Question

Let $f$ be a real-valued function defined on the interval $(0,\infty )$ by $f(x)=\ell nx+\underset{0}{\overset{x}{\int }}\sqrt{1+\sin t}dt$. 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 $f^{\prime }$ 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 $|f(x)|+\left|f^{\prime }(x)\right|\le \beta$ for all $x\in (0,\infty )$

Answer 1

Answer: (B), (C)

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