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.

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

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "M.C.Q (1 Marks)" from STD 12 - 5. continuity and differentiation→STD 12 - 5. continuity and differentiation — all question types→Mathematics STD 12 Science question papers→CBSE Board STD 12 Science question papers→CBSE Board question paper generator→

Similar questions

Choose the correct answer
If $\theta$ is the angle between any two vectors $\vec{\text{a}}\ \text{and}\ \vec{\text{b}}, \text{then}\ |\vec{\text{a}}\cdot\vec{\text{b}}|=|\vec{\text{a}}\times\vec{\text{b}}|\ \text{when}\ \theta$ is equal to

If $\int\limits^1_0\text{f}(\text{x})\text{dx}=1,\int\limits^1_0\text{x}\text{f}(\text{x})\text{dx}=\text{a},\int\limits^1_0\text{x}^2\text{f}(\text{x})\text{dx}=\text{a}^2,$ then $\int\limits^1_0(\text{a}-\text{x})^2\text{f(x)}\text{dx}$ equals :

$\cos \left[ {{{\tan }^{ - 1}}\frac{1}{3} + {{\tan }^{ - 1}}\frac{1}{2}} \right] = $

How many lines through the origin in make equal angles with the coordinate axis:

He area of the region bounded by the parabola $y = x^2$ and $y = |x|$ is:

Simplify $ {\cot ^{ - 1}}\frac{1}{{\sqrt {{\text{x}^2} - 1} }} $ for $\text{ x} < - 1$:

The value of $\int\limits^\frac{\pi}{4}_0\cos\text{x}\text{ e}^{\sin\text{x}}\text{ dx}$ is:

Let $f(x+y)=f(x).f(y)$ for all $x, y$ where $ f(0) \ne 0$. If $f(5) = 2$ and $f '(0) = 3,$ then $f '(5)$ is equal to

Let $\theta$ be the angle between the vectors $\vec{a}$ and $\vec{b}$, where $|\vec{a}|=4,|\vec{b}|=3 \quad \theta \in\left(\frac{\pi}{4}, \frac{\pi}{3}\right)$. Then $|(\vec{a}-\vec{b}) \times(\vec{a}+\vec{b})|^{2}+4(\vec{a} \cdot \vec{b})^{2}$ is equal to

The vectors $a, b $ and $ a + b $ are