The integral _ 0 ^ 1 1 _ 7 ^ [1 x ] d x= where [.] denotes the gr… - Vidyadip

MCQ

The integral $\int_{0}^{1} \frac{1}{{ }_{7}^{\left[\frac{1}{x}\right]}} d x=$ where [.] denotes the greatest integer function is equal to

✓
$1+6 \log _{e}\left(\frac{6}{7}\right)$
B
$1-6 \log _{e}\left(\frac{6}{7}\right)$
C
$\log _{e}\left(\frac{7}{6}\right)$
D
$1-7 \log _{ e }\left(\frac{6}{7}\right)$

✓

Answer

Correct option: A.

$1+6 \log _{e}\left(\frac{6}{7}\right)$

a
$\int\limits_{0}^{1} \frac{1}{7\left[\frac{1}{x}\right]} dx =-\int\limits_{1}^{0} \frac{1}{\left.7^{\left[\frac{1}{x}\right.}\right]} dx$

$=(-1)\left[\int\limits_{1}^{1 / 2} \frac{1}{7} dx +\int\limits_{1 / 2}^{1 / 3} \frac{1}{7^{2}} dx +\int\limits_{1 / 3}^{1 / 4} \frac{1}{7^{3}} dx +\ldots \ldots \infty\right]$

$=\left(\frac{1}{7}+\frac{1}{2 \cdot 7^{2}}+\frac{1}{3 \cdot 7^{3}}+\ldots \infty\right)-\left(\frac{1}{7 \cdot 2}+\frac{1}{7^{2} \cdot 3}+\frac{1}{7^{2} \cdot 4} \ldots \infty\right)$

$=-\ln \left(1-\frac{1}{7}\right)-7\left(\frac{1}{7^{2} \cdot 2}+\frac{1}{7^{3} \cdot 3}+\frac{1}{7^{4} \cdot 4}+\ldots . \ldots\right)$

$= {\left[\ln \frac{6}{7}-7\left(-\ln \left(1-\frac{1}{7}\right)-\frac{1}{7}\right)\right.}$

$=6 \ln \frac{6}{7}+1$

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 7.2 definite integral→7.2 definite integral — all question types→Maths STD 12 Science question papers→Gujarat Board STD 12 Science question papers→Gujarat Board question paper generator→

Similar questions

The area bounded by y –1 = |x|, y = 0 and |x| $\frac{1}{2}$ will be:

$\frac{3}{4}$
$\frac{3}{2}$
$\frac{5}{4}$
$\text{None of these}$

$\tan\Big(\frac{\pi}{4}+\frac{1}{2}\cos^{-1}\text{x}\Big)+\tan\Big(\frac{\pi}{4}-\frac{1}{2}\cos^{-1}\text{x}\Big)=$

$\text{x}$
$\frac{1}{\text{x}}$
$2\text{x}$
$\frac{2}{\text{x}}$

The total surface area $S$ can be expressed in terms of $V$ and $r$ as

Let A = {1, 2, 3}. Then, the number of equivalence relations containing (1, 2) is:

1
2
3
4

$\int\limits^\frac{\pi}{2}_0\frac{1}{2+\cos\text{x}}\text{ dx}$ equals:

$\frac{1}{3}\tan^{-1}\Big(\frac{1}{\sqrt{3}}\Big)$
$\frac{2}{\sqrt{3}}\tan^{-1}\Big(\frac{1}{\sqrt{3}}\Big)$
${\sqrt{3 }}\tan^{-1}\big({\sqrt{3}}\big)$
$2{\sqrt{3 }}\tan^{-1}{\sqrt{3}}$

The sum of two forces is $18\, N$ and resultant whose direction is at right angles to the smaller force is $12\,N.$ The magnitude of the two forces are

The function $\text{f(x)}=\frac{\text{x}^3+\text{x}^2-16\text{x}+20}{\text{x}-2}$ is not defind for x = 2. in order to make f(x) continuous at x = 2, here f(2) should be defined as:

0
1
2
3

He area bounded by y = x², x = y²is:

$1$
$\frac{1}{6}$
$\frac{3}{4}$
$\text{None}\text{ of}\text{ these}$

Assume that in a family, each chold is equally likely to be a boy or a girl. A family with tree cgildren is chosen at random. Tere probability that the eldest child is a girl given that the family has at least oe girl.

$\frac{1}{2}$
$\frac{1}{3}$
$\frac{2}{3}$
$\frac{4}{7}$

Two points are randomly chosen on the circumference of a circle of radius $r$. The probability that the distance between the two points is at least $r$ is equal to