Q121 / 200 · Manipal (MET) 2014 · Mathematics

The range of $f(x)=\sec \left(\frac{\pi}{4} \cos ^2 x\right),-\infty \lt x \lt \infty$ is

  • Correct answer
    A
    $[1, \sqrt{2}]$
  • B
    $[1, \infty]$
  • C
    $[-\sqrt{2},-1] \cup[1, \sqrt{2}]$
  • D
    $[-\infty,-1] \cup[1, \infty]$

Solution

Given, $f(x)=\sec \left(\frac{\pi}{4} \cos ^2 x\right)$ We know that, $0 \leq \cos ^2 x \leq 1$ $\begin{aligned} & \therefore f(x)=\sec \left(\frac{\pi}{4} \times 1\right)=\sqrt{2} \\ & f(x)=\sec \left(\frac{\pi}{4} \times 0\right)=1 \quad \therefore \text { Range } \in[1, \sqrt{2}] \end{aligned}$

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

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