If $f(θ) =\begin{vmatrix}1&\tan θ&1\\-\tan θ&1&\tan θ\\-1&-\tan θ&1\end{vmatrix}$, then the set $\{f(θ):0≤θ≤\frac{π}{2}\}$ is
Answer & explanation
Correct answer: option 2
Applying $R_1 → R_1 + R_3$, we get
$f(θ) =\begin{vmatrix}θ&0&2\\-\tan θ&1&\tan θ\\-1&-\tan θ&1\end{vmatrix}$
$=2 (\tan^2θ+1)=2 \sec^2 θ>2$
$∴f(θ) ∈[2,∞)$.