Question

(理)已知:f_n(x)=a₁x+a₂x²+…+a_nxⁿ,f_n(-1)=(-1)ⁿ·n,n=1,2,3,….

(1)求a₁、a₂、a₃;

(2)求數(shù)列{a_n}的通項(xiàng)公式;

(3)求證:f_n()＜1.

(文)設(shè)函數(shù)f(x)=2ax³-(6a+3)x²+12x(a∈R),

(1)當(dāng)a=1時,求函數(shù)f(x)的極大值和極小值;

(2)若函數(shù)f(x)在區(qū)間(-∞,1)上是增函數(shù),求實(shí)數(shù)a的取值范圍.

Answer 1

答案：(理)解：(1)由已知f₁(-1)=-a₁=-1,所以a₁=1.

f₂(-1)=-a₁+a₂=2,所以a₂=3.f₃(-1)=-a₁+a₂-a₃=-3,所以a₃=5.

(2)∵(-1)ⁿ⁺¹·a_n+1=f_n+1(-1)-f_n(-1)=(-1)ⁿ⁺¹·(n+1)-(-1)ⁿ·n,∴a_n+1=(n+1)+n,

即a_n+1=2n+1.所以對于任意的n=1,2,3,…,a_n=2n-1.

(3)f_n(x)=x+3x²+5x³+…+(2n-1)xⁿ,∴f_n()=+3()²+5()³+…+(2n-1)()ⁿ.①

·f_n()=()²+3()³+5()⁴+…+(2n-1)()ⁿ⁺¹.②

①-②,得f_n()=+2()²+2()³+…+2()ⁿ-(2n-1)()ⁿ⁺¹

-(2n-1)()ⁿ⁺¹.∴f_n()=1-,

又n=1,2,3,…,故f_n()＜1.

(文)解：(1)當(dāng)a=1時,f(x)=2x³-9x²+12x.

∴f′(x)=6x²-18x+12=6(x²-3x+2).

令f′(x)=0,得x₁=1,x₂=2.

列表

x	(-∞,1)	1	(1,2)	2	(2,+∞)
f′(x)	+	0	-	0	+
f(x)	?	極大值		極小值

∴f(x)的極大值為f(1)=5,f(x)的極小值為f(2)=4.

(2)f′(x)=6ax²-(12a+6)x+12=6[ax²-(2a+1)x+2]=6(ax-1)(x-2).

①若a=0,則f(x)=-3x²+12x,此函數(shù)在(-∞,2)上單調(diào)遞增,滿足題意.

②若a≠0,則令f′(x)=0,得x₁=2,x₂=,由已知,f(x)在區(qū)間(-∞,1)上是增函數(shù),即當(dāng)x＜1時,f′(x)≥0恒成立,

若a＞0,則只需≥1,即0＜a≤1,

若a＜0,則＜0,當(dāng)x＜時,f′(x)＜0,則f(x)在區(qū)間(-∞,1)上不是增函數(shù).綜上所述,實(shí)數(shù)a的取值范圍是[0,1].