Question

If $\Delta =\left|\begin{array}{ccc}2& -\sin \theta & 1\\ -\sin \theta & 2& \sin \theta \\ -1& -\sin \theta & 2\end{array}\right|$ then

• A. Minimum of $\Delta =8$
• B. Maximum of $\Delta =12$
• C. Minimum of $\Delta =10$
• D. Minimum of $\Delta =$ Minimum of $\sin \theta$

Answer 1

Answer: (B), (C)

The correct options are
B Maximum of $\Delta =12$
C Minimum of $\Delta =10$

$\Delta =\left|\begin{array}{ccc}2& -\sin \theta & 1\\ -\sin \theta & 2& \sin \theta \\ -1& -\sin \theta & 2\end{array}\right|$
$\Rightarrow \Delta =2{\sin }^2\theta +10$
Now, $\because -1\le \sin \theta \le 1$

$\Rightarrow 0\le {\sin }^2\theta \le 1$

$\Rightarrow 0\le 2{\sin }^2\theta \le 2$

$\Rightarrow 10\le 2{\sin }^2\theta +10\le 12$

Hence, $10\le \Delta \le 12$

Hence, minimum value is $10$ and maximum value is $12.$