Application Of Derivatives Question 9
Question 9 - 01 February - Shift 1
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} $,
$x \in[\frac{\pi}{6}, \frac{\pi}{3}]$. If $\alpha$ a $\beta$ 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) $\alpha^{2}+\beta^{2}=\frac{9}{2}$
Show Answer
Answer: (1)
Solution:
Formula: Maximum of function and minima of a function, Properties of determinant
$C_1 \to C_1+C_2+C_3$
$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} $
$ \begin{aligned} & R_2 \to R_2-R_1 \\ & R_3 \to R_3-R_1 \end{aligned} $
$f(x)=2+\sin 2 x) \begin{vmatrix} 1 & \cos ^{2} x & \sin 2 x \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{vmatrix} $
$=(2+\sin 2 x)(1)=2+\sin 2 x$
$=\sin 2 x \in[\frac{\sqrt{3}}{2}, 1]$
Hence $2+\sin 2 x \in[2+\frac{\sqrt{3}}{2}, 3]$
