Q89 / 160 · WBJEE 2015 · Mathematics

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

  • Correct answer
    A
    $(2, \infty)$
  • B
    $(-\infty,-2]$
  • C
    $[2, \infty)$
  • D
    $(-\infty, 2]$

Solution

We have, $f(\theta)=\left|\begin{array}{ccc}1 & \tan \theta & 1 \\ -\tan \theta & 1 & \tan \theta \\ -1 & -\tan \theta & 1\end{array}\right|$ $=1\left(1+\tan ^{2} \theta\right)-\tan \theta$ $(-\tan \theta+\tan \theta)+1\left(\tan ^{2} \theta+1\right)$ $=2\left(1+\tan ^{2} \theta\right)-0=2 \sec ^{2} \theta$ $\therefore$ Range of $f=(2, \infty)$

Hence, the correct answer is option: 'A'.

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