Let $D= \begin{vmatrix}1&sin \theta & 1\\-sin \theta & 1& sin \theta \\-1& -sin \theta & 1\end {vmatrix}; 0≤\theta ≤ 2\pi , $ then
Answer & explanation
Correct answer: option 3
The correct answer is option (3) : $D \in [2, 4]$
We have,
$D=\begin{vmatrix}1&sin \theta & 1\\-sin \theta & 1& sin \theta \\-1& -sin \theta & 1\end {vmatrix}$
$⇒D=\begin{vmatrix}2 &sin \theta & 1\\0 & 1 & sin \theta \\0 & -sin \theta & 1\end {vmatrix}$ [Applying $C_1→C_1+C_3$]
$⇒D= 2(1+ sin^2 \theta )$
Now, $0≤sin^2\theta ≤ 1$
$⇒2≤2(1+sin^2 \theta ) ≤ 4 $
$⇒ D \in [2, 4]$