Maths Olympiad Prep

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Algebra Difficulty 5.2 AIME, harder Find the answer

Example 4 Given that aa is an integer, the equation concerning xx
x2x2+14xx2+1+2a=0 \frac{x^{2}}{x^{2}+1}-\frac{4|x|}{\sqrt{x^{2}+1}}+2-a=0

has real roots. Then the possible values of aa are \qquad
(2008, I Love Mathematics Junior High School Summer Camp Mathematics Competition)

A number or a short expression. Spacing and $ signs are ignored.

Solution

【Analysis】Take any real root x0x_{0} of the equation. Then
x02x02+14x0x02+1+2=a \frac{x_{0}^{2}}{x_{0}^{2}+1}-\frac{4\left|x_{0}\right|}{\sqrt{x_{0}^{2}+1}}+2=a \text {. }

Notice that x02x02+1=(x0x02+1)2\frac{x_{0}^{2}}{x_{0}^{2}+1}=\left(\frac{\left|x_{0}\right|}{\sqrt{x_{0}^{2}+1}}\right)^{2}.
Let x0x02+1=t\frac{\left|x_{0}\right|}{\sqrt{x_{0}^{2}+1}}=t.
By x02+1>x02=x0\sqrt{x_{0}^{2}+1}>\sqrt{x_{0}^{2}}=\left|x_{0}\right|
x0x02+1<10t<1\Rightarrow \frac{\left|x_{0}\right|}{\sqrt{x_{0}^{2}+1}}<1 \Rightarrow 0 \leqslant t<1
a=t24t+2=(t2)22\Rightarrow a=t^{2}-4 t+2=(t-2)^{2}-2.
Also 0t<11<(t2)2220 \leqslant t<1 \Rightarrow-1<(t-2)^{2}-2 \leqslant 2
1<a2a\Rightarrow-1<a \leqslant 2 \Rightarrow a's possible values are 0,1,20, 1, 2.

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.