Question

15．已知函數(shù)f（x）=x²+$\frac{2}{x}$+alnx．
（Ⅰ）若f（x）在區(qū)間[2，3]上單調(diào)遞增，求實(shí)數(shù)a的取值范圍；
（Ⅱ）設(shè)f（x）的導(dǎo)函數(shù)f′（x）的圖象為曲線C，曲線C上的不同兩點(diǎn)A（x₁，y₁）、B（x₂，y₂）所在直線的斜率為k，求證：當(dāng)a≤4時(shí)，|k|＞1．

Answer 1

分析 （1）由函數(shù)單調(diào)性，知其導(dǎo)函數(shù)≥0在[2，3]上恒成立，將問(wèn)題轉(zhuǎn)化為$g（x）=\frac{2}{x}-2{x}^{2}$在[2，3]上單調(diào)遞減即可求得結(jié)果；
（2）根據(jù)題意，將$|\begin{array}{l}{f′（{x}_{1}）-f′（{x}_{2}）}\end{array}|$寫(xiě)成$|\begin{array}{l}{{x}_{1}-{x}_{2}}\end{array}|•|\begin{array}{l}{2+\frac{2（{x}_{1}+{x}_{2}）}{{{x}_{1}}^{2}{{x}_{2}}^{2}}-\frac{a}{{x}_{1}{x}_{2}}}\end{array}|$，利用不等式的性質(zhì)證明$|\begin{array}{l}{2+\frac{2（{x}_{1}+{x}_{2}）}{{{x}_{1}}^{2}{{x}_{2}}^{2}}-\frac{a}{{x}_{1}{x}_{2}}}\end{array}|＞1$，所以$|\begin{array}{l}{f′（{x}_{1}）-f′（{x}_{2}）}\end{array}|$＞$|\begin{array}{l}{{x}_{1}-{x}_{2}}\end{array}|$，即得$|\begin{array}{l}{k}\end{array}|＞1$．

解答 解：（1）由$f（x）={x}^{2}+\frac{2}{x}+alnx$，得$f′（x）=2x-\frac{2}{{x}^{2}}+\frac{a}{x}$．
因?yàn)閒（x）在區(qū)間[2，3]上單調(diào)遞增，
所以$f′（x）=2x-\frac{2}{{x}^{2}}+\frac{a}{x}$≥0在[2，3]上恒成立，
即$a≥\frac{2}{x}-2{x}^{2}$在[2，3]上恒成立，
設(shè)$g（x）=\frac{2}{x}-2{x}^{2}$，則$g′（x）=-\frac{2}{{x}^{2}}-4x＜0$，
所以g（x）在[2，3]上單調(diào)遞減，
故g（x）_max=g（2）=-7，
所以a≥-7；
（2）對(duì)于任意兩個(gè)不相等的正數(shù)x₁、x₂有
${x}_{1}{x}_{2}+\frac{2（{x}_{1}+{x}_{2}）}{{x}_{1}{x}_{2}}$＞${x}_{1}{x}_{2}+\frac{4}{\sqrt{{x}_{1}{x}_{2}}}$
=${x}_{1}{x}_{2}+\frac{2}{\sqrt{{x}_{1}{x}_{2}}}+\frac{2}{\sqrt{{x}_{1}{x}_{2}}}$
$≥3\root{3}{{x}_{1}{x}_{2}×\frac{2}{\sqrt{{x}_{1}{x}_{2}}}×\frac{2}{\sqrt{{x}_{1}{x}_{2}}}}$
=$3\root{3}{4}＞4.5＞a$，
∴$|\begin{array}{l}{2+\frac{2（{x}_{1}+{x}_{2}）}{{{x}_{1}}^{2}{{x}_{2}}^{2}}-\frac{a}{{x}_{1}{x}_{2}}}\end{array}|＞1$，
而$f′（x）=2x-\frac{2}{{x}^{2}}+\frac{a}{x}$，
∴$|\begin{array}{l}{f′（{x}_{1}）-f′（{x}_{2}）}\end{array}|$=$|\begin{array}{l}{（2{x}_{1}-\frac{2}{{{x}_{1}}^{2}}+\frac{a}{{x}_{1}}）-（2{x}_{2}-\frac{2}{{{x}_{2}}^{2}}+\frac{a}{{x}_{2}}）}\end{array}|$
=$|\begin{array}{l}{{x}_{1}-{x}_{2}}\end{array}|•|\begin{array}{l}{2+\frac{2（{x}_{1}+{x}_{2}）}{{{x}_{1}}^{2}{{x}_{2}}^{2}}-\frac{a}{{x}_{1}{x}_{2}}}\end{array}|$＞$|\begin{array}{l}{{x}_{1}-{x}_{2}}\end{array}|$，
故：$|\begin{array}{l}{f′（{x}_{1}）-f′（{x}_{2}）}\end{array}|$＞$|\begin{array}{l}{{x}_{1}-{x}_{2}}\end{array}|$，即$|\begin{array}{l}{\frac{f′（{x}_{1}）-f′（{x}_{2}）}{{x}_{1}-{x}_{2}}}\end{array}|$＞1，
∴當(dāng)a≤4時(shí)，$|\begin{array}{l}{k}\end{array}|＞1$．

點(diǎn)評(píng) 本題考查導(dǎo)數(shù)及基本不等式的應(yīng)用，解題的關(guān)鍵是利用不等式得到函數(shù)值的差的絕對(duì)值要大于自變量的差的絕對(duì)值．