Let f(x) = max {sin^{-1}x, cos^{-1}x}. Then,t | vedclass

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(A) The functions $y = \sin^{-1}x$ and $y = \cos^{-1}x$ intersect where $\sin^{-1}x = \cos^{-1}x$. Since $\sin^{-1}x + \cos^{-1}x = \frac{\pi}{2}$,we

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$\frac{3\pi}{2} - \sqrt{2}$

Vedclass · Answer

(A) The functions $y = \sin^{-1}x$ and $y = \cos^{-1}x$ intersect where $\sin^{-1}x = \cos^{-1}x$. Since $\sin^{-1}x + \cos^{-1}x = \frac{\pi}{2}$,we have $2\sin^{-1}x = \frac{\pi}{2}$,which implies $\sin^{-1}x = \frac{\pi}{4}$,so $x = \frac{1}{\sqrt{2}}$. For $x \in [-1, \frac{1}{\sqrt{2}}]$,$\cos^{-1}x \ge \sin^{-1}x$,so $f(x) = \cos^{-1}x$. For $x \in [\frac{1}{\sqrt{2}}, 1]$,$\sin^{-1}x \ge \cos^{-1}x$,so $f(x) = \sin^{-1}x$. The area $A$ is given by $\int_{-1}^{1/\sqrt{2}} \cos^{-1}x \, dx + \int_{1/\sqrt{2}}^{1} \sin^{-1}x \, dx$. Using $\int \cos^{-1}x \, dx = x\cos^{-1}x - \sqrt{1-x^2}$ and $\int \sin^{-1}x \, dx = x\sin^{-1}x + \sqrt{1-x^2}$: $A = [x\cos^{-1}x - \sqrt{1-x^2}]_{-1}^{1/\sqrt{2}} + [x\sin^{-1}x + \sqrt{1-x^2}]_{1/\sqrt{2}}^{1}$ $A = (\frac{1}{\sqrt{2}} \cdot \frac{\pi}{4} - \sqrt{1 - \frac{1}{2}}) - (-1 \cdot \pi - \sqrt{1 - 1}) + (1 \cdot \frac{\pi}{2} + 0) - (\frac{1}{\sqrt{2}} \cdot \frac{\pi}{4} + \sqrt{1 - \frac{1}{2}})$ $A = \frac{\pi}{4\sqrt{2}} - \frac{1}{\sqrt{2}} + \pi + \frac{\pi}{2} - \frac{\pi}{4\sqrt{2}} - \frac{1}{\sqrt{2}} = \frac{3\pi}{2} - \frac{2}{\sqrt{2}} = \frac{3\pi}{2} - \sqrt{2}$.

Let $f(x) = \max \{\sin^{-1}x, \cos^{-1}x\}$. Then,the area bounded by $x = -1$,$x = 1$,$y = f(x)$,and $y = 0$ is:

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