theta + 4\cos\theta)$$

and

$$\omega_2 = (\sin\theta + 4\cos\theta) + i(8\sin\theta + 7\cos\theta)$$

Let

$$A = 8\sin\theta + 7\cos\theta, \qquad B = \sin\theta + 4\cos\theta$$

Then

$$\omega_1 = A + iB, \qquad \omega_2 = B + iA$$

So,

$$\omega_1\omega_2 = (A + iB)(B + iA) = AB - AB + i(A^2 + B^2) = i(A^2 + B^2)$$

Hence $\alpha = 0$ and

$$\alpha + \beta = A^2 + B^2$$

Now,

$$A^2 + B^2 = (8\sin\theta + 7\cos\theta)^2 + (\sin\theta + 4\cos\theta)^2$$

Expanding,

$$= 65\sin^2\theta + 65\cos^2\theta + 120\sin\theta\cos\theta$$

Using $\sin^2\theta + \cos^2\theta = 1$ and $2\sin\theta\cos\theta = \sin 2\theta$,

$$\alpha + \beta = 65 + 60\sin 2\theta$$

Since $-1 \leq \sin 2\theta \leq 1$,

$$\max(\alpha + \beta) = 65 + 60 = 125$$

and

$$\min(\alpha + \beta) = 65 - 60 = 5$$

Therefore,

$$p + q = 125 + 5 = 130$$

Thus, the correct option is B.

The solution indicates the key simplification is to rewrite the numbers in the form

$$\omega_1 = A + iB, \qquad \omega_2 = B + iA$$

with

$$A = 8\sin\theta + 7\cos\theta, \qquad B = \sin\theta + 4\cos\theta$$

This gives

$$\omega_1\omega_2 = (A + iB)(B + iA)$$

Now expand carefully:

$$\begin{aligned} (A + iB)(B + iA) &= AB + iA^2 + iB^2 + i^2AB \\ &= AB + i(A^2 + B^2) - AB \\ &= i(A^2 + B^2) \end{aligned}$$

So the real part is $\alpha = 0$ and the imaginary part is $\beta = A^2 + B^2$. Hence

$$\alpha + \beta = A^2 + B^2$$

Now expand the two squares:

$$(8\sin\theta + 7\cos\theta)^2 = 64\sin^2\theta + 49\cos^2\theta + 112\sin\theta\cos\theta$$ $$(\sin\theta + 4\cos\theta)^2 = \sin^2\theta + 16\cos^2\theta + 8\sin\theta\cos\theta$$

Adding,

$$\alpha + \beta = 65\sin^2\theta + 65\cos^2\theta + 120\sin\theta\cos\theta$$ $$= 65 + 120\sin\theta\cos\theta = 65 + 60\sin 2\theta$$

Therefore the extreme values are $125$ and $5$, whose sum is $130$.

The solution also contains inconsistent intermediate wording, but its final conclusion and the correct option shown are both B, corresponding to $130$.