jee-main 2023 Q74

jee-main · India · session1_01feb_shift1 Matrices Determinant and Rank Computation
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
(1) $\beta^2 - 2\sqrt{\alpha} = \frac{19}{4}$
(2) $\beta^2 + 2\sqrt{\alpha} = \frac{19}{4}$
(3) $\alpha^2 - \beta^2 = 4\sqrt{3}$
(4) $\alpha^2 + \beta^2 = \frac{9}{2}$
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\\
(1) $\beta^2 - 2\sqrt{\alpha} = \frac{19}{4}$\\
(2) $\beta^2 + 2\sqrt{\alpha} = \frac{19}{4}$\\
(3) $\alpha^2 - \beta^2 = 4\sqrt{3}$\\
(4) $\alpha^2 + \beta^2 = \frac{9}{2}$