F.4 Maths

1a. f(a) = f(b)==> 3a^2 - 9a + 4 = 3b^2 - 9b + 4==> 3(a^2 - b^2) - 9(a - b) = 0==> 3(a - b)(a + b - 3) = 0==> a - b = 0 (rej as a is not equal to b) or a + b = 3Therefore, a + b = 31b. f(a + b)= f(3)= 3*3^2 - 9*3 + 4= 27 - 27 + 4= 4
2. f(0) = c = 3 .................... (i)f(2) = 4a + 2b + c = 13 ..... (ii)f(4) = 16a + 4b + c = 55 ... (iii)Solve (i), (ii), (iii), we get,a = 4, b = -3, c = 3

[ 被檢舉為違規，請看我 ]，你個新名好「得意」。

同邊個有仇呀？

第一題其實是好的題目～

努力呀～ smile~

2013-10-30 18:21:17 補充：
如果公開了，可能會引起更大的廻響～

我只想息事寧人，希望不要再被胡亂檢舉～

^__^

1.Consider a function f(x)=3x^2－9x+4 such that f(a)=f(b)，where a is not equal b
(a)Find the value of a+b
Sol
a<>b
設 f(x)=3x^2－9x+4=3[x－(a+b)/2]^2+p
－9=3*[－2(a+b)/2]
a+b=3
(b)Hence find f(a+b)
f(3)=27－27+4=4

2.Consider a function f(x)=ax^2+bx+c，If f(0)=3，f(2)=13 andf(4)=55，find the values
of a，b and c
Sol
設 f(x)=a(x－0)(x－2)+p(x－0)+3
f(2)=2p+3=13
p=5
f(x)=a(x－0)(x－2)+5(x－0)+3
f(4)=a*4*2+5*4+3=55
8a=32
a=4
f(x)=4(x－0)(x－2)+5(x－0)+3=(4x^2－8x)+5x+3=4x^2－3x+3
a=4，b=－3，c=3