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(D) We are given the integral equation $\int_{\sqrt{2}}^x \frac{dt}{|t| \sqrt{t^2-1}} = \frac{\pi}{12}$.Recall the standard integral formula $\int \fr

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$2$

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(D) We are given the integral equation $\int_{\sqrt{2}}^x \frac{dt}{|t| \sqrt{t^2-1}} = \frac{\pi}{12}$.Recall the standard integral formula $\int \frac{dt}{|t| \sqrt{t^2-1}} = \sec^{-1}(t) + C$.Applying the limits of integration,we get:$\left[ \sec^{-1}(t) \right]_{\sqrt{2}}^x = \frac{\pi}{12}$.Substituting the limits,we have $\sec^{-1}(x) - \sec^{-1}(\sqrt{2}) = \frac{\pi}{12}$.Since $\sec^{-1}(\sqrt{2}) = \frac{\pi}{4}$,the equation becomes $\sec^{-1}(x) - \frac{\pi}{4} = \frac{\pi}{12}$.Adding $\frac{\pi}{4}$ to both sides,we get $\sec^{-1}(x) = \frac{\pi}{12} + \frac{\pi}{4} = \frac{\pi + 3\pi}{12} = \frac{4\pi}{12} = \frac{\pi}{3}$.Therefore,$x = \sec\left(\frac{\pi}{3}\right) = 2$.

The value of $x$ that satisfies the equation $\int_{\sqrt{2}}^x \frac{dt}{|t| \sqrt{t^2-1}} = \frac{\pi}{12}$ is

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