2. 利用奇偶性
例4 已知函数f(x)=ax5+bsinx+3,且f(-3)=7,求f(3)的值.
分析 f(x)的解析式含有两个参数a、b,却只有一个条件f(-3)=7,无法确定a、b的值,因此f(x)仍是抽象函数,但我们注意到g(x)=ax5+bsinx是奇函数,有g(-3)=-g(3).
解 设g(x)=ax5+bsinx,显然g(x)是奇函数,
∵ f(-3)=7,
∴ f(-3)=g(-3)+3=-g(3)+3=7 g(3)=-4,
∴ f(3)=g(3)+3=-4+3=-1.
函数的特征是通过函数的性质反映出来的,抽象函数也不例外,只有充分利用题设条件所表明的函数的性质,灵活进行等价转化,抽象函数问题才能峰回路
转、化难为易.
1. 利用单调性
例3 设f(x)是定义在(0,+∞)上的增函数,满足f(xy)=f(x)+f(y), f(3)=1,解不等式f(x)+f(x-8)≤2.
解 ∵ 函数f(x)满足f(xy)=f(x)+f(y), f(3)=1,
∴ 2=1+1=f(3)+f(3)=f(9),
由f(x)+f(x-8)≤2,得 f[x(x-8)]≤f(9),
∵ 函数f(x)是定义在(0,+∞)上的增函数,
则
∴ 不等式解集为 {x|8<x≤9}.
有些抽象函数问题,用常规解法很难解决,但与具体函数“对号入座”后,问题容易迎刃而解.这种方法多用于解填空题、选择题、解答题的解题后的检验,但解答题的解答书写过程一般不能用此法.
例1 若函数f(x)与g(x)在R上有定义,且f(x-y)=f(x)g(y)-g(x)f(y),
f(-2)=f(1)≠0,则g(1)+g(-1)= .
解 因为 f(x-y)=f(x)g(y)-g(x)f(y), 这是两角差的正弦公式模型,
又f(-2)=f(1)≠0,
则可取
于是 f(-1-1)=f(-1)g(1)-g(-1)f(1)
例2 设函数f(x)是定义在R上的减函数,且满足f(x+y)=f(x)f(y),
f(-3)=8,则不等式f(x)f(x-2)< 的解集为 .
解 因为函数f(x)满足f(x+y)=f(x)f(y),这是指数函数模型,
又 f(-3)=8,
则可取
∵f(x)f(x-2)<
∴<, 即<,
∴ 2x-2 >8, 解不等式,得 x>5,
∴ 不等式的解集为 {x|x>5}.
8.(★★★★★)The old man was once________ sing the song.
A.listened to B.listening to
C.listened to to D.listen to
7.(★★★★)-Have you anything________ there.
-No.Thank you just the same.
A.should be taken B.to be taken
C.to take D.which should be taken
6.(★★★★★)Every minute is made full________ of________ our lessons well.
A.to use;study B.use;studying
C.use;to study D.used;studying
5.(★★★★★)The way they talked________ the problem seemed impossible.
A.about settling B.to settle
C.of settling D.about to settle
4.(★★★★★)Tom did whatever he could________ those who were in trouble.
A.to help B.help C.helping D.do help
3.(★★★★)They had nothing else to do but________ a doctor.
A.call for B.to send C.to call on D.send for
2.(★★★★)I lost my way in complete darkness and,________ matters worse,it began to rain.
A.made B.having made C.making D.to make
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