The value of the greatest integer k satisfyin | vedclass

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

Question

(B) Let $m = n+4$. Since $n \in \mathbb{N}$,$n \geq 1$,so $m \geq 5$.The given inequality becomes $2^m + 12 \geq km$,which implies $k \leq \frac{2^m +

Vedclass · Accepted Answer

$8$

Vedclass · Answer

(B) Let $m = n+4$. Since $n \in \mathbb{N}$,$n \geq 1$,so $m \geq 5$.The given inequality becomes $2^m + 12 \geq km$,which implies $k \leq \frac{2^m + 12}{m}$ for all $m \geq 5$.To find the greatest integer $k$,we need to find the minimum value of the function $f(m) = \frac{2^m + 12}{m}$ for $m \in \{5, 6, 7, \dots\}$.For $m = 5$: $f(5) = \frac{2^5 + 12}{5} = \frac{32 + 12}{5} = \frac{44}{5} = 8.8$.For $m = 6$: $f(6) = \frac{2^6 + 12}{6} = \frac{64 + 12}{6} = \frac{76}{6} \approx 12.66$.For $m = 7$: $f(7) = \frac{2^7 + 12}{7} = \frac{128 + 12}{7} = \frac{140}{7} = 20$.As $m$ increases,$f(m)$ increases for $m \geq 5$.Thus,the minimum value is $8.8$ at $m = 5$.Since $k \leq f(m)$ for all $m$,$k$ must be less than or equal to the minimum value of $f(m)$.Therefore,$k \leq 8.8$.The greatest integer $k$ is $8$.

The value of the greatest integer $k$ satisfying the inequation $2^{n+4} + 12 \geq k(n+4)$ for all $n \in \mathbb{N}$ is

Similar Questions

Find the sum of all integral values of $a$ such that the equation $||x - 2| - |3 - x|| = 2 - a$ has a solution.

The number of elements in the set $S = \{x \in \mathbb{Z} : x^2 - 7x + 6 \leq 0 \text{ and } x^2 - 3x > 0\}$ is

The solution set contained in $R$ of the inequation $3^x+3^{1-x}-4 < 0$ is:

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

The number of points $P(x, y)$ with natural numbers as coordinates that lie inside the quadrilateral formed by the lines $2x + y = 2$,$x = 0$,$y = 0$,and $x + y = 5$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module