Question

If the maximum and minimum values of the determinant
$\begin{vmatrix} 1+\sin^{2}x & \cos^{2}x & \sin2x \\ \sin^{2}x & 1+\cos^{2}x & \sin2x \\ \sin^{2}x & \cos^{2}x & 1+\sin2x \end{vmatrix}$ are $\alpha $ and $\beta $ , respectively, then $\alpha + 2 \beta $ is equal to

Answer 1

Given determinant is
$\Delta =\begin{vmatrix} 1+\sin^{2}x & \cos^{2}x & \sin2x \\ \sin^{2}x & 1+\cos^{2}x & \sin2x \\ \sin^{2}x & \cos^{2}x & 1+\sin2x \end{vmatrix}$
Applying the operation $\text{C}_{1} \rightarrow \text{C}_{1} + \text{C}_{2}$ , we get
$\Delta =\begin{vmatrix} 2 & \cos^{2}x & \sin2x \\ 2 & 1+\cos^{2}x & \sin2x \\ 1 & \cos^{2}x & 1+\sin2x \end{vmatrix}$
Applying the operations $R_{2} \rightarrow R_{2}-R_{1}$ and then $R_{3} \rightarrow R_{3}-R_{1}$ , we get
$\Delta =\begin{vmatrix} 2 & \cos^{2}x & \sin2x \\ 0 & 1 & 0 \\ -1 & 0 & 1 \end{vmatrix}$
Now, expanding the above determinant along $R_{2}$ , we get
$\Delta =-0+\left(\right.2+\sin2x\left.\right)-0=2+\sin2x$
Since, the maximum value of $\sin 2x$ is $1$ and minimum value of $\sin 2x$ is $-1$ .
Therefore, $\alpha =$ maximum value of $\Delta = 2 + 1 = 3$ and $\beta =$ minimum value of $\Delta =2-1=1$ .
$\Rightarrow \alpha =3$ and $\beta = 1$
$\therefore \alpha +2\beta =3+2=5$