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$f(a)=a$.

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