Question

【題目】設(shè)函數(shù)f(x)=ex-1-x-ax2.

(1)若a=0,求f(x)的單調(diào)區(qū)間;

(2)若當(dāng)x≥0時(shí),f(x)≥0,求a的取值范圍.

Answer 1

【答案】（1）f(x)在(-∞,0)內(nèi)單調(diào)遞減,在(0,+∞)內(nèi)單調(diào)遞增；（2）a的取值范圍為.

【解析】試題分析：（1）求導(dǎo)數(shù)，再求導(dǎo)函數(shù)零點(diǎn)，根據(jù)導(dǎo)函數(shù)符號(hào)確定單調(diào)區(qū)間，（2）當(dāng)自變量大于零時(shí)分離變量： ，再利用導(dǎo)數(shù)求函數(shù)單調(diào)性，根據(jù)單調(diào)性確定最值取法，利用洛必達(dá)法則求函數(shù)最小值，即得a的取值范圍

試題解析：(1)當(dāng)a=0時(shí),f(x)=ex-1-x,f'(x)=ex-1.

當(dāng)x∈(-∞,0)時(shí),f'(x)<0;當(dāng)x∈(0,+∞)時(shí),f'(x)>0.

故f(x)在(-∞,0)內(nèi)單調(diào)遞減,在(0,+∞)內(nèi)單調(diào)遞增.

(2)f'(x)=ex-1-2ax.

由(1)知f(x)≥f(0),即ex≥1+x,當(dāng)且僅當(dāng)x=0時(shí)等號(hào)成立,故f'(x)≥x-2ax=(1-2a)x.

當(dāng)a≤時(shí),1-2a≥0,f'(x)≥0(x≥0),f(x)在[0,+∞)上是增函數(shù),

因?yàn)?/span>f(0)=0,于是當(dāng)x≥0時(shí),f(x)≥0.符合題意.

當(dāng)a>時(shí),由ex>1+x(x≠0)可得e-x>1-x(x≠0).

所以f'(x)<ex-1+2a(e-x-1)=e-x(ex-1)(ex-2a),

故當(dāng)x∈(0,ln 2a)時(shí),f'(x)<0,而f(0)=0,于是當(dāng)x∈(0,ln 2a)時(shí),f(x)<0.不符合題意.

綜上可得a的取值范圍為.