Let f be a real-valued function defined on the interval (0, ) by … - Vidyadip

MCQ

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?

$(A)$ $f^{\prime \prime}(x)$ 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 $|f(x)|+\left|f^{\prime}(x)\right| \leq \beta$ for all $x \in(0, \infty)$

✓
$(B,C)$
B
$(B,D)$
C
$(A,D)$
D
$(A,B)$

✓

Answer

Correct option: A.

$(B,C)$

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

$f^{\prime}(x)$ is not differentiable at $\sin x=-1$ or $x=2 n \pi-\frac{\pi}{2}, n \in N$

$\text { In } x \in(1, \infty) f(x)>0, f^{\prime}(x)>0$

Consider $f(x)-f^{\prime}(x)$

$ =\ln x+\int_0^x \sqrt{1+\sin t} d t-\frac{1}{x}-\sqrt{1+\sin x} $

$ =\left(\int_0^x \sqrt{1+\sin t} d t-\sqrt{1+\sin x}\right)+\ln x-\frac{1}{x}$

Consider $\mathrm{g}(\mathrm{x})=\int_0^{\mathrm{x}} \sqrt{1+\sin \mathrm{t}} d t-\sqrt{1+\sin \mathrm{x}}$

It can be proved that $\mathrm{g}(\mathrm{x}) \geq 2 \sqrt{2}-\sqrt{10} \quad \forall \mathrm{x} \in(0, \infty)$

Now there exists some $\alpha>1$ such that $\frac{1}{x}-\ln x \leq 2 \sqrt{2}-\sqrt{10}$ for all $x \in(\alpha, \infty)$ as $\frac{1}{x}-\ln x$ is strictly decreasing function.

$\Rightarrow g(x) \geq \frac{1}{x}-\ln x$

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