Using the principal values of the inverse tri | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

$22$

Vedclass · Answer

(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 $\pi^2$)

Similar Questions

$\sin \left(\tan ^{-1} \frac{4}{5}+\tan ^{-1} \frac{4}{3}+\tan ^{-1} \frac{1}{9}-\tan ^{-1} \frac{1}{7}\right) = $

$\cot^{-1}(1) + \cot^{-1} (\frac{1}{2}) + \cot^{-1}(\frac{1}{3}) =$

If for some $\alpha, \beta$ such that $\alpha \leq \beta$ and $\alpha+\beta=8$,the equation $\sec^2(\tan^{-1} \alpha) + \operatorname{cosec}^2(\cot^{-1} \beta) = 36$ holds,then the value of $\alpha^2+\beta$ is . . . . . .

If the value of $x$ satisfying the equation $\sin^{-1} \sqrt{1-x^2} = \tan^{-1} \sqrt{\frac{2}{x}-1}$ is $\frac{a}{b}$ (where $a$ and $b$ are coprime),then the value of $a^2 + b^2$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module