Question

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

Answer 1

Answer: 5

Given determinant is
$\Delta =\left|\begin{array}{ccc}1+{\sin }^{2}x& {\cos }^{2}x& \sin 2x\\ {\sin }^{2}x& 1+{\cos }^{2}x& \sin 2x\\ {\sin }^{2}x& {\cos }^{2}x& 1+\sin 2x\end{array}\right|$
Applying the operation ${\text{C}}_1\to {\text{C}}_1+{\text{C}}_2$ , we get
$\Delta =\left|\begin{array}{ccc}2& {\cos }^{2}x& \sin 2x\\ 2& 1+{\cos }^{2}x& \sin 2x\\ 1& {\cos }^{2}x& 1+\sin 2x\end{array}\right|$
Applying the operations $R_2\to R_2-R_1$ and then $R_3\to R_3-R_1$ , we get
$\Delta =\left|\begin{array}{ccc}2& {\cos }^{2}x& \sin 2x\\ 0& 1& 0\\ -1& 0& 1\end{array}\right|$
Now, expanding the above determinant along $R_2$ , we get
$\Delta =-0+(2+\sin 2x)-0=2+\sin 2x$
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 $\beta =1$
$\therefore \alpha +2\beta =3+2=5$