規(guī)定A
 
m
x
=x(x-1)…(x-m+1),其中x∈R,m為正整數(shù),且
A
0
x
=1,這是排列數(shù)A
 
m
n
(n,m是正整數(shù),n≤m)的一種推廣.
(Ⅰ) 求A
 
3
-9
的值;
(Ⅱ)排列數(shù)的兩個(gè)性質(zhì):①A
 
m
n
=nA
 
m-1
n-1
,②A
 
m
n
+mA
 
m-1
n
=A
 
m
n+1
(其中m,n是正整數(shù)).是否都能推廣到A
 
m
x
(x∈R,m是正整數(shù))的情形?若能推廣,寫出推廣的形式并給予證明;若不能,則說明理由;
(Ⅲ)已知函數(shù)f(x)=A
 
3
x
-4lnx-m,試討論函數(shù)f(x)的零點(diǎn)個(gè)數(shù).
分析:(Ⅰ)直接代入定義求解;
(Ⅱ)利用新定義,結(jié)合排列數(shù)的兩個(gè)性質(zhì)即可證明推廣的結(jié)論;
(Ⅲ)由新定義展開函數(shù)f(x),求導(dǎo)后得其導(dǎo)函數(shù)的零點(diǎn),得其在各區(qū)間段內(nèi)的單調(diào)性,然后對m進(jìn)行討論得其零點(diǎn)個(gè)數(shù).
解答:解:(Ⅰ)
A
3
-9
=-9×(-10)×(-11)=-990

(Ⅱ)性質(zhì)①、②均可推廣,推廣的形式分別是①
A
m
x
=x
A
m-1
x-1
,②
A
m
x
+m
A
m-1
x
=
A
m
x+1
(x∈R,m∈N*
證明:①當(dāng)m=1時(shí),左邊=
A
1
x
=x
,右邊=x
A
0
x
=x
,等式成立;
當(dāng)m≥2時(shí),
左邊=x(x-1)…(x-m+1)=x{(x-1)(x-2)…[(x-1)-(m-1)+1]}=x
A
m-1
x-1

因此,
A
m
x
=x
A
m-1
x-1
(x∈R,m∈N*)成立.
②當(dāng)m=1時(shí),左邊=
A
1
x
+
A
0
x
=x+1=
A
1
x+1
=右邊,等式成立;
當(dāng)m≥2時(shí),左邊x(x-1)…(x-m+1)+mx(x-1)…(x-m+2)
=x(x-1)…(x-m+2)(x-m+1+m)
=(x+1)x(x-1)…(x-m+2)
=(x+1)x(x-1)…[(x+1)-m=1]
=
A
m
x+1
=右邊
因此,
A
m
x
+m
A
m-1
x
=
A
m
x+1
(x∈R,m∈N*)成立.
(Ⅲ)f(x)=
A
3
x
-4lnx-m=x(x-1)(x-2)-4lnx-m=x3-3x2+2x-4lnx-m

設(shè)函數(shù)g(x)=x3-3x2+2x-4lnx,
函數(shù)f(x)零點(diǎn)的個(gè)數(shù)等價(jià)于函數(shù)g(x)與y=m公共點(diǎn)的個(gè)數(shù).
f(x)的定義域?yàn)椋?,+∞)
g(x)=3x2-6x+2-
4
x
=
3x3-6x2+2x-4
x
=
(x-2)(3x2+2)
x

令g(x)=0,得x=2
x                                                      (0,2)                      2 (2,+∞)
g(x) - 0 +
g(x) -4ln2
∴當(dāng)m<-4ln2時(shí),函數(shù)g(x)與y=m沒有公共點(diǎn),即函數(shù)f(x)不存在零點(diǎn),
當(dāng)m=-4ln2時(shí),函數(shù)g(x)與y=m有一個(gè)公共點(diǎn),即函數(shù)f(x)有且只有一個(gè)零點(diǎn),
當(dāng)m>-4ln2時(shí),函數(shù)g(x)與y=m有兩個(gè)公共點(diǎn),即函數(shù)f(x)有且只有兩個(gè)零點(diǎn).
點(diǎn)評:本題考查了排列及排列數(shù)公式,考查了利用導(dǎo)函數(shù)判斷原函數(shù)的單調(diào)性,考查了分類討論的數(shù)學(xué)思想方法,解答的關(guān)鍵是對新定義的理解與運(yùn)用,是中檔題.
練習(xí)冊系列答案
相關(guān)習(xí)題

同步練習(xí)冊答案