# Recent questions tagged inequality 1
Let $\begin{array}{} V_1 & = & \frac{7^2+8^2+15^2+23^2}{4} – \left( \frac{7+8+15+23}{4} \right) ^2, \\ V_2 & = & \frac{6^2+8^2+15^2+24^2}{4} – \left( \frac{6+8+15+24}{4} \right) ^2 , \\ V_3 & = & \frac{5^2+8^2+15^2+25^2}{4} – \left( \frac{5+8+15+25}{4} \right) ^2 . \end{array}$ Then $V_3<V_2<V_1$ $V_3<V_1<V_2$ $V_1<V_2<V_3$ $V_2<V_3<V_1$
2
Let $n$ be a positive real number and $p$ be a positive integer. Which of the following inequalities is true? $n^p > \frac{(n+1)^{p+1} – n^{p+1}}{p+1}$ $n^p < \frac{(n+1)^{p+1} – n^{p+1}}{p+1}$ $(n+1)^p < \frac{(n+1)^{p+1} – n^{p+1}}{p+1}$ none of the above
3
The smallest positive number $K$ for which the inequality $\mid \sin ^2 x – \sin ^2 y \mid \leq K \mid x-y \mid$ holds for all $x$ and $y$ is $2$ $1$ $\frac{\pi}{2}$ there is no smallest positive value of $K$; any $K>0$ will make the inequality hold.
1 vote
4
Let $S=\{6,10,7,13,5,12,8,11,9\},$ and $a=\sum_{x\in S}(x-9)^{2}\:\&\: b=\sum_{x\in S}(x-10)^{2}.$ Then $a<b$ $a>b$ $a=b$ None of these
5
For natural numbers $n,$ the inequality $2^{n}>2n+1$ is valid when $n\geq 3$ $n<3$ $n=3$ None of these
1 vote
6
Let $0.01^x+0.25^x=0.7$ . Then $x\geq1$ $0\lt x\lt1$ $x\leq0$ no such real number $x$ is possible.
7
Let $n \geq 3$ be an integer. Then the statement $(n!)^{1/n} \leq \dfrac{n+1}{2}$ is true for every $n \geq 3$ true if and only if $n \geq 5$ not true for $n \geq 10$ true for even integers $n \geq 6$, not true for odd $n \geq 5$
1 vote
8
Let $A=\{1, 2, 3, 4, 5, 6, 7, 8 \}$. How many functions $f: A \rightarrow A$ can be defined such that $f(1)< f(2) < f(3)$? $\begin{pmatrix} 8 \\ 3 \end{pmatrix}$ $\begin{pmatrix} 8 \\ 3 \end{pmatrix} 5^8$ $\begin{pmatrix} 8 \\ 3 \end{pmatrix} 8^5$ $\frac{8!}{3!}$
1 vote
The inequality $\frac{2-gx+x^{2}}{1-x+x^{2}}\leq 3$ is true for all the value of $x$ if and only if $1\leq g\leq 7$ $-1\leq g\leq 1$ $-6\leq g\leq 7$ $-1\leq g\leq 7$
I find alpha <x+y which gives me x+y<2. But the answer is A. Can someone please help. Consider the statement$:$ $x(\alpha-x)<y(\alpha-y)$ for all $x,y$ with $0<x<y<1.$ The statement is true if and only if $\alpha\geq 2$ if and only if $\alpha >2$ if and only if $\alpha <-1$ for no values of $\alpha$
If $a, b,$ and $c$ are constants, which of the following is a linear inequality? $ax+bcy=0$ $ax^{2}+cy^{2}=21$ $abx+a^{2}y \geq 15$ $xy+ax \geq 20$