Question

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

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

Answer 1

Answer: (B), (C)

The correct options are
B $f\left(x\right)$ exists for all $x\in (0,\infty )$ and $f^{\prime }$ is continuous on $(0,\infty )$, but not differentiable on $(0,\infty )$
C There exists $\alpha >1$ such that $|f^{\prime }\left(x\right)|<|f\left(x\right)|$ for all $x\in (\alpha ,\infty )$

Option A:

$f\left(x\right)=\ln x+{\int }_0^1\sqrt{1+\sin t}dx$

$f^{\prime }\left(x\right)=\frac{1}{x}+\sqrt{1+\sin x}$

$f^{\prime \prime }\left(x\right)=-\frac{1}{x^2}-\frac{\cos x}{{2(1+\sin x)}^{1.5}}$

$f^{\prime \prime }\left(x\right)$ doesnt exist for values in which $\sin x=-1$

Option B:

It can be seen from given definition that $f\left(x\right)$ exists for the given interval.

Also $f^{\prime }\left(x\right)$ is continuous in the given interval, it can be observed by above $f^{\prime }\left(x\right)$ calculations.
$f^{\prime }\left(x\right)$ is not differentiable because $f^{\prime }\left(x\right)$ doesn't exist at points where $\sin x=-1$ in the given interval.

Option C:
Now consider

$f\left(x\right)-f^{\prime }\left(x\right)=[{\int }_0^x\sqrt{1+\sin t}dt-\sqrt{1+\sin x}]+\ln x-\frac{1}{x}$
$f^{\prime }\left(x\right)=\frac{1}{x}+\sqrt{1+\sin x}\le \sqrt{2}$

The maximum value of $f^{\prime }\left(x\right)$ is $\sqrt{2}$.
So, there exists some $\alpha$ for which $f^{\prime }\left(x\right)<f\left(x\right)$
$f^{\prime }\left(x\right)$ is bounded where as $f\left(x\right)$ is an increasing function (unbounded), after some value of $x$, $f\left(x\right)$ will be dominating $f^{\prime }\left(x\right)$

Hence, option B and C.