Question

Let $f(x)= \begin{vmatrix} 1+\sin ^2 x & \cos ^2 x & \sin 2 x \\ \sin ^2 x & 1+\cos ^2 x & \sin 2 x \\ \sin ^2 x & \cos ^2 x & 1+\sin 2 x\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

• A. $\beta^2+2 \sqrt{\alpha}=\frac{19}{4}$
• B. $\alpha^2+\beta^2=\frac{9}{2}$
• C. $\alpha^2-\beta^2=4 \sqrt{3}$
• D. $\beta^2-2 \sqrt{\alpha}=\frac{19}{4}$

Answer 1

The Correct Option is D

$C_1\to C_1+C_2+C_3$
$f(x)=\left|\begin{array}{ccc}2+\sin 2x& {\cos }^{2}x& \sin 2x\\ 2+\sin 2x& 1+{\cos }^{2}x& \sin 2x\\ 2+\sin 2x& {\cos }^{2}x& 1+\sin 2x\end{array}\right|$
$f(x)=(2+\sin 2x)\left|\begin{array}{ccc}1& {\cos }^{2}x& \sin 2x\\ 1& 1+{\cos }^{2}x& \sin 2x\\ 1& {\cos }^{2}x& 1+\sin 2x\end{array}\right|$
$R_2\to R_2-R_1$
$R_3\to R_3-R_1$
$f(x)=2+\sin 2x)\left|\begin{array}{ccc}1& {\cos }^{2}x& \sin 2x\\ 0& 1& 0\\ 0& 0& 1\end{array}\right|$
$=(2+\sin 2x)(1)=2+\sin 2x$
$=\sin 2x\in \left[\frac{\sqrt{3}}{2},1\right]$
Hence $2+\sin 2x\in \left[2+\frac{\sqrt{3}}{2},3\right]$

Answer 2

Applying column operations: \[ C_1 \to C_1 + C_2 + C_3 \] 

\[f(x) = \begin{bmatrix} 2 + \sin 2x & \cos^2 x & \sin 2x \\ 2 + \sin 2x & 1 + \cos^2 x & \sin 2x \\ 2 + \sin 2x & \cos^2 x & 1 + \sin 2x \end{bmatrix}\]

 Subtracting rows: \[ R_2 \to R_2 - R_1, \quad R_3 \to R_3 - R_1 \]

\[f(x) = (2 + \sin 2x) \begin{bmatrix} 1 & \cos^2 x & \sin 2x \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}\]

\[ = (2 + \sin 2x) (1) \] \[ = 2 + \sin 2x \] For given range \( x \in \left[ \frac{\pi}{6}, \frac{\pi}{3} \right] \): \[ \sin 2x \in \left[ \frac{\sqrt{3}}{2}, 1 \right] \] Thus, \[ 2 + \sin 2x \in \left[ 2 + \frac{\sqrt{3}}{2}, 3 \right] \] \[ \alpha = 3, \quad \beta = 2 + \frac{\sqrt{3}}{2} \] \[ \beta^2 - 2\sqrt{\alpha} = \frac{19}{4} \]