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Thread: Inequalities-related maths problems

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    Exclamation 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 + 1
    x2 + 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
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    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.
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    Re: Inequalities-related maths problems

    Quote Originally Posted by Cactus View Post
    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.
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