1. ## Inequalities-related maths problems

Hi everyone, I found some inequalities problems that I have no idea how to solve. There aren't any answers online or any working, so I've got no idea how to approach these. All of these are multi-choice and I'll list the possible answers below (except from the first one). It would be great if someone could jot down a bit of working out to some of these questions!
The expression -1 <
 1-x2 1+x2
< 1 is satisfied by what values of x?
I'm 99% sure that the answer is "all values of x satisfy this expression" just from plugging in a bunch of numbers, but I can't prove it, so any working out/explanation would be amazing.

The inequality:

 x2- 2x + 1x2 + x + 1 > 0

is equivalent to
a) x>1
b) x is not equal to 1
c) x<0
d) x>0
e) x<1

Thanks again

2. ## Re: Inequalities-related maths problems

$-1<\dfrac{1-x^2}{1+x^2}<1 \implies -1 < \dfrac{1-x^2}{1+x^2} \text{ and } \dfrac{1-x^2}{1+x^2}<1$

Working each inequality separately ...

$\dfrac{1-x^2}{1+x^2}<1 \implies \dfrac{1-x^2}{1+x^2}-\dfrac{1+x^2}{1+x^2}<0 \implies \dfrac{-2x^2}{1+x^2}<0$

the last inequality is true for all $x \in \mathbb{R}$ except $x=0$

$-1 < \dfrac{1-x^2}{1+x^2} \implies 0 < \dfrac{1+x^2}{1+x^2}+\dfrac{1-x^2}{1+x^2} \implies 0<\dfrac{2}{1+x^2}$

the last inequality is true for all $x \in \mathbb{R}$

The intersection of the two separate solution sets is the solution set for the original inequality, namely $(-\infty,0) \cup (0,\infty)$

Note the rational expression's range in the attached graph ...

For the multiple choice question, note the denominator is always positive (why?) ... look for the values of x where the numerator is always positive.

3. ## Re: Inequalities-related maths problems

Originally Posted by Cactus
Hi everyone, I found some inequalities problems that I have no idea how to solve. There aren't any answers online or any working, so I've got no idea how to approach these. All of these are multi-choice and I'll list the possible answers below (except from the first one). It would be great if someone could jot down a bit of working out to some of these questions! $-1<\dfrac{1-x^2}{1+x^2}<1$
Here is another solution. Because $(\forall x\in\mathbb{R}\setminus\{0\})[x^2+1>1>0]$ there is no danger is multiplying getting:
\begin{align*}-1-x^2 &<1-x^2<1+x^2\\-1<1&<1+2x^2\\-2<0&<2x^2 \end{align*} So the solution set is all non-zero real numbers.