每日一题[854]三个角度看问题

已知$f\left( x \right)$为一元二次函数，且$a,f\left( a \right),f\left( {f\left( a \right)} \right),f\left( {f\left( {f\left( a \right)} \right)} \right)$为正项等比数列，求证：$f\left( a \right) = a$．

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分析与解　证法一　设$f\left( x \right) = m{x^2} + nx + t\left( {m \ne 0} \right)$，数列$a,f\left( a \right),f\left( {f\left( a \right)} \right),f\left( {f\left( {f\left( a \right)} \right)} \right)$的公比为$q\left( {q > 0} \right)$，则

$$\begin{split} f\left( a \right) = &aq,\\f\left( {f\left( a \right)} \right) =& f\left( {aq} \right) = a{q^2},\\f\left( {f\left( {f\left( a \right)} \right)} \right) =&f\left( {a{q^2}} \right) = a{q^3},\end{split}$$ 所以 $$\begin{cases} m{a^2} + na + t = aq\cdots (1) \\ m{(aq)^2} + naq + t = a{q^2}\cdots (2) \\ m{(a{q^2})^2} + na{q^2} + t = a{q^3}\cdots(3)\end{cases}$$
(1)$-$(2)得 $$m{a^2}\left( {1 - {q^2}} \right) + na\left( {1 - q} \right) = aq\left( {1 - q} \right),$$ (2)$-$(3)得 $$m{a^2}{q^2}\left( {1 - {q^2}} \right) + naq\left( {1 - q} \right) = a{q^2}\left( {1 - q} \right).$$
若$q = 1$，则$f\left( a \right) = a$；

若$q \ne 1$，则$m{a^2}\left( {1 + q} \right) + na = aq$与$m{a^2}q\left( {1 + q} \right) + na = aq$矛盾．所以$f\left( a \right) = a$．

证法二　由$a,f\left( a \right),f\left( {f\left( a \right)} \right),f\left( {f\left( {f\left( a \right)} \right)} \right)$成等比数列得

$$\dfrac{{f\left( a \right)}}{a} = \dfrac{{f\left( {f\left( a \right)} \right)}}{{f\left( a \right)}} = \dfrac{{f\left( {f\left( {f\left( a \right)} \right)} \right)}}{{f\left( {f\left( a \right)} \right)}},$$
下面用反证法，若$f(a)\ne a$，则有 $$\dfrac{{f\left( {f\left( a \right)} \right) - f\left( a \right)}}{{f\left( a \right) - a}} = \dfrac{{f\left( {f\left( {f\left( a \right)} \right)} \right) - f\left( {f\left( a \right)} \right)}}{{f\left( {f\left( a \right)} \right) - f\left( a \right)}}.$$
所以，三点$A\left( {a,f\left( a \right)} \right)$，$B\left( {f\left( a \right),f\left( {f\left( a \right)} \right)} \right)$，$C\left( {f\left( {f\left( a \right)} \right),f\left( {f\left( {f\left( a \right)} \right)} \right)} \right)$满足 $${k_{AB}} = {k_{BC}},$$ 所以$A,B,C$三点共线，与$A,B,C$三点在抛物线上矛盾，所以$f\left( a \right) = a$．

证法三　设数列$a,f\left( a \right),f\left( {f\left( a \right)} \right),f\left( {f\left( {f\left( a \right)} \right)} \right)$的公比为$q\left( {q > 0} \right)$，考虑方程$f(x)=qx$，则$a,qa,q^2a$都是它的解．

因为方程$f(x)=qx$为一元二次方程，所以$a,qa,q^2a$中至少有两个数相等，又因为$q>0$，所以只可能有$q=1$，从而$f(a)=a$．