Question

Let f be a real-valued function defined on the interval $(0,\infty )$ by $f\left(x\right)={\log }_{e}x+{\int }_0^x\sqrt{1+\sin t}dt.$ Which of the following statement (s) is (are) true?

• A. $f\left(x\right)={\log }_{e}x+{\int }_0^x\sqrt{1+\sin t}dt.$ for all $xϵ(0,\infty )$
• B. $f^{\prime }\left(x\right)$ exists for all $xϵ(0,\infty )$ and $f^{\prime }\left(x\right)$ is continuous on $(0,\infty )$ but not differentiable on $(0,\infty )$
• C. there exists $\alpha >1$ such that $\left|f^{\prime }\left(x\right)\right|<\left|f\left(x\right)\right|$ for all
  $xϵ(\alpha ,\infty )$
• D. there exists $\beta >0$ such that $\left|f\left(x\right)\right|+\left|f^{\prime }\left(x\right)\right|\le \beta$ for all $xϵ(\alpha ,\infty )$

Answer 1

Answer: (A), (B), (C)

The correct options are
A $f^{\prime }\left(x\right)$ exists for all $xϵ(0,\infty )$ and $f^{\prime }\left(x\right)$ is continuous on $(0,\infty )$ but not differentiable on $(0,\infty )$
B $f\left(x\right)={\log }_{e}x+{\int }_0^x\sqrt{1+\sin t}dt.$ for all $xϵ(0,\infty )$
C there exists $\alpha >1$ such that $\left|f^{\prime }\left(x\right)\right|<\left|f\left(x\right)\right|$ for all

$xϵ(\alpha ,\infty )$
We have $f^{\prime }\left(x\right)=\frac{1}{x}+\sqrt{1+\sin x}$
Consider $f\left(x\right)-f^{\prime }\left(x\right)$
$=1nx+{\int }_0^x\sqrt{1+\sin t}dt-\frac{1}{x}-\sqrt{1+\sin x}$
$=({\int }_0^x\sqrt{1+\sin t}dt-\sqrt{1+\sin x})+1nx-\frac{1}{x}$
Consider $g\left(x\right)={\int }_0^x\sqrt{1+\sin t}dt-\sqrt{1+\sin x}$
it can be proved that $g\left(x\right)\ge 2\sqrt{2}-\sqrt{10}\forall x\in (0,\infty )$
Now there exist some $\alpha >1$ such that
$\frac{1}{x}-1nx\le 2\sqrt{2}-\sqrt{10}$ for all $x\in (0,\infty )$ as $\frac{1}{x}-1nx$ is strictly decreasing function.
$\Rightarrow g\left(x\right)\ge \frac{1}{x}-1nx$