Question

If $f:[0, \pi / 2) \rightarrow R$ is defined as $f(\theta)=\begin{vmatrix}1 & \tan \theta & 1 \\\ -\tan \theta & 1 & \tan \theta \\\ -1 & -\tan \theta & 1\end{vmatrix}$. Then, the range of $f$ is

• A. $\left(2, \infty\right)$
• B. $(-\:\infty ,\:-2]$
• C. $[2, \infty)$
• D. $(-\infty, 2]$

Answer 1

Answer: (C)

$[2, \infty)$

Answer 2

The correct answer is C:$[2,\infin]$
We have, $f(\theta)=\begin{vmatrix}1 & \tan \theta & 1 \\\ -\tan \theta & 1 & \tan \theta \\\ -1 & -\tan \theta & 1\end{vmatrix}$
For, $f:[0,\frac{\pi}{2}]\rightarrow{R}$
=1\left(1+\tan ^{2} \theta\right)-\tan \theta$$(-\tan \theta+\tan \theta)+1\left(\tan ^{2} \theta+1\right)
$=2\left(1+\tan ^{2} \theta\right)-0=2 \sec ^{2} \theta$
$\therefore$ Range of $f=(2, \infty)$