Question

已知函數(shù)f(x)和g(x)的圖象關(guān)于原點(diǎn)對(duì)稱,且f(x)=x²+2x.

(1)求函數(shù)g(x)的解析式;

(2)解不等式g(x)≥f(x)-|x-1|;

(3)(文)若h(x)=g(x)-λf(x)+1在［-1,1］上是增函數(shù),求實(shí)數(shù)λ的取值范圍.

Answer 1

解析:(1)設(shè)函數(shù)y=f(x)的圖象上任一點(diǎn)Q(x₀,y₀)關(guān)于原點(diǎn)的對(duì)稱點(diǎn)為P(x,y),??

則

即

∵點(diǎn)Q(x₀,y₀)在函數(shù)y=f(x)的圖象上,?

∴-y=x²-2x,即y=-x²+2x.?

故g(x)=-x²+2x.?

(2)由g(x)≥f(x)-|x-1|可得?

2x²-|x-1|≤0.?

當(dāng)x≥1時(shí),2x²-x+1≤0,此時(shí)不等式無解.?

當(dāng)x＜1時(shí),2x²+x-1≤0,?

∴-1≤x≤.?

因此,原不等式的解集為［-1, ］.?

(3)(文)h(x)=-(1+λ)x²+2(1-λ)x+1.??

①當(dāng)λ=-1時(shí),h(x)=4x+1在［-1,1］上是增函數(shù),∴λ=-1.?

②當(dāng)λ≠-1時(shí),對(duì)稱軸的方程為x=.?

(ⅰ)當(dāng)λ＜-1時(shí), ≤-1,?

解得λ＜-1.?

(ⅱ)當(dāng)λ＞-1時(shí), ≥1,?

解得-1＜λ≤0.?

綜上,λ≤0.