Download App

STD 12 - 7.2 definite integral — Mathematics STD 12 Science — Question

CBSE BoardEnglish MediumSTD 12 ScienceMathematicsSTD 12 - 7.2 definite integral1 Mark
MCQ
For each positive integer $n$, defined $f_n(x)=$ minimum $\left(\frac{x^n}{n !}, \frac{(1-x)^n}{n !}\right)$, for $0 \leq x \leq 1 .$ Let $I_n=\int \limits_0^1 f_n(x) d x, n \geq 1$. Then, $\sum \limits_{n=1}^{\infty} I_n$ is equal to
  • ✓
    $2 \sqrt{e}-3$
  • B
    $2 \sqrt{e}-2$
  • C
    $2 \sqrt{e}-1$
  • D
    $2 \sqrt{e}$
✓

Answer

Correct option: A.
$2 \sqrt{e}-3$
a
(a)

We have,

$f_n(x) =\operatorname{minimum}\left(\frac{x^n}{n !}, \frac{(1-x)^n}{n !}\right), x \in[0,1]$

$I_n =\int \limits_0^1 f_n(x) d x$

$I_n =\left|\int \limits_0^{1 / 2} \frac{x^n}{n !} d x\right|+\left|\int_{1 / 2}^1 \frac{(1-x)^n}{n !} d x\right|$

$I_n =\left|\left[\frac{x^{n+1}}{(n+1) n !}\right]_0^{1 / 2}\right|+\mid\left[\left.\frac{(1-x)^{n+1}}{(n+1) n !}\right|_{1 / 2} ^1 \mid\right.$

$I_n =\frac{2}{(n+1) !}\left[\left(\frac{1}{2}\right)^{n+1}\right]$

$\sum \limits_{n=1}^{\infty} I_n =2\left[\frac{1}{2 !}\left(\frac{1}{2}\right)^2+\frac{1}{3 !}\left(\frac{1}{2}\right)^3\right]$

$\sum \limits_{n=1}^{\infty} I_n =2\left[e^{1 / 2}-\left(1+\frac{1}{2}\right)\right]$

$=2 \sqrt{e}-3$

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 - 7.2 definite integral→STD 12 - 7.2 definite integral — all question types→Mathematics STD 12 Science question papers→CBSE Board STD 12 Science question papers→CBSE Board question paper generator→

Similar questions

The shortest distance between the lines $\frac{\text{x}-3}{3}=\frac{\text{y}-8}{-1}=\frac{\text{z}-3}{1}$ and, $\frac{\text{x}+3}{-3}=\frac{\text{y}+7}{2}=\frac{\text{z}-6}{4}$ is:
→
Let the co-ordinates of one vertex of $\triangle ABC$ be $A (0,2, \alpha)$ and the other two vertices lie on the line $\frac{x+\alpha}{5}=\frac{y-1}{2}=\frac{z+4}{3}$. For $\alpha \in Z$, if the area of $\triangle ABC$ is $21$ sq. units and the line segment $BC$ has length $2 \sqrt{21}$ units, then $\alpha^2$ is equal to $...........$.
→
At present, a firm is manufacturing $2000$ items. It is estimated that the rate of change of production $P$ w.r.t. additional number of workers $x$ is given by $\frac{{dp}}{{dx}} = 100 - 12\sqrt x $ . If the firm employs $25 $ more workers, then the new level of production of itmes is 
→
The differential equation $\frac{\text{d}^2\text{y}}{\text{dx}^2}+\frac{\text{dy}}{\text{dx}}+\text{y}+\text{x}^2=0$ of the following type:
→
The solution of the differential equation $2\text{x}\frac{\text{dy}}{\text{dx}}-\text{y}=3$ resresents:
→
If the mean and variance of a binomial distribution are 4 and 3, respectively, the probability of getting exactly six successes in this distribution is:
→
The distance moved by a particle travelling in straight line in $t$ seconds is given by $s = 45t + 11t^2- t^3$. The time taken by the particle to come to rest is :
→
Differential coefficient of $\sec(\tan^{-1}\text{x})$ is:
→
The value of $\theta$ lying between $- \,\frac{\pi }{4}$ and $\frac{\pi }{2}$ and $0 \le A \le \frac{\pi }{2}$ and satisfying the equation $\left| {\begin{array}{*{20}{c}}{1\, + \,{{\sin }^2}A}&{{{\cos }^2}A}&{2\,\sin \,4\theta }\\{{{\sin }^2}A}&{1\, + \,{{\cos }^2}A}&{2\,\sin \,4\theta }\\{{{\sin }^2}A}&{{{\cos }^2}A}&{1\, + \,2\,\sin \,4\theta }\end{array}} \right|$ $= 0$ are :
→
The value of $\frac{d}{{d(\ln x)}}({e^x}{\ln ^2}x)$ at $x=e$, is
→