Question

Let $f$ be a real-valued function defined on the interval $(0, \infty)$ by $f(x)={\ell} \,n \,x+\int\limits_{0}^{x} \sqrt{1+\sin \,t}\, d t$. Then which of the following statement(s) is (are) true ?

• A. $f''(x)$ exists for all $x \in(0, \infty)$
• B. $f '( x )$ exists for all $x \in(0, \infty)$ and $f '$ is continuous on $(0, \infty)$, but not differentiable on $(0, \infty)$
• C. there exists $\alpha>1$ such that $\left|f'(x)\right|<|f(x)|$ 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 1

Answer: (B), (C)

$f '( x )=\frac{1}{ x }+\sqrt{1+\sin x }$
$f '( x )$ is not differentiable at $\sin x =-1$
or $x =2 n \pi-\frac{\pi}{2}, n \in N$
$\text{In} x \in(1, \infty) f ( x )>0, f '( x )>0$
Consider $f(x)-f'(x)$
$=\ln x+\int\limits_{0}^{x} \sqrt{1+\sin \,t} \,d t-\frac{1}{x}-\sqrt{1+\sin x}$
$=\left(\int\limits_{0}^{x} \sqrt{1+\sin \,t} \,d t-\sqrt{1+\sin x}\right)+\ln x-\frac{1}{x} $
Consider $g(x)=\int\limits_{0}^{x} \sqrt{1+\sin t} d t-\sqrt{1+\sin x}$
It can be proved that $g(x) \geq 2 \sqrt{2}-\sqrt{10}\forall x \in(0, \infty)$
Now there exists some $\alpha>1$ such that $\frac{1}{x}-\ln x \leq 2 \sqrt{2}-\sqrt{10}$ for all $x \in(\alpha, \infty)$ as $\frac{1}{x}-\ln x$ is strictly decreasing function.
$\Rightarrow g ( x ) \geq \frac{1}{ x }-\ln x$