Search any question & find its solution
Question:
Answered & Verified by Expert
If , then for all , lies in the interval :
Options:
Solution:
2964 Upvotes
Verified Answer
The correct answer is:
$A=\left[\begin{array}{ccc}1 & \sin \theta & 1 \\ -\sin \theta & 1 & \sin \theta \\ -1 & -\sin \theta & 1\end{array}\right]$
$\Rightarrow \operatorname{det}(A)=\left(11+\sin ^2 \theta\right)-\sin \theta(-\sin \theta+\sin \theta)+1\left(\sin ^2 \theta+1\right)$
$=2+2 \sin ^2 \theta$
$\because \theta \in\left(\frac{3 \pi}{4}, \frac{5 \pi}{4}\right)$
$\therefore \sin 2 \theta \in\left[0, \frac{1}{2}\right]$
$\therefore \operatorname{det}(A) \in[2,3]$ which is a subset of $\left(\frac{3}{2}, 3\right]$
$\Rightarrow \operatorname{det}(A)=\left(11+\sin ^2 \theta\right)-\sin \theta(-\sin \theta+\sin \theta)+1\left(\sin ^2 \theta+1\right)$
$=2+2 \sin ^2 \theta$
$\because \theta \in\left(\frac{3 \pi}{4}, \frac{5 \pi}{4}\right)$
$\therefore \sin 2 \theta \in\left[0, \frac{1}{2}\right]$
$\therefore \operatorname{det}(A) \in[2,3]$ which is a subset of $\left(\frac{3}{2}, 3\right]$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.