题160907(14分)$fleft( x ight)={{x}^{2}}+px+q$,若$fleft( fleft( x ight) ight)=0$只有一个实数根,求证:$p,qge 0$.
题目来源:2011年北大保送生
解:$fleft( fleft( x ight) ight)=0$,即$f{{left( x ight)}^{2}}+pfleft( x ight)+q=0$,
$Delta ={{p}^{2}}-4qge 0$,
(1)若$Delta ={{p}^{2}}-4q>0$,设$fleft( x ight)={{x}_{1}}$,$fleft( x ight)={{x}_{2}}$,${{x}_{1}} e {{x}_{2}}$,
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${{Delta }_{1}}={{p}^{2}}-4left( q-{{x}_{1}} ight)<0$,${{Delta }_{1}}={{p}^{2}}-4left( q-{{x}_{2}} ight)=0$,
故${{x}_{1}}<q-dfrac{{{p}^{2}}}{4}<0$,${{x}_{2}}<q-dfrac{{{p}^{2}}}{4}<0$.
再由韦达定理知$q={{x}_{1}}cdot {{x}_{2}}>0$,$p=-left( {{x}_{1}}+{{x}_{2}} ight)<0$.
(2)若$Delta ={{p}^{2}}-4q=0$,设$fleft( x ight)={{x}_{0}}$,
即${{x}^{2}}+px+q-{{x}_{0}}=0$,
依题意${{Delta }_{3}}={{p}^{2}}-4left( q-{{x}_{0}} ight)=0$,所以${{x}_{0}}=0$.
所以$q={{x}_{0}}cdot {{x}_{0}}=0$,$p=-left( {{x}_{0}}+{{x}_{0}} ight)=0$.
综上,$pge 0$,$qge 0$.
法2:$fleft( fleft( x ight) ight)=0$,即$f{{left( x ight)}^{2}}+pfleft( x ight)+q=0$,
由求根公式知$fleft( x ight)=dfrac{-ppm sqrt{{{p}^{2}}-4q}}{2}$,
$fleft( fleft( x ight) ight)=0$有且仅有一个实数解,且$fleft( x ight)in left[ q-dfrac{{{p}^{2}}}{4},+infty ight)$,
所以$fleft( x ight)=dfrac{-p+sqrt{{{p}^{2}}-4q}}{2}$($fleft( x ight)=dfrac{-p-sqrt{{{p}^{2}}-4q}}{2}$舍去),
且$fleft( x ight)={{x}^{2}}+px+q=dfrac{-p+sqrt{{{p}^{2}}-4q}}{2}$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}}-4left( q-dfrac{-p+sqrt{{{p}^{2}}-4q}}{2} ight)=0].
所以${{p}^{2}}-4q-2p+2sqrt{{{p}^{2}}-4q}=0$,
可得$p=dfrac{1}{2}left( {{p}^{2}}-4q ight)+sqrt{{{p}^{2}}-4q}=dfrac{1}{2}{{left( sqrt{{{p}^{2}}-4q}+1 ight)}^{2}}-dfrac{1}{2}ge 0$,
即$pge 0$,
又$p=dfrac{1}{2}left( {{p}^{2}}-4q ight)+sqrt{{{p}^{2}}-4q}ge sqrt{{{p}^{2}}-4q}$,
从而${{p}^{2}}ge {{p}^{2}}-4q$,即$qge 0$.