Question

If $f$ be a real valued function defined on the interval $(0,\infty )$ by $f\left(x\right)=\ln x+$ integral from $0$ to $x\sqrt{1+\sin t}dt$. Then which of the following statement (s) is (are) correct?

• A. $f\text{'}\text{'}\left(x\right)$ exists for all $x\in (0,\infty )$
• B. $f\text{'}\left(x\right)$ exists for all $x\in (0,\infty )$ and $f\text{'}$ is continuous on $(0,\infty )$ but not differentiable on $(0,\infty )$
• C. there exist $ɑ>1$ such that $|f\text{'}(x)<1|f\left(x\right)|$ for all $x\in (ɑ,\infty )$
• D. there exists $>0$ such that $|f(x\left)\right|<1|f\text{'}(x\left)\right|\le$ form all $x\in (0,\infty )$ but not differentiable on $(0,\infty )$

Answer 1

Answer: (B), (C)

there exist $ɑ>1$ such that $|f\text{'}(x)<1|f\left(x\right)|$ for all $x\in (ɑ,\infty )$

Explanation for the correct option:

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

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

$\therefore$$f\prime \left(x\right)$ exists for all $x\in (0,\infty )$) and $f\prime$ is continuous on $(0,\infty )$, but not differentiable on $(0,\infty )$ as $\sin x$ may be $-1$ then $f\text{'}\text{'}\left(x\right)$ cannot exist.

$f\text{'}\text{'}\left(x\right)=-\frac{1}{x^2}-\frac{\cos x}{2\sqrt{1+\sin x}}$

Now, in the interval $(0,\infty )$ , the value of $\sin x=-1$ with some particular values, so $\sqrt{1+\sin x}=0$

$\Rightarrow$$\frac{1}{\sqrt{1+\sin x}}$ is not defined in the interval $(0,\infty )$

So, $f\text{'}\text{'}\left(x\right)$ is not differentiable in the interval $(0,\infty )$

Step 2. Now Consider,

$f\left(x\right)-f\text{'}\left(x\right)=\left[{\int }_0^x\sqrt{1+\sin t}dt-\sqrt{1+\sin x}\right]+\ln x-\frac{1}{x}$

As we know that $-1\le \sin x\le 1$ so the maximum value of

$f\left(x\right)=\frac{1}{x}+\sqrt{1+\sin x}\ge \sqrt{2}$

So in the interval there exist $\alpha$ value $\alpha >1$, such that $\left|f\left(x\right)\right|-\left|f\text{'}\left(x\right)\right|>0$

$\therefore \left|f\text{'}\left(x\right)\right|<\left|f\left(x\right)\right|$ for all $x\in \left(\alpha ,\infty \right)$

Hence, Option ‘B’ and ‘C’ is Correct.