Properties of Real Valued Integral Function and Differentiability

To determine which statement(s) are true, we analyze the function $f(x)$ given by:
$f(x) = \ln x + \int_0^x \sqrt{1+\sin t} \, dt, \quad x \in (0, \infty)$

1. Differentiability of $f(x)$ and $f'(x)$ (Statements A and B)

By the Fundamental Theorem of Calculus, since $g(t) = \sqrt{1+\sin t}$ is continuous on $[0, \infty)$, the function $G(x) = \int_0^x \sqrt{1+\sin t} \, dt$ is differentiable on $(0, \infty)$ with derivative:
$G'(x) = \sqrt{1+\sin x}$

Thus, $f(x)$ is differentiable on $(0, \infty)$ and its first derivative is given by:
$f'(x) = \frac{1}{x} + \sqrt{1+\sin x}$

Since both $\frac{1}{x}$ and $\sqrt{1+\sin x}$ are continuous functions on $(0, \infty)$, $f'(x)$ exists and is continuous on $(0, \infty)$.

Now, let us examine the differentiability of $f'(x)$ (i.e., existence of $f''(x)$):
Notice that $\sqrt{1+\sin x} = \sqrt{\left(\sin\frac{x}{2} + \cos\frac{x}{2}\right)^2} = \left| \sin\frac{x}{2} + \cos\frac{x}{2} \right|$.

At $x = \frac{3\pi}{2} + 2k\pi$ for $k \in \mathbb{Z}_{\ge 0}$:
$\sin x = -1 \implies 1+\sin x = 0$

Let $u = x - \frac{3\pi}{2}$. For $u$ near $0$:
$\sqrt{1+\sin x} = \sqrt{1 - \cos u} = \sqrt{2\sin^2\frac{u}{2}} = \sqrt{2}\left|\sin\frac{u}{2}\right|$

Evaluating the one-sided derivatives of $\sqrt{1+\sin x}$ at $x = \frac{3\pi}{2}$:

• Right-hand derivative: $\lim_{u \to 0^+} \frac{\sqrt{2}\sin\frac{u}{2}}{u} = \frac{1}{\sqrt{2}}$
• Left-hand derivative: $\lim_{u \to 0^-} \frac{-\sqrt{2}\sin\frac{u}{2}}{u} = -\frac{1}{\sqrt{2}}$

Since the left-hand and right-hand derivatives are not equal at $x = \frac{3\pi}{2} + 2k\pi$, $f'(x)$ is not differentiable at these points.

Hence, $f''(x)$ does not exist for all $x \in (0, \infty)$.
Therefore, Statement A is false and Statement B is true.

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2. Behavior of $|f'(x)|$ and $|f(x)|$ as $x \to \infty$ (Statement C)

For $x \ge 1$:
$|f'(x)| = \left| \frac{1}{x} + \sqrt{1+\sin x} \right| \le \frac{1}{x} + \sqrt{2} \le 1 + \sqrt{2}$
This shows that $f'(x)$ is bounded above by a constant for all $x \ge 1$.

For $f(x)$, since $\sqrt{1+\sin t} \ge 0$:
$f(x) = \ln x + \int_0^x \sqrt{1+\sin t} \, dt \ge \ln x$

As $x \to \infty$, $\lim_{x \to \infty} f(x) = \infty$.

Since $f(x) \to \infty$ while $f'(x)$ remains bounded, there exists a sufficiently large real number $\alpha > 1$ such that for all $x \in (\alpha, \infty)$:
$f(x) > 1 + \sqrt{2} \ge |f'(x)| \implies |f'(x)| < |f(x)|$

Therefore, Statement C is true.

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3. Boundedness of $|f(x)| + |f'(x)|$ (Statement D)

Since $\lim_{x \to \infty} f(x) = \infty$, we have:
$\lim_{x \to \infty} (|f(x)| + |f'(x)|) = \infty$

Thus, there cannot exist any real constant $\beta > 0$ such that $|f(x)| + |f'(x)| \le \beta$ for all $x \in (0, \infty)$.

Therefore, Statement D is false.

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Conclusion

The correct statements are B and C.