Using the principal values of the inverse trigonometric functions, the sum of the maximum and the minimum values of $16\left((\sec^{-1}x)^{2}+(cosec^{-1}x)^{2}\right)$ is:
- (1) $24\pi^{2}$
- (2) $18\pi^{2}$
- (3) $31\pi^{2}$
- (4) $22\pi^{2}$
Solution:
We know that $\sec^{-1}x + cosec^{-1}x = \frac{\pi}{2}$ for $|x| \ge 1$. Let $\sec^{-1}x = \theta$, where $\theta \in [0, \pi] – \{\frac{\pi}{2}\}$.
Then $cosec^{-1}x = \frac{\pi}{2} – \theta$. The given expression becomes:
$$ E = 16\left[ \theta^2 + \left( \frac{\pi}{2} – \theta \right)^2 \right] $$
$$ E = 16\left[ \theta^2 + \frac{\pi^2}{4} + \theta^2 – \pi\theta \right] = 16\left[ 2\theta^2 – \pi\theta + \frac{\pi^2}{4} \right] $$
To find the minimum value, we complete the square or find the vertex ($ \theta = -b/2a $):
$$ \theta = \frac{\pi}{4} \quad (\text{which is in the domain}) $$
$$ E_{\min} = 16\left[ \left(\frac{\pi}{4}\right)^2 + \left(\frac{\pi}{4}\right)^2 \right] = 16\left[ \frac{2\pi^2}{16} \right] = 2\pi^2 $$
To find the maximum value, we check the boundary points of $\theta \in [0, \pi]$:
At $\theta = 0$: $E = 16[0 + \pi^2/4] = 4\pi^2$.
At $\theta = \pi$: $E = 16[\pi^2 + (\pi/2 – \pi)^2] = 16[\pi^2 + \pi^2/4] = 20\pi^2$.
So, $E_{\max} = 20\pi^2$.
Sum of maximum and minimum values = $20\pi^2 + 2\pi^2 = 22\pi^2$.
Ans. (4)