Using the principal values of the inverse trigonometric functions, the sum of the maximum …

Question

Using the principal values of the inverse trigonometric functions, the sum of the maximum and the minimum values of $16\left(\left(\sec ^{-1} x\right)^2+\left(\operatorname{cosec}^{-1} x\right)^2\right)$ is :

Accepted Answer

Let $ f(x) = 16\left[\left(\sec^{-1} xight)^2 + \left(\operatorname{cosec}^{-1} xight)^2ight] $.

We can express $ f(x) $ as:

$ f(x) = 16\left[\left(\sec^{-1} x + \operatorname{cosec}^{-1} xight)^2 - 2\left(\sec^{-1} xight)\left(\frac{\pi}{2} - \sec^{-1} xight)ight] $

This simplifies to:

$ f(x) = 16\left[\frac{\pi^2}{4} - \pi \sec^{-1} x + 2 \left(\sec^{-1} xight)^2ight], \quad 	ext{where } \sec^{-1} x \in [0, \pi] - \left\{\frac{\pi}{2}ight\} $

Further simplification gives:

$ f(x) = 16\left[2\left(\sec^{-1} x - \frac{\pi}{4}ight)^2 + \frac{\pi^2}{4} - \frac{\pi^2}{8}ight] $

For the maximum value when $ \sec^{-1} x = \pi $:

$ \max = 16\left[2\pi^2 - \pi^2 + \frac{\pi^2}{4}ight] = 20\pi^2 $

For the minimum value when $ \sec^{-1} x = \frac{\pi}{4} $:

$ \min = 16\left[\frac{2 	imes \pi^2}{16} - \frac{\pi^2}{4} + \frac{\pi^2}{4}ight] = 2\pi^2 $

Therefore, the sum of the maximum and minimum values is:

$ 	ext{Sum} = 22\pi^2 $

Solutions · 1