Question

P₁是橢圓+y²=1(a＞0且a≠1)上不與頂點(diǎn)重合的任一點(diǎn),P₁P₂是垂直于x軸的弦,A₁(-a,0),A₂(a,0)是橢圓的兩個(gè)頂點(diǎn),直線(xiàn)A₁P₁與直線(xiàn)A₂P₂的交點(diǎn)為P.

(1)求點(diǎn)P的軌跡曲線(xiàn)C的方程;

(2)設(shè)曲線(xiàn)C與直線(xiàn)l:x+y=1相交于兩個(gè)不同的點(diǎn)A、B,求曲線(xiàn)C的離心率e的取值范圍;

(3)設(shè)曲線(xiàn)C與直線(xiàn)l:x+y=1相交于兩個(gè)不同的點(diǎn)A、B,O為坐標(biāo)原點(diǎn),且=-3,求a的值.

(文)(本小題滿(mǎn)分12分)設(shè)函數(shù)f(x)=x³+2ax²-3a²x+a(0＜a＜1).

(1)求函數(shù)f(x)的單調(diào)區(qū)間;

(2)若當(dāng)x∈［a,2］時(shí),恒有f(x)≤0,試確定實(shí)數(shù)a的取值范圍.

Answer 1

(理)解析：(1)設(shè)P₁(m,n)(mn≠0),則P₂(m,-n),直線(xiàn)A₁P₁:y=(x+a);①

直線(xiàn)A₂P₂:y=(x-a);②                                                   

設(shè)P點(diǎn)坐標(biāo)為(x,y),

由①②得m=,n=,                                                      

       

∵點(diǎn)P₁(m,n)在橢圓+y²=1上,

∴有m²+a²n²=a²,

即()²+a²()²=a²,整理得-y²=1(y≠0),

∴直線(xiàn)A₁P₁與直線(xiàn)A₂P₂交點(diǎn)P的軌跡方程是雙曲線(xiàn)-y²=1(y≠0).                

(2)由C與l相交于兩個(gè)不同的點(diǎn),故知方程組有兩個(gè)不同的實(shí)數(shù)解.消去y并整理得(1-a²)x²+2a²x-2a²=0.

又∵a＞0且a≠1,

∴4a⁴+8a²(1-a²)＞0.

∴0＜a²＜2且a²≠1.                                                         

雙曲線(xiàn)的離心率e=.

∴＜e＜或e＞,即e∈()∪(,+∞).                        

(3)設(shè)A(x₁,y₁)、B(x₂,y₂),

則-3=

=x₁x₂+y₁y₂

=x₁x₂+(1-x₁)(1-x₂)

=2x₁x₂-(x₁+x₂)+1

=+1,

即=-4,由a＞0,得a=.                                               

(文)解：(1)∵f(x)=x³+2ax²-3a²x+a(0＜a＜1),

∴f′(x)=-x²+4ax-3a²=-(x-a)(x-3a).                                              

∵0＜a＜1,∴f′(x)＞0a＜x＜3a,f′(x)＜0x＜a或x＞3a.                      

∴函數(shù)f(x)的遞增區(qū)間為［a,3a］;遞減區(qū)間為(-∞,a］,［3a,+∞).                   

(2)∵x∈［a,2］,

①當(dāng)2≤3a,即≤a＜1時(shí),f(x)在區(qū)間［a,2］內(nèi)是增函數(shù).

∴f(x)_max=f(2)=a-6a².

又當(dāng)x∈［a,2］時(shí),恒有f(x)≤0,

∴.                                       

②當(dāng)2＞3a即0＜a＜時(shí),則f(x)在［a,3a］上單調(diào)遞增;在［3a,2］上單調(diào)遞減,

∴f(x)_max=f(3a)=a.

又當(dāng)x∈［a,2］時(shí),恒有f(x)≤0,

∴(無(wú)解).

綜上所述,a的取值范圍是≤a＜1.