Using the principal values of the inverse tri | vedclass

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(D) Let $f(x) = 16((\sec^{-1} x)^2 + (\operatorname{cosec}^{-1} x)^2)$.We know that $\sec^{-1} x + \operatorname{cosec}^{-1} x = \frac{\pi}{2}$ for al

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$22$

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(D) Let $f(x) = 16((\sec^{-1} x)^2 + (\operatorname{cosec}^{-1} x)^2)$.We know that $\sec^{-1} x + \operatorname{cosec}^{-1} x = \frac{\pi}{2}$ for all $x \in (-\infty, -1] \cup [1, \infty)$.Let $a = \sec^{-1} x$. Then $\operatorname{cosec}^{-1} x = \frac{\pi}{2} - a$.The domain of $\sec^{-1} x$ is $a \in [0, \pi] \setminus \{\frac{\pi}{2}\}$.Substituting this into the expression, we get $f(a) = 16(a^2 + (\frac{\pi}{2} - a)^2) = 16(a^2 + \frac{\pi^2}{4} - \pi a + a^2) = 16(2a^2 - \pi a + \frac{\pi^2}{4}) = 32a^2 - 16\pi a + 4\pi^2$.This is a quadratic in $a$ opening upwards. The vertex is at $a = \frac{-(-16\pi)}{2(32)} = \frac{\pi}{4}$.Since $a \in [0, \pi] \setminus \{\frac{\pi}{2}\}$, the minimum value occurs at $a = \frac{\pi}{4}$.$	ext{Min} = f(\frac{\pi}{4}) = 32(\frac{\pi^2}{16}) - 16\pi(\frac{\pi}{4}) + 4\pi^2 = 2\pi^2 - 4\pi^2 + 4\pi^2 = 2\pi^2$.The maximum value occurs at the boundaries of the interval $[0, \pi] \setminus \{\frac{\pi}{2}\}$, which are $a=0$ or $a=\pi$.$f(0) = 16(0^2 + (\frac{\pi}{2} - 0)^2) = 16(\frac{\pi^2}{4}) = 4\pi^2$.$f(\pi) = 16(\pi^2 + (\frac{\pi}{2} - \pi)^2) = 16(\pi^2 + \frac{\pi^2}{4}) = 16(\frac{5\pi^2}{4}) = 20\pi^2$.Thus, the maximum value is $20\pi^2$ and the minimum value is $2\pi^2$.The sum is $20\pi^2 + 2\pi^2 = 22\pi^2$.

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

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