Let $f$ be a real-valued function defined on the interval $(0, \infty)$, by $f(x)=\ln x+\int_0^x…
Let $f$ be a real-valued function defined on the interval $(0, \infty)$, by $f(x)=\ln x+\int_0^x \sqrt{1+\sin t} d t$. Then which of the following statement(s) is (are) true ?
$f^{\prime \prime}(x)$ exists for all $x \in(0, \infty)$
$f^{\prime}(x)$ exists for all $x \in(0, \infty)$ and $f^{\prime}$ is continuous on $(0, \infty)$, but not differentiable on $(0, \infty)$
there exists $\alpha>1$ such that $\left|f^{\prime}(x)\right| < |f(x)|$ for all $x \in(\alpha, \infty)$
there exists $\beta > 0$ such that $|f(x)|+\left|f^{\prime}(x)\right| \leq \beta$ for all $x \in(0, \infty)$
Solution
Here, $f^{\prime}(x)=\frac{1}{x}+\sqrt{1+\sin x}, x>0$ but $f(x)$ is not differentiable in $(0, \infty)$ as $\sin x$ may be $-1$ and then $f^{\prime \prime}(x)=-\frac{1}{x^2}+\frac{\cos x}{2 \sqrt{1+\sin x}}$ will not exists.
$\Rightarrow f^{\prime}(x)$ is continuous for all $x \in(0, \infty)$ but $f^{\prime}(x)$ is not differentiable on $(0, \infty)$.
$\therefore$ Option (b) is true.
Also, $\quad f^{\prime}(x) \leq 3$, if $x>1$ and $\quad f(x)>3$, if $x>e^3$
$\therefore$ Let $\alpha=e^3$
$\Rightarrow$ Option (c) is true.
(d) is not possible as $f(x) \rightarrow \infty$ when $x \rightarrow \infty$.
Hence, (b, c) is the correct option.