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(A) Given $f(x) = \ln x + \int_0^x \sqrt{1+\sin t} \, dt$.By the Fundamental Theorem of Calculus,$f^{\prime}(x) = \frac{1}{x} + \sqrt{1+\sin x}$.Note

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$(B, C)$

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(A) Given $f(x) = \ln x + \int_0^x \sqrt{1+\sin t} \, dt$.By the Fundamental Theorem of Calculus,$f^{\prime}(x) = \frac{1}{x} + \sqrt{1+\sin x}$.Note that $\sqrt{1+\sin x} = \sqrt{(\sin(x/2) + \cos(x/2))^2} = |\sin(x/2) + \cos(x/2)|$.This expression is not differentiable where $\sin(x/2) + \cos(x/2) = 0$,i.e.,$	an(x/2) = -1$,which implies $x/2 = n\pi - \pi/4$,or $x = 2n\pi - \pi/2$.Since these points exist in $(0, \infty)$,$f^{\prime}(x)$ is not differentiable at these points. Thus,$(B)$ is true and $(A)$ is false.For $(C)$,as $x 	o \infty$,$f(x) \approx \int_0^x \sqrt{1+\sin t} \, dt \approx \frac{2\sqrt{2}}{\pi} x$ and $f^{\prime}(x) \approx \sqrt{1+\sin x}$. Since $f(x)$ grows linearly and $f^{\prime}(x)$ is bounded,$|f^{\prime}(x)| $(D)$ is false because $f(x) 	o \infty$ as $x 	o \infty$.

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