For a positive integer n,if 10^{n-2} 91n,th | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) We test values for $n$ to check the inequality $10^{n-2} > 91n$: For $n=1$: $10^{-1} = 0.1$,$91(1) = 91$. $0.1 > 91$ is false. For $n=2$: $10^{0}

Vedclass · Accepted Answer

$\left\{ 5, 6, 7, 8, \dots \right\}$

Vedclass · Answer

(C) We test values for $n$ to check the inequality $10^{n-2} > 91n$: For $n=1$: $10^{-1} = 0.1$,$91(1) = 91$. $0.1 > 91$ is false. For $n=2$: $10^{0} = 1$,$91(2) = 182$. $1 > 182$ is false. For $n=3$: $10^{1} = 10$,$91(3) = 273$. $10 > 273$ is false. For $n=4$: $10^{2} = 100$,$91(4) = 364$. $100 > 364$ is false. For $n=5$: $10^{3} = 1000$,$91(5) = 455$. $1000 > 455$ is true. For $n=6$: $10^{4} = 10000$,$91(6) = 546$. $10000 > 546$ is true. Since the exponential function $10^{n-2}$ grows much faster than the linear function $91n$,the inequality will hold for all $n \ge 5$. Thus,the set of values is $\left\{ 5, 6, 7, 8, \dots \right\}$.

For a positive integer $n$,if $10^{n-2} > 91n$,then the complete set of values of $n$ is:

Similar Questions

Find the solution set of the inequality: $4x + 3 \geq 2x + 17$ and $3x - 5 < -2$.

$\left\{x \in R : \frac{\sqrt{6+x-x^2}}{2x+5} \geq \frac{\sqrt{6+x-x^2}}{x+4}\right\}=$

If $\frac{2x}{2x^2 + 5x + 2} > \frac{1}{x + 1}$,then

The number of elements in the set $\{n \in \mathbb{Z} : |n^2 - 10n + 19| < 6\}$ is $...........$

In a manufacturing company,the cost and revenue functions of a product are given by $C(x) = 500 + \frac{5}{2}x$ and $R(x) = 3x$,where $x$ is the number of items produced and sold. How many items must be sold to realize no profit or loss? Is there any profit?

Vedclass Test Series

Exam Paper Generator

Online Exam Module