学数答题160903-函数方程
题160907(14分)$f\left( x \right)={{x}^{2}}+px+q$,若$f\left( f\left( x \right) \right)=0$只有一个实数根,求证:$p,q\ge 0$.
题目来源:2011年北大保送生
解:$f\left( f\left( x \right) \right)=0$,即$f{{\left( x \right)}^{2}}+pf\left( x \right)+q=0$,
$\Delta ={{p}^{2}}-4q\ge 0$,
(1)若$\Delta ={{p}^{2}}-4q>0$,设$f\left( x \right)={{x}_{1}}$,$f\left( x \right)={{x}_{2}}$,${{x}_{1}}\ne {{x}_{2}}$,
即${{x}^{2}}+px+q-{{x}_{1}}=0$或${{x}^{2}}+px+q-{{x}_{2}}=0$.
依题意,两方程中仅有一解,
不妨设${{\Delta }_{1}}={{p}^{2}}-4\left( q-{{x}_{1}} \right)<0$,${{\Delta }_{1}}={{p}^{2}}-4\left( q-{{x}_{2}} \right)=0$,
故${{x}_{1}}<q-\dfrac{{{p}^{2}}}{4}<0$,${{x}_{2}}<q-\dfrac{{{p}^{2}}}{4}<0$.
再由韦达定理知$q={{x}_{1}}\cdot {{x}_{2}}>0$,$p=-\left( {{x}_{1}}+{{x}_{2}} \right)<0$.
(2)若$\Delta ={{p}^{2}}-4q=0$,设$f\left( x \right)={{x}_{0}}$,
即${{x}^{2}}+px+q-{{x}_{0}}=0$,
依题意${{\Delta }_{3}}={{p}^{2}}-4\left( q-{{x}_{0}} \right)=0$,所以${{x}_{0}}=0$.
所以$q={{x}_{0}}\cdot {{x}_{0}}=0$,$p=-\left( {{x}_{0}}+{{x}_{0}} \right)=0$.
综上,$p\ge 0$,$q\ge 0$.
法2:$f\left( f\left( x \right) \right)=0$,即$f{{\left( x \right)}^{2}}+pf\left( x \right)+q=0$,
由求根公式知$f\left( x \right)=\dfrac{-p\pm \sqrt{{{p}^{2}}-4q}}{2}$,
$f\left( f\left( x \right) \right)=0$有且仅有一个实数解,且$f\left( x \right)\in \left[ q-\dfrac{{{p}^{2}}}{4},+\infty \right)$,
所以$f\left( x \right)=\dfrac{-p+\sqrt{{{p}^{2}}-4q}}{2}$($f\left( x \right)=\dfrac{-p-\sqrt{{{p}^{2}}-4q}}{2}$舍去),
且$f\left( x \right)={{x}^{2}}+px+q=\dfrac{-p+\sqrt{{{p}^{2}}-4q}}{2}$有且仅有一解,
即${{x}^{2}}+px+q-\dfrac{-p+\sqrt{{{p}^{2}}-4q}}{2}=0$有且仅有一解,
\[\Delta ={{p}^{2}}-4\left( q-\dfrac{-p+\sqrt{{{p}^{2}}-4q}}{2} \right)=0\].
所以${{p}^{2}}-4q-2p+2\sqrt{{{p}^{2}}-4q}=0$,
可得$p=\dfrac{1}{2}\left( {{p}^{2}}-4q \right)+\sqrt{{{p}^{2}}-4q}=\dfrac{1}{2}{{\left( \sqrt{{{p}^{2}}-4q}+1 \right)}^{2}}-\dfrac{1}{2}\ge 0$,
即$p\ge 0$,
又$p=\dfrac{1}{2}\left( {{p}^{2}}-4q \right)+\sqrt{{{p}^{2}}-4q}\ge \sqrt{{{p}^{2}}-4q}$,
从而${{p}^{2}}\ge {{p}^{2}}-4q$,即$q\ge 0$.