Mathematics · 2019
JEE Main · 12 January 2019, Shift 2 · Q64
If A = [1, sin θ, 1; - sin θ, 1, sin θ; -1, - sin θ, 1]; then for all θ ∈((3 π)/4, (5 π)/4), det ( A ) lies in the interval:
If $\displaystyle \mathrm{A}=\left[\begin{array}{ccc}1 & \sin \theta & 1 \\ -\sin \theta & 1 & \sin \theta \\ -1 & -\sin \theta & 1\end{array}\right]$; then for all $\displaystyle \theta \in\left(\frac{3 \pi}{4}, \frac{5 \pi}{4}\right), \operatorname{det}(\mathrm{A})$ lies in the interval :
Official answer
(3)
$\displaystyle \left(\frac{3}{2}, 3\right]$
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JEE Main 2019 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.