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If $f$ is an even function defined on the interval $(-5,5)$ then four real values of $x$ satisfying the equation $f(x)=f(\frac{x+1}{x+2})$ are?

I thought that $(x+1)/(x+2)=-x$.But I'm getting only two values by solving this.How do I get the other two values?

And if i put $(x+1)/(x+2)=x$ then the answer is not matching.I dont know why.

1 Answers1

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Don't forget the possibility $(x+1)/(x+2)=+x$.

Setting $\frac{x+1}{x+2}=-x$ leads to $x+1=-x^2-2x$, i.e., $x^2+3x+1=0$, $x=\frac{-3\pm\sqrt{5}}2$. Setting $\frac{x+1}{x+2}=+x$ leads to $x+1=x^2+2x$, i.e., $x^2+x-1=0$, $x=\frac{-1\pm\sqrt{5}}2$. All four values are well within $(-5,5)$.

Hagen von Eitzen
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  • But that is giving the wrong answer.I dont know why.In my book the other condition they used is $f(x)=f(-x)=f((-x+1)/(-x+2))$ and so $x=(-x+1)/(-x+2)$.I'm a bit confused. –  Apr 14 '16 at 17:13
  • @SanchayanDutta Well, that would lead to $-x^2+2x=-x+1$, i.e., $x^2-3x+1=0$, $x=\frac{3\pm\sqrt{5}}{2}$. – Hagen von Eitzen Apr 14 '16 at 17:16
  • No.See https://www.wolframalpha.com/input/?i=solve+(-x%2B1)%2F(-x%2B2)%3Dx –  Apr 14 '16 at 17:17
  • You made a calculation error. $x^2-2x=x-1$ gives $x^2-3x+1=0$ –  Apr 14 '16 at 17:20
  • Nevertheless, $x=\frac{-x+1}{-x+2}$ does not entail $f(x)=f(\frac{x+1}{x+2})$. – Hagen von Eitzen Apr 14 '16 at 17:22
  • Well i dont know why they solved it using that in my book.BTW the question is from IITJEE 1996.Well i'll overlook it for the time being. –  Apr 14 '16 at 17:25