Let $f$ be a real-valued function defined on the interval $(0, \infty)$ by $f(x)=\ln x+\int\limits_0^x \sqrt{1+\sin t} d t$. Then which of the following statement(s) is(are) true?
(a) $f^{\prime \prime}(x)$ exists for all $(x \in(0, \infty)$
(b) $f^{\prime}(x)$ exists for all $x \in(0, \infty)$ and $f^{\prime}(x)$ is continuous on $(0, \infty)$, but not differentiable on $(0, \infty)$
(c) there exists $\alpha>1$ such that $\left|f^{\prime}(x)\right|<f(x)$ for all $x \in(\alpha, \infty)$
(d) there exists $\beta>0$ such that $|f(x)|+\left|f^{\prime}(x)\right| \leq \beta$ for all $x \in(0, \infty)$
Answer & explanation
Correct answer: option 2
We have,
$f(x) =\ln x+\int\limits_0^x \sqrt{1+\sin t} d t$
$\Rightarrow f^{\prime}(x) =\frac{1}{x}+\sqrt{1+\sin x}$
$\Rightarrow f^{\prime}(x)$ exists for all $x \in(0, \infty)$ and is continuous on $(0, \infty)$
Clearly, $f^{\prime}(x)$ is not differentiable at $x=2 n \pi-\frac{\pi}{2}, n \in N$. So, $f^{\prime}(x)$ is not differentiable on $(0, \infty)$.