Question

【題目】已知函數(shù)f(x)=ex-ax(a為常數(shù))的圖象與y軸交于點(diǎn)A,曲線y=f(x)在點(diǎn)A處的切線斜率為-1.

(1)求a的值及函數(shù)f(x)的極值;

(2)證明:當(dāng)x>0時(shí),x2<ex.

Answer 1

【答案】（1）a=2，極小值為f(ln 2)=2-2ln 2=2-ln 4,f(x)無(wú)極大值；（2）見解析.

【解析】試題分析：（1）根據(jù)導(dǎo)數(shù)幾何意義得f'(0)=-1，解得a，再根據(jù)導(dǎo)函數(shù)零點(diǎn)列表分段討論導(dǎo)函數(shù)符號(hào)，根據(jù)符號(hào)變化規(guī)律確定極值，（2）作差函數(shù)，利用二次求導(dǎo)，確定函數(shù)單調(diào)性，根據(jù)單調(diào)性證明不等式.

試題解析：(1)解 由f(x)=ex-ax,得f'(x)=ex-a.

因?yàn)?/span>f'(0)=1-a=-1,所以a=2.

所以f(x)=ex-2x,f'(x)=ex-2.

令f'(x)=0,得x=ln 2.

當(dāng)x<ln 2時(shí),f'(x)<0,f(x)單調(diào)遞減;當(dāng)x>ln 2時(shí),f'(x)>0,f(x)單調(diào)遞增,

所以當(dāng)x=ln 2時(shí),f(x)取得極小值,極小值為f(ln 2)=2-2ln 2=2-ln 4,f(x)無(wú)極大值.

(2)證明 令g(x)=ex-x2,則g'(x)=ex-2x.由(1),得g'(x)=f(x)≥f(ln 2)=2-ln 4>0,

故g(x)在R上單調(diào)遞增.

因?yàn)?/span>g(0)=1>0,所以當(dāng)x>0,g(x)>g(0)>0,即x2<ex.