Question

已知函數(shù)f(x)=x²,g(x)=|x-2|.

(1)解不等式f(x)＞g(x);

(2)設(shè)函數(shù)f(x)的圖象為C₁,g(x)的圖象為C₂,l是和曲線C₁相切且與曲線C₂無公共點(diǎn)的直線,求直線l的斜率的取值范圍.

Answer 1

解法一:(1)∵f(x)=x²,g(x)=|x-2|,在同一直角坐標(biāo)系中畫出函數(shù)f(x)和g(x)的圖象,

?

解方程組得A(-2,4),B(1,1),                                                            ?

∴不等式f(x)＞g(x)的解集為函數(shù)f(x)的圖象C₁位于函數(shù)g(x)的圖象C₂的上方時(shí),自變量x的取值范圍,即為{x|x＜-2或x＞1}.?

綜上可得,原不等式的解集為{x|x＜-2或x＞1}.                                                 

?

(2)如圖所示,設(shè)直線l為曲線C₁的切線,切點(diǎn)為C.當(dāng)切線l的斜率為-1時(shí),切線l與曲線C₂

無公共點(diǎn),                                                                                                               ?

此時(shí),若讓切線l按逆時(shí)針方向沿曲線C₁運(yùn)動(dòng)至坐標(biāo)原點(diǎn)O時(shí),切線l與曲線C₂恰有一個(gè)公共點(diǎn)D,由此可得,當(dāng)切線l的斜率k∈［-1,0）時(shí),直線為l與曲線C₂無公共點(diǎn).    ?

解法二:(1)∵f(x)=x²,g(x)=|x-2|,f(x)≥g(x),x²＞|x-2|,??

當(dāng)x≤2時(shí),得x²＞2-x,解得x＜-2或1＜x≤2;                                                               ?

當(dāng)x＞2時(shí),得x²＞x-2,解得x＞2.                                                                              ?

綜上可得,原不等式的解集為{x|x＜-2或x＞1}.                                                  ?

(2)∵f′(x)=2x,直線l與曲線C₁相切時(shí)的切點(diǎn)為M(x₀,x₀²),則直線l的方程為y-x₀²=2x₀

(x-x₀),即y=2x₀x-x₀²,又直線l與曲線C₂無公共點(diǎn),則需滿足方程組無解,

即方程2x₀x-x₀²=|x-2|無解.                                                                                 ?

當(dāng)x≥2時(shí),方程2x₀x-x₀²=|x-2|可化為2x₀x-x₀²=x-2,即(2x₀-1)x=x₀²-2,               ①?

若x₀=,方程①無解;?

若x₀≠,則方程①無解需滿足不等式＜2,由此解得x₀＜0或＜x₀＜4.?

∴方程①無解的條件是x₀＜0或≤x₀＜4.?

當(dāng)x＜2時(shí),方程2x₀x-x₀²=|x-2|可化為2x₀x-x₀²=2-x,即(2x₀+1)x=x₀²+2,            ②?

若x₀=-,方程②無解;?

若x₀≠-,則方程②無解需滿足不等式≥2,由此解得-＜x₀≤0或x₀≥4.?

∴方程②無解的條件是-≤x₀≤0或x₀≥4.?

綜上可得,方程2x₀x-x₀²=|x-2|無解的條件是-≤x₀＜0.?

故f′(x₀)=2x₀的取值范圍是［-1,0）,即直線l的斜率的取值范圍是［-1,0）.