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

(A) $\beta^{2}-2 \sqrt{\alpha}=\frac{19}{4}$

(B) $\beta^{2}+2 \sqrt{\alpha}=\frac{19}{4}$

(C) $\alpha^{2}-\beta^{2}=4 \sqrt{3}$

(D) $\alpha^{2}+\beta^{2}=\frac{9}{2}$

JEE Main · 2023 · 01 Feb 2023 · Shift 1 · Determinants

Let

f (x) = 1 + sin^{2} x sin^{2} x sin^{2} x cos^{2} x 1 + cos^{2} x cos^{2} x sin 2 x sin 2 x 1 + sin 2 x

x \in [\frac{π}{6}, \frac{π}{3}]

. If

α

and

β

respectively are the maximum and the minimum values of f , then

Answer: (A)

Applying

C_{1} \to C_{1} + C_{2} + C_{3}

makes every entry of the first column

2 + sin 2 x

; taking this factor out and then using

R_{2} \to R_{2} - R_{1}, R_{3} \to R_{3} - R_{1}

reduces the determinant to the diagonal product

1 \cdot 1 \cdot 1

, so

f (x) = 2 + sin 2 x

. For

x \in [\frac{π}{6}, \frac{π}{3}]

2 x \in [\frac{π}{3}, \frac{2 π}{3}]

and

sin 2 x \in [\frac{3}{2}, 1]

, giving maximum

α = 3

and minimum

β = 2 + \frac{3}{2}

. Then

β^{2} - 2 α = (2 + \frac{3}{2})^{2} - 23 = 4 + 23 + \frac{3}{4} - 23

, which equals

\frac{19}{4}

JEE Main · 2023 · 01 Feb 2023 · Shift 1 · Single correct option