The number of solutions of the equation frac{ | vedclass

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(D) Let $f(x) = \int_{\cos x}^{\sin x} \frac{dt}{\sqrt{1 - t^2}}$.Using Leibniz's rule,$\frac{d}{dx} f(x) = \frac{1}{\sqrt{1 - (\sin x)^2}} \cdot \fra

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$0$

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(D) Let $f(x) = \int_{\cos x}^{\sin x} \frac{dt}{\sqrt{1 - t^2}}$.Using Leibniz's rule,$\frac{d}{dx} f(x) = \frac{1}{\sqrt{1 - (\sin x)^2}} \cdot \frac{d}{dx}(\sin x) - \frac{1}{\sqrt{1 - (\cos x)^2}} \cdot \frac{d}{dx}(\cos x)$.Since $\sqrt{1 - \sin^2 x} = |\cos x|$ and $\sqrt{1 - \cos^2 x} = |\sin x|$,the expression becomes $\frac{\cos x}{|\cos x|} - \frac{-\sin x}{|\sin x|} = 	ext{sgn}(\cos x) + 	ext{sgn}(\sin x)$.In the interval $[0, \pi]$,$\sin x$ is always positive,so $	ext{sgn}(\sin x) = 1$.Thus,the equation is $	ext{sgn}(\cos x) + 1 = 2\sqrt{2}$.Since $	ext{sgn}(\cos x)$ can only be $1, -1,$ or $0$,the sum $	ext{sgn}(\cos x) + 1$ can only be $2, 0,$ or $1$.Since $2\sqrt{2} \approx 2.828$,there is no value of $x$ such that $	ext{sgn}(\cos x) + 1 = 2\sqrt{2}$.Therefore,the number of solutions is $0$.

The number of solutions of the equation $\frac{d}{dx} \int_{\cos x}^{\sin x} \frac{dt}{\sqrt{1 - t^2}} = 2\sqrt{2}$ in the interval $[0, \pi]$ is:

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