(1.1)^ 10000 is .......... 1000: - Vidyadip

MCQ

$(1.1)^{10000}$ is $.......... 1000:$

✓
greater than
B
less than
C
equal
D
None of these

✓

Answer

Correct option: A.

greater than

Given, $ (1.1)^{10000 }= (1 + 0.1)^{10000}$
$ 10000\text{C}_0 + 10000\text{C}_1 \times (0.1) + 10000\text{C}_2 \times (0.1)^2 +$ other $+ve$ terms
$= 1 + 10000 \times (0.1) +$ other $+ ve$ terms
$= 1 + 1000 +$ other $+ ve$ terms
$> 1000$
So, $(1.1)^{10000}$ is greater than $1000$

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 Limits and Derivatives→Limits and Derivatives — all question types→Maths STD 11 Science question papers→CBSE Board STD 11 Science question papers→CBSE Board question paper generator→

Similar questions

Let $P=\{\theta: \sin \theta-\cos \theta=\sqrt{2} \cos \theta\}$ and $Q=\{\theta: \sin \theta+\cos \theta=\sqrt{2} \sin \theta\}$ be two sets. Then

The product of $r$ consecutive positive integers is divisible by:

What is the approximate radius of the circle whose equation is $\text{(x}-\sqrt{3)}^2+(\text{y}+2)^2=11?$

Middle point of the chord of the circle ${x^2} + {y^2} = 25$ intercepted on the line $x - 2y = 2$ is

Given $a_1,a_2,a_3.....$ form an increasing geometric progression with common ratio $r$ such that $log_8a_1 + log_8a_2 +.....+ log_8a_{12} = 2014,$ then the number of ordered pairs of integers $(a_1, r)$ is equal to

If $\alpha $ is the interior angle of a regular octagon, then $\mathop {\lim }\limits_{\theta \to {\alpha ^ + }} \frac{{\tan \theta - 1}}{{\left[ {\sin \theta + \cos \theta } \right]}}$ is equal to (Note : $[k]$ denotes greatest integer less than or equal to $k$ )

The mean age of a combined group of men and women is $30$ years. If the means of the age of men and women are respectively $32$ and $27$, then the percentage of women in the group is

If $f(x)=x \sin x$, then $f^{\prime}\left(\frac{\pi}{2}\right)$ is equal to

The value of $\cos \left(-660^{\circ}\right)$ is :

The eccentricity of the hyperbola $\frac{{{x^2}}}{{16}} - \frac{{{y^2}}}{{25}} = 1$ is