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 $2+\sin 2x\in \left[2+\frac{\sqrt{3}}{2},3\right]$