Question

Let $f$ be a real-valued function defined on the interval $(0,\infty )$ by $f\left(x\right)=\ln x+\underset{0}{\overset{x}{\int }}\sqrt{1+\sin t}dt$. Then which of the following statement(s) is/are true?

• A. There exists $\alpha >1$ such that $|f^{\prime }\left(x\right)|<|f\left(x\right)|$ for all $x\in (\alpha ,\infty )$
• B. $f^{\prime }\left(x\right)$exists for all $x\in (0,\infty )$ and $f^{\prime }$ is continuous on $(0,\infty )$, but not differentiable on $(0,\infty )$
• C. There exists $\beta >0$ such that $|f\left(x\right)|+|f^{\prime }\left(x\right)|\le \beta$ for all $x\in (0,\infty )$
• D. $f^{\prime \prime }\left(x\right)$exists for all $x\in (0,\infty )$

Answer 1

Answer: (A), (B)

$f^{\prime }\left(x\right)$exists for all $x\in (0,\infty )$ and $f^{\prime }$ is continuous on $(0,\infty )$, but not differentiable on $(0,\infty )$

$f^{\prime }\left(x\right)=\frac{1}{x}+\sqrt{1+\sin x},x>0$ but not diffrentiable in $(0,\infty )$ as $\sin x$ may be $-1$ and then $f^{\prime \prime }\left(x\right)$ can't exist.
Clearly $f^{\prime }\left(x\right)$ is continuous on $(0,\infty )$
(4) Is not true because $f\left(x\right)$ tends to $\infty$ as $x$ tends to $\infty$
(3) Is true because $|f^{\prime }\left(x\right)\le 3|$ if $x>1$
But $|f\left(x\right)|>3$, if $x>e^3$
So we may take $\alpha =e^3$