Q66 / 125 · BITSAT 2010 · Mathematics

The function \(\mathrm{f}(\mathrm{x})=\tan \mathrm{x}-4 \mathrm{x}\) is strictly decreasing on

  • Correct answer
    A
    \(\left(-\frac{\pi}{3}, \frac{\pi}{3}\right)\)
  • B
    \(\left(\frac{\pi}{3}, \frac{\pi}{2}\right)\)
  • C
    \(\left(-\frac{\pi}{3}, \frac{\pi}{2}\right)\)
  • D
    \(\left(\frac{\pi}{2}, \pi\right)\)

Solution

\(f(x)=\tan x-4 x \Rightarrow f^{\prime}(x)=\sec ^2 x-4\) When \(\frac{-\pi}{3} < \mathrm{x} < \frac{\pi}{3}, 1 < \sec \mathrm{x} < 2\) Therefore, \(1 < \sec ^2 x < 4\) \(\Rightarrow-3 < \left(\sec ^2 x-4\right) < 0\) Thus, for \(\frac{-\pi}{3} < \mathrm{x} < \frac{\pi}{3}, \mathrm{f}^{\prime}(\mathrm{x}) < 0\) Hence, \(\mathrm{f}\) is strictly decreasing on \(\left(\frac{-\pi}{3}, \frac{\pi}{3}\right)\)

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

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