1. Consider a function f(x)= x^2 - 9x +4 such that f(a) = f(b), where a is not equal to b.
(a) Find the value of a+b
(b) Hence find f(a+b)
2. Consider h(x) = x^2 - 3x - 6
(a) Express h(a-1) - h(a+1) in terms of a.
(b) If h(a+1)=4, find the values of a.
F.4 Maths
2009-11-08 11:08 pm
回答 (2)
2009-11-08 11:16 pm
✔ 最佳答案
1(a) f(x)=x^2 - 9x +4,f(a)=f(b)a^2-9a+4=b^2-9b+4
(a-b)(a+b)=9(a-b)
a+b=9
(b) f(a+b)=f(9)=81-81+4=4
2 h(a-1) - h(a+1)
=[(a-1)^2-3(a-1)-6]-[(a+1)^2-3(a+1)-6]
=[a^2-2a+1-3a+3-6]-[a^2+2a+1-3a-3-6]
=-4a
(b) h(a+1)=4
(a+1)^2-3(a+1)-6=4
Let x=a+1, x^2-3x-10=0
(x-5)(x+2)=0
x=5 or -2
a=4 or -3
2009-11-08 11:50 pm
1a.
f(a)=a^2-9a+4
f(b)=b^2-9b+4
a^2-9a+4=b^2-9b+4
since f(a)=f(b),but a not equal to b,let b be a negative no.
a^2-9a+4=(-b)^2+9b+4
a^2-b^2=9a+9b
(a-b)(a+b)=9(a+b)
-(a-b)(a+b)=-9(-a-b)
a+b=-9
b.
f(a+b)=(a+b)^2-9(a+b)+4
=(-9)^2-9(-9)+4
=166
2a.
h(a-1)-h(a+1)=(a-1)^2-3(a-1)-6-[(a+1)^2-3(a+1)-6]
=a^2-2a+1-3a+3-6-[a^2+2a+2-3a-3-6]
=a^2-2a+1-3a+3-6-a^2-2a-2+3a+3+6
=-4a+5
b.
h(a+1)=4
(a+1)^2-3(a+1)-6=4
a^2+2a+12-3a-3-6-4=0
a^2-a-11=0
a=4 or -3
f(a)=a^2-9a+4
f(b)=b^2-9b+4
a^2-9a+4=b^2-9b+4
since f(a)=f(b),but a not equal to b,let b be a negative no.
a^2-9a+4=(-b)^2+9b+4
a^2-b^2=9a+9b
(a-b)(a+b)=9(a+b)
-(a-b)(a+b)=-9(-a-b)
a+b=-9
b.
f(a+b)=(a+b)^2-9(a+b)+4
=(-9)^2-9(-9)+4
=166
2a.
h(a-1)-h(a+1)=(a-1)^2-3(a-1)-6-[(a+1)^2-3(a+1)-6]
=a^2-2a+1-3a+3-6-[a^2+2a+2-3a-3-6]
=a^2-2a+1-3a+3-6-a^2-2a-2+3a+3+6
=-4a+5
b.
h(a+1)=4
(a+1)^2-3(a+1)-6=4
a^2+2a+12-3a-3-6-4=0
a^2-a-11=0
a=4 or -3
參考: ME
收錄日期: 2021-04-26 13:59:33
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