Mathematics - Continuity - Differentiability Question with Solution | TestHub

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, $f^{\prime }(x)\le 3$, if $x>1$ and $f(x)>3$, if $x>e^3$ $\therefore$ Let $\alpha =e^3$ $\Rightarrow$ Option (c) is true. (d) is not possible as $f(x)\to \infty$ when $x\to \infty$. Hence, (b, c) is the correct option.