JEE 2023 — Mathematics PYQ

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Question

JEE 2023 — Mathematics PYQ

JEE | Mathematics | 2023

Let $f(x) = \begin{vmatrix} 1 + \sin^2 x & \cos^2 x & \sin 2x \\ \sin^2 x & 1 + \cos^2 x & \sin 2x \\ \sin^2 x & \cos^2 x & 1 + \sin 2x \end{vmatrix}, x \in \left[\frac{\pi}{6}, \frac{\pi}{3}\right]$ . If $\alpha$ and $\beta$ respectively are the maximum and the minimum values of $f$ , then:

Choose the correct answer:

Explanation

Solution

Step 1: Simplify the determinant

Apply the operation $R_1 \to R_1 - R_2$ and $R_2 \to R_2 - R_3$ :

f(x) = \begin{vmatrix} 1 & -1 & 0 \\ 0 & 1 & -1 \\ \sin^2 x & \cos^2 x & 1 + \sin 2x \end{vmatrix}

Expanding along $R_1$ :

f(x) = 1[(1 + \sin 2x) + \cos^2 x] - (-1)[0 + \sin^2 x]

f(x) = 1 + \sin 2x + \cos^2 x + \sin^2 x

Since $\cos^2 x + \sin^2 x = 1$ :

f(x) = 2 + \sin 2x

Step 2: Find the range for $x \in [\frac{\pi}{6}, \frac{\pi}{3}]$

If $\frac{\pi}{6} \leq x \leq \frac{\pi}{3}$ , then $\frac{\pi}{3} \leq 2x \leq \frac{2\pi}{3}$ .

In this interval, the range of $\sin 2x$ is:

\sin\left(\frac{\pi}{3}\right) \leq \sin 2x \leq \sin\left(\frac{\pi}{2}\right)

\frac{\sqrt{3}}{2} \leq \sin 2x \leq 1

Step 3: Determine $\alpha$ (max) and $\beta$ (min)

Maximum value $\alpha$ : $2 + 1 = 3$
Minimum value $\beta$ : $2 + \frac{\sqrt{3}}{2}$

Step 4: Verify the options

Let's check option (4): $\beta^2 + 2\sqrt{\alpha} = \frac{19}{4}$

\beta^2 = \left(2 + \frac{\sqrt{3}}{2}\right)^2 = 4 + \frac{3}{4} + 2\sqrt{3} = \frac{19}{4} + 2\sqrt{3}

Then, $\beta^2 - 2\sqrt{\alpha} = \frac{19}{4} + 2\sqrt{3} - 2\sqrt{3} = \frac{19}{4}$ .

(Wait, let's re-verify the substitution).

Actually, looking at the calculated values: $\beta^2 - 2\sqrt{3} = \frac{19}{4}$ .

Since $\alpha = 3$ , $2\sqrt{\alpha} = 2\sqrt{3}$ .

So, $\beta^2 - 2\sqrt{\alpha} = \frac{19}{4}$ .

Correct Option: (2)