If alpha and beta are the least and the great | vedclass

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

Question

(B) Given $f(x)=(\sin ^{-1} x)^2+(\cos ^{-1} x)^2$. Since $\sin ^{-1} x + \cos ^{-1} x = \frac{\pi}{2}$,we have $\cos ^{-1} x = \frac{\pi}{2} - \sin ^

Vedclass · Accepted Answer

$11 \pi^2$

Vedclass · Answer

(B) Given $f(x)=(\sin ^{-1} x)^2+(\cos ^{-1} x)^2$. Since $\sin ^{-1} x + \cos ^{-1} x = \frac{\pi}{2}$,we have $\cos ^{-1} x = \frac{\pi}{2} - \sin ^{-1} x$. Let $t = \sin ^{-1} x$. Since $x \in [-1, 1]$,$t \in [-\frac{\pi}{2}, \frac{\pi}{2}]$. Then $f(t) = t^2 + (\frac{\pi}{2} - t)^2 = t^2 + \frac{\pi^2}{4} - \pi t + t^2 = 2t^2 - \pi t + \frac{\pi^2}{4}$. This is a parabola opening upwards. The vertex is at $t = -\frac{-\pi}{2(2)} = \frac{\pi}{4}$. Since $\frac{\pi}{4} \in [-\frac{\pi}{2}, \frac{\pi}{2}]$,the minimum value occurs at $t = \frac{\pi}{4}$. $\alpha = f(\frac{\pi}{4}) = 2(\frac{\pi^2}{16}) - \pi(\frac{\pi}{4}) + \frac{\pi^2}{4} = \frac{\pi^2}{8} - \frac{\pi^2}{4} + \frac{\pi^2}{4} = \frac{\pi^2}{8}$. The maximum value occurs at the endpoints of the interval $[-\frac{\pi}{2}, \frac{\pi}{2}]$. At $t = -\frac{\pi}{2}$,$f(-\frac{\pi}{2}) = 2(-\frac{\pi}{2})^2 - \pi(-\frac{\pi}{2}) + \frac{\pi^2}{4} = \frac{\pi^2}{2} + \frac{\pi^2}{2} + \frac{\pi^2}{4} = \frac{5\pi^2}{4}$. At $t = \frac{\pi}{2}$,$f(\frac{\pi}{2}) = 2(\frac{\pi}{2})^2 - \pi(\frac{\pi}{2}) + \frac{\pi^2}{4} = \frac{\pi^2}{2} - \frac{\pi^2}{2} + \frac{\pi^2}{4} = \frac{\pi^2}{4}$. Thus,$\beta = \frac{5\pi^2}{4}$. Finally,$8(\alpha + \beta) = 8(\frac{\pi^2}{8} + \frac{5\pi^2}{4}) = 8(\frac{\pi^2 + 10\pi^2}{8}) = 11\pi^2$.

If $\alpha$ and $\beta$ are the least and the greatest values of $f(x)=(\sin ^{-1} x)^2+(\cos ^{-1} x)^2$ for all $x \in [-1, 1]$ respectively,then $8(\alpha+\beta)=$

Similar Questions

Evaluate: $\cot ^{ - 1}\left(\frac{xy + 1}{x - y}\right) + \cot ^{ - 1}\left(\frac{yz + 1}{y - z}\right) + \cot ^{ - 1}\left(\frac{zx + 1}{z - x}\right)$

$\sin ^{-1} \frac{4}{5} + 2 \tan ^{-1} \frac{1}{3}$ is equal to

If ${\sin ^{ - 1}}\left( {\frac{{2a}}{{1 + {a^2}}}} \right) + {\sin ^{ - 1}}\left( {\frac{{2b}}{{1 + {b^2}}}} \right) = 2{\tan ^{ - 1}}x,$ then $x = $

If ${\cos ^{ - 1}}x + {\cos ^{ - 1}}y + {\cos ^{ - 1}}z = 3\pi ,$ then $xy + yz + zx = $

If $\tan ^{-1}\left[\frac{1}{1+1 \cdot 2}\right]+\tan ^{-1}\left[\frac{1}{1+2 \cdot 3}\right]+\cdots+\tan ^{-1}\left[\frac{1}{1+n(n+1)}\right]=\tan ^{-1}[x]$,then $x=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module