Question

The function $f(x)=\tan x-4x$ is strictly decreasing on

• 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. $\left(\frac{\pi }{2},\pi \right)$

Answer 1

Answer: (A)

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