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原文:中值定理"下嫁"高考
拉格朗日中值定理:若函数\(f\)满足如下条件:
(i)\(f\)在闭区间\([ a,b]\)上连续;
(ii)\(f\)在开区间\((a,b)\)内可导;
则在\((a,b)\)内至少存在一点\(\xi\),使得\(f'(\xi) = \frac{f(b) - f(a)}{b - a}\).
一,证明\(\frac{f(x)}{x} > a\)或\(\frac{f(x)}{x} < a\)成立(其中\(x > 0\))
例:(2007年高考全国卷I第20题)
设函数\(f(x) = e^{x} - e^{- x}\).
(Ⅰ)证明:\(f(x)\)的导数\(f'(x) \geq 2\);
(Ⅱ)证明:若对所有\(x \geq 0\),都有\(f(x) \geq ax\),则\(a\)的取值范围是\(( - \infty,2]\).
(Ⅰ)略.
(Ⅱ)证明:(i)当\(x = 0\)时,对任意的\(a\),都有\(f(x) \geq ax\)
(ii)当\(x > 0\)时,问题即转化为\(a \leq \frac{e^{x} - e^{- x}}{x}\)对所有\(x > 0\)恒成立.
令\(G(x) = \frac{e^{x} - e^{- x}}{x} = \frac{f(x) - f(0)}{x - 0}\),由拉格朗日中值定理知\((0,x)\)内至少存在一点\(\xi\)(从而\(\xi > 0\)),使得\(f'(\xi) = \frac{f(x) - f(0)}{x - 0}\),即\(G(x) = f'(\xi) = e^{\xi} + e^{- \xi}\),由于\(f''(\xi) = e^{\xi} - e^{- \xi}>0(\xi > 0)\),故\(f'(\xi)\)在\((0,x)\)上是增函数,让\(x \rightarrow 0\)得\(G(x)_{\min}=f'(\xi)=e^{\xi}+e^{-\xi}≥f'(0)=2\),所以\(a\)的取值范围是\(( - \infty,2]\).
评注:第(2)小题提供的参考答案用的是初等数学的方法.即令\(g(x) = f(x) - ax\),再分\(a \leq 2\)和\(a > 2\)两种情况讨论.其中,\(a > 2\)又要去解方程\(g'(x) = 0\).但这有两个缺点:首先,为什么\(a\)的取值范围要以2为分界展开.其次,方程\(g'(x) = 0\)求解较为麻烦.但用拉格朗日中值定理求解就可以避开讨论,省去麻烦.
二,证明\(g(a) + g(b) - 2g\left( \frac{a + b}{2} \right) < \lambda(b - a),(b > a)\)成立
例:(2004年四川卷第22题)
已知函数\(f(x) = \ln(1 + x) - x,g(x) = x\ln x\).
(Ⅰ)求函数\(f(x)\)的最大值;
(Ⅱ)设\(0 < a < b < 2a\),证明:\(g(a) + g(b) - 2g\left( \frac{a + b}{2} \right) < (b - a)\ln 2\).
(Ⅰ)略;
(Ⅱ)证明:依题意,有\(g'(x) = \ln x + 1\)
\(g(a) + g(b) - 2g\left( \frac{a + b}{2} \right) = g(b) - g\left( \frac{a + b}{2} \right) - \left( g\left( \frac{a + b}{2} \right) - g(a) \right)\)
由拉格朗日中值定理得,存在\(\lambda \in \left( a,\frac{a + b}{2} \right),\mu \in \left( \frac{a + b}{2},b \right)\),使得
\(g(b) - g\left( \frac{a + b}{2} \right) - \left( g\left( \frac{a + b}{2} \right) - g(a) \right) = \left( g'(\mu) - g'(\lambda) \right) \cdot \frac{b - a}{2} = \left( \ln\mu - \ln\lambda \right) \cdot \frac{b - a}{2}\)
\(= \ln\frac{\mu}{\lambda} \cdot \frac{b - a}{2} < \ln\frac{b}{a} \cdot \frac{b - a}{2} < \ln\frac{4a}{a} \cdot \frac{b - a}{2} = (b - a)\ln 2\)
评注:对于不等式中含有\(g(a),g(b),g\left( \frac{a + b}{2} \right)(a < b)\)的形式,我们往往可以把\(g\left( \frac{a + b}{2} \right) - g(a)\)和\(g(b) - g\left( \frac{a + b}{2} \right)\),分别对\(g\left( \frac{a + b}{2} \right) - g(a)\)和\(g(b) - g\left( \frac{a + b}{2} \right)\)两次运用拉格朗日中值定理.
三,证明\(\left| f\left( x_{1} \right) - f\left( x_{2} \right) \right| > \left| \lambda\left( x_{1} - x_{2} \right) \right|\)成立
例: (2006年四川卷理第22题)
已知函数\(f(x) = x^{2} + \frac{2}{x} + a\ln x(x > 0)\),对任意两个不相等的正数$x_1,x_2$,证明:
(1)当\(a \leq 0\)时,\(\frac{f\left( x_{1} \right) + f\left( x_{2} \right)}{2} > f\left( \frac{x_{1} + x_{2}}{2} \right)\);
(2)当\(a \leq 4\)时,\(|f'(x_1)-f'(x_2)|>|x_1-x_2|\).
证明:(1)不妨设\(x_{1} < x_{2}\),即证\(f\left( x_{2} \right) - f\left( \frac{x_{1} + x_{2}}{2} \right) > f\left( \frac{x_{1} + x_{2}}{2} \right) - f\left( x_{1} \right)\)．由拉格朗日中值定理知,存在\(\xi_{1} \in \left( x_{1},\frac{x_{1} + x_{2}}{2} \right),\xi_{2} \in \left( \frac{x_{1} + x_{2}}{2},x_{2} \right)\),则\(\xi_{1} < \xi_{2}\)且
\(f\left( x_{2} \right) - f\left( \frac{x_{1} + x_{2}}{2} \right) = f'\left( \xi_{2} \right) \cdot \frac{x_{2} - x_{1}}{2}\),\(f\left( \frac{x_{1} + x_{2}}{2} \right) - f\left( x_{1} \right) = f'\left( \xi_{1} \right) \cdot \frac{x_{2} - x_{1}}{2}\).又\(f'(x) = 2x - \frac{2}{x^{2}} + \frac{a}{x}\),\(f''(x) = 2 + \frac{4}{x^{3}} - \frac{a}{x^{2}}\).当\(a \leq 0\)时,\(f''(x) \geq 0\).所以\(f'(x)\)是一个单调递增函数,故\(f'\left( \xi_{1} \right) < f'\left( \xi_{2} \right)\).从而\(f\left( x_{2} \right) - f\left( \frac{x_{1} + x_{2}}{2} \right) > f\left( \frac{x_{1} + x_{2}}{2} \right) - f\left( x_{1} \right)\)成立,因此命题获证．
(2)由\(f(x) = x^{2} + \frac{2}{x} + a\ln x\)得,\(f'(x) = 2x - \frac{2}{x^{2}} + \frac{a}{x}\),令\(g(x) = f'(x)\)则由拉格朗日中值定理得:$\left|g\left(x_{1}\right)-g\left(x_{2}\right)\right|=\left|g^{\prime}(\lambda)\left(x_{1}-x_{2}\right)\right|$
下面只要证明:当\(a \leq 4\)时,任意\(\lambda > 0\),都有\(g'(\lambda) > 1\),则有\(g'(x) = 2 + \frac{4}{x^{3}} - \frac{a}{x^{2}} > 1\),即证\(a \leq 4\)时,\(a < x^{2} + \frac{4}{x}\)恒成立.这等价于证明\(x^{2} + \frac{4}{x}\)的最小值大于4.
由于\(x^{2} + \frac{4}{x} = x^{2} + \frac{2}{x} + \frac{2}{x} \geq 3\sqrt[3]{4}\),当且仅当\(x = \sqrt[3]{2}\)时取到最小值,又\(a \leq 4 < 3\sqrt[3]{4}\),故\(a \leq 4\)时,\(2 + \frac{4}{x^{3}} - \frac{a}{x^{2}} > 1\)恒成立.
所以由拉格朗日定理得:$\left|g\left(x_{1}\right)-g\left(x_{2}\right)\right|=\left|g^{\prime}(\lambda)\left(x_{1}-x_{2}\right)\right|$.
评注:这道题用初等数学的方法证明较为冗长,而且技巧性较强.因而思路较为突兀,大多数考生往往难以想到.相比之下,用拉格朗日中值定理证明,思路较为自然,流畅.体现了高观点解题的优越性,说明了学习高等数学的重要性.
四,证明\(f\left( x_{1} \right) - f\left( x_{2} \right) > \lambda\left( x_{1} - x_{2} \right)\)或\(\left| f\left( x_{1} \right) - f\left( x_{2} \right) \right| > \left| \lambda\left( x_{1} - x_{2} \right) \right|\)成立
例:(2008年全国卷Ⅱ22题)设函数\(f(x) = \frac{\sin x}{2 + \cos x}\).
(Ⅰ)求\(f(x)\)的单调区间;
(Ⅱ)如果对任何\(x \geq 0\),都有\(f(x) \leq ax\),求\(a\)的取值范围.
(Ⅰ)略;
(Ⅱ)证明:当\(x = 0\)时,显然对任何\(a\),都有\(f(x) \leq ax\);当\(x > 0\)时,\(\frac{f(x)}{x} = \frac{f(x) - f(0)}{x - 0}\)
由拉格朗日中值定理知存在\(\xi \in (0,x)\),使得\(\frac{f(x)}{x} = \frac{f(x) - f(0)}{x - 0} = f'(\xi)\).由(Ⅰ)知\(f'(x) = \frac{2\cos x + 1}{\left( 2 + \cos x \right)^{2}}\),从而\(f''(x) = \frac{2\sin x\left( 2 + \cos x \right)\left( \cos x - 1 \right)}{\left( 2 + \cos x \right)^{2}}\).令\(f''(x) \geq 0\)得,\(x \in \left[ (2k + 1)\pi,(2k + 2)\pi \right]\);令\(f''(x) \leq 0\)得,\(x \in \left[ 2k\pi,(2k + 1)\pi \right]\).所以在\(\left[ (2k + 1)\pi,(2k + 2)\pi \right]\)上,\(f'(x)\)的最大值\(f'(x)_{\max}=f'\left( (2k + 2)\pi \right)=\frac13\);在\(\left[ 2k\pi,(2k + 1)\pi \right]\)上,\(f'(x)\)的最大值\(f'(x)_{\max}=f'(2k\pi)=\frac{1}{3}\).从而函数\(f'(x)\)在\(\left[ 2k\pi,(2k + 2)\pi \right]\)上的最大值是\(f'(x)_{\max}=\frac{1}{3}\).由\(k \in N\)知,当\(x > 0\)时,\(f'(x)\)的最大值为\(f'(x)_{\max}=\frac{1}{3}\).所以,\(f'(\xi)\)的最大值\(f'(\xi)_{\max}=\frac{1}{3}\).为了使\(f'(\xi) \leq a\)恒成立,应有\(f'(\xi)_{\max}≤a\).所以\(a\)的取值范围是\(\left[\frac{1}{3}, + \infty \right)\).
评注:这道题的参考答案的解法是令\(g(x) = ax - f(x)\),再去证明函数\(g(x)\)的最小值\(g(x)_{\min}\).这与上述的思路是一样的.但首先参考答案的解法中有个参数\(a\),要对参数\(a\)进行分类讨论;其次为了判断\(g(x)\)的单调性,还要求\(g'(x) \geq 0\)和\(g'(x) \leq 0\)的解,这个求解涉及到反余弦\(\arccos 3a\),较为复杂.而用拉格朗日中值定理就可以避开麻烦,省去讨论.再次体现了高观点解题的优越性.
五,证明\(f(x) > 0,(x > a)\)成立,(其中\(f(a) = 0\))
例:(2007年安徽卷18题)
设\(a \geq 0,f(x) = x - 1 - \ln^{2}x + 2a\ln x(x > 0)\).
(Ⅰ)令\(F(x) = xf'(x)\),讨论\(F(x)\)在\((0, + \infty)\)内的单调性并求极值;
(Ⅱ)求证:当\(x > 1\)时,恒有\(x > \ln^{2}x - 2a\ln x + 1\).
(Ⅰ)略;
(Ⅱ)证明:即证\(f(x) > 0\),由于\(x > 1\),则\(\frac{f(x)}{x - 1} = \frac{f(x) - f(1)}{x - 1}\).由拉格朗日中值定理得,存在\(\xi \in (1,x)\),使得\(\frac{f(x) - f(1)}{x - 1} = f'(\xi)\).由(Ⅰ)的解题过程知\(f'(x) = 1 - \frac{2}{x}\ln x + \frac{2a}{x}\),所以\(f''(x) = - \frac{2}{x^{2}} + \frac{2}{x^{2}}\ln x - \frac{2a}{x^{2}} = \frac{2}{x^{2}}\left( \ln x - 1 - a \right)\).令\(f''(x) \geq 0\)得,\(x \geq e^{1 + a}\).令\(f''(x) \leq 0\)得,\(1<x \leq e^{1 + a}\).故\(f'(x)\)在\(x \in (1, + \infty)\)上最小值\(f'(x)_{\min}=f'\left( e^{1 + a} \right)=1 - \frac{2(1 + a)}{e^{1 + a}} + \frac{2a}{e^{1 + a}} = \frac{e^{1 + a} - 2}{e^{1 + a}} > 0\).所以\(f'(\xi) \geq f'(x)_{\min}\).从而\(\frac{f(x)}{x - 1} > 0\).又\(x > 1\),则\(f(x) > 0\)成立,从而当\(x > 0\)时,\(x > \ln^{2}x - 2a\ln x + 1\)成立.
评注:这道题的参考答案是用(Ⅰ)中\(F(x)\)在\((0, + \infty)\)内的极小值\(F(2) > 0\)得到\(F(x) = xf'(x) > 0\).又\(x > 1\),所以\(f'(x) > 0\).从而\(f(x)\)在\((1, + \infty)\)上单调递增,故\(f(x)\)的最小值\(f(x)_{\min}>f(1)=0\),所以\(x > \ln^{2}x - 2a\ln x + 1\).但是如果没有(Ⅰ),很难想到利用\(F(x) = xf'(x)\)来判断\(f(x)\)的单调性.而用拉格朗日中值定理证明,就不存在这个问题.
六,证明\(\frac{f\left( x_{1} \right) - f\left( x_{2} \right)}{x_{1} - x_{2}} > \lambda\)或\(\frac{f\left( x_{1} \right) - f\left( x_{2} \right)}{x_{1} - x_{2}} < \lambda\)(其中\(x_{1} \neq x_{2}\))
例:(2009年辽宁卷理21题)
已知函数$f(x)=\frac{1}{2} x^{2}-a x+(a-1) \ln x, a>1$
(Ⅰ)讨论函数\(f(x)\)的单调性;
(Ⅱ)证明:若\(a < 5\),则对任意\(x_{1},x_{2} \in (0, + \infty)\),\(x_{1} \neq x_{2}\),有\(\frac{f(x_{1}) - f(x_{2})}{x_{1} - x_{2}} > - 1\).
(Ⅰ)略;
(Ⅱ)\(\frac{f(x_{1}) - f(x_{2})}{x_{1} - x_{2}} = f'(\xi)\).由(Ⅰ)得,\(f'(x) = x - a + \frac{a - 1}{x}\).所以要证\(\frac{f(x_{1}) - f(x_{2})}{x_{1} - x_{2}} > - 1\)成立,即证\(f'(\xi) = \xi - a + \frac{a - 1}{\xi} > - 1\).下面即证之.
令\(g(\xi) = \xi^{2} - (a - 1)\xi + a - 1\),则\(\Delta = (a - 1)^{2} - 4(a - 1) = (a - 1)(a - 5)\).由于\(1 < a < 5\),所以\(\Delta < 0\).从而\(g(\xi) > 0\)在\(R\)恒成立.也即\(\xi^{2} - a\xi + a - 1 > - \xi\).又\(\xi \in \left( x_{1},x_{2} \right)\),\(x_{1},x_{2} \in (0, + \infty)\),故\(\xi > 0\).则\(\frac{\xi^{2} - a\xi + a - 1}{\xi} > - 1\),即\(f'(\xi) = \xi - a + \frac{a - 1}{\xi} > - 1\),也即\(\frac{f(x_{1}) - f(x_{2})}{x_{1} - x_{2}} > - 1\).
评注:这道题(Ⅱ)小题存在两个难点:首先有两个变量\(x_{1},x_{2}\);其次\(a\)的值是变化的.参考答案的解法是考虑函数\(g(x) = f(x) + x\).为什么考虑函数\(g(x) = f(x) + x\)?很多考生一下子不易想到.而且\(g'(x)\)的放缩也不易想到. |
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