Let A be the area bounded by the curve y = co | vedclass

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(C) First,simplify the curve equation: $y = \cos^{-1}\sqrt{1 - x^2} = \sin^{-1}x$ for $x \in [0, 1]$.The tangent to $y = \sin^{-1}x$ at $x = 0$ is fou

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$\pi - 1$

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(C) First,simplify the curve equation: $y = \cos^{-1}\sqrt{1 - x^2} = \sin^{-1}x$ for $x \in [0, 1]$.The tangent to $y = \sin^{-1}x$ at $x = 0$ is found by $y' = \frac{1}{\sqrt{1-x^2}}$. At $x = 0$,$y' = 1$. The tangent line is $y - 0 = 1(x - 0)$,which is $y = x$.The area $A$ is bounded by $y = \sin^{-1}x$,$y = x$,and $x = 1$. Thus,$A = \int_0^1 (\sin^{-1}x - x) dx$.Using integration by parts for $\int \sin^{-1}x dx = x\sin^{-1}x + \sqrt{1-x^2}$,we get:$A = [x\sin^{-1}x + \sqrt{1-x^2} - \frac{x^2}{2}]_0^1$$A = (1 \cdot \frac{\pi}{2} + 0 - \frac{1}{2}) - (0 + 1 - 0) = \frac{\pi}{2} - \frac{1}{2} - 1 = \frac{\pi - 3}{2}$.Since $0 Therefore,$2(\{A\} + 	ext{sgn}(A)) = 2(\frac{\pi - 3}{2} + 1) = \pi - 3 + 2 = \pi - 1$.

Let $A$ be the area bounded by the curve $y = \cos^{-1}\sqrt{1 - x^2}$,the tangent to the curve $y = \sin^{-1}x$ at $x = 0$,and the line $x = 1$. Then the value of $2(\{A\} + \text{sgn}(A))$ is (where $\{.\}$ is the fractional part function and $\text{sgn}(x)$ is the signum function).

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