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