If A = 1 sin θ 1 - sin θ 1 sin θ - 1 - sin θ 1 , then for all θ ∈ 3 π 4 …

If A=1sinθ1-sinθ1sinθ-1-sinθ1, then for all θ∈3π4,5π4, detA lies in the interval :
  1. 1, 52
  2. 52, 4
  3. 32, 3
  4. 0, 32

Solution

$A=\left[\begin{array}{ccc}1 & \sin \theta & 1 \\ -\sin \theta & 1 & \sin \theta \\ -1 & -\sin \theta & 1\end{array}\right]$
$\Rightarrow \operatorname{det}(A)=\left(11+\sin ^2 \theta\right)-\sin \theta(-\sin \theta+\sin \theta)+1\left(\sin ^2 \theta+1\right)$
$=2+2 \sin ^2 \theta$
$\because \theta \in\left(\frac{3 \pi}{4}, \frac{5 \pi}{4}\right)$
$\therefore \sin 2 \theta \in\left[0, \frac{1}{2}\right]$
$\therefore \operatorname{det}(A) \in[2,3]$ which is a subset of $\left(\frac{3}{2}, 3\right]$

Asked in: JEE Main 2019 (12 Jan Shift 2)

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