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

Answer: (D)

$C_1 \rightarrow C _1+ C _2+ C _3 $
$f ( x ) =\begin{vmatrix}2+\sin 2 x & \cos ^2 x & \sin 2 x \\ 2+\sin 2 x & 1+\cos ^2 x & \sin 2 x \\ 2+\sin 2 x & \cos ^2 x & 1+\sin 2 x\end{vmatrix}$
$f(x)=(2+\sin 2 x)\begin{vmatrix}1 & \cos ^2 x & \sin 2 x \\ 1 & 1+\cos ^2 x & \sin 2 x \\ 1 & \cos ^2 x & 1+\sin 2 x\end{vmatrix}$
$ R_2 \rightarrow R_2-R_1$
$ R_3 \rightarrow R_3-R_1$
$f(x)=2+\sin 2 x)\begin{vmatrix}1 & \cos ^2 x & \sin 2 x \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{vmatrix}$
$=(2+\sin 2 x)(1)=2+\sin 2 x$
$=\sin 2 x \in\left[\frac{\sqrt{3}}{2}, 1\right]$
Hence $2+\sin 2 x \in\left[2+\frac{\sqrt{3}}{2}, 3\right]$