幾條F4maths 問題

1 f(x) has roots 1 and 4

Let f(x) = K(x - 1)(x - 4)(x - a)

Sub. x = 2 and x = -1

f(2) = -2K(2 - a) = -8

f(-1) = -10K(1 + a) = 10

So, (1 + a)/(2 - a) = -1/4

a - 2 = 4 + 4a

-3a = 6

a = -2, K = 1

f(x) = (x - 1)(x - 4)(x + 2)

2 a^2 + 4a + k = b^2 + 4b + k = 0

b^2 = -k - 4b

4a - b^2 = 4a + 4b + k = k - 16

3 Let the workload is 1

Then the rate of Amy is 1/x and the rate of Billy is 1/(x + 6)

As 4[1/x + 1/(x + 6)] = 1

4(2x + 6) = x(x + 6)

x^2 - 2x - 24 = 0

(x - 6)(x + 4) = 0

x = 6

4 f(2x - 1) = Q(x)(4x - 3)

Let y = 2x - 1 => x = (y + 1)/2 => 4x - 3 = 2(y + 1) - 3 = 2y - 1

Then f(y) = T(y)(2y - 1), So, f(x) is divisible by 2x - 1

1. By remainder theorem, f(2)=-8 and f(-1)=10
Since x^2-5x+4=(x-4)(x-1)
Thus, f(x)=(x-1)(x-4)(ax+b)
Sub f(2)=-8,
-8=(2-1)(2-4)(2a+b)
4=2a+b................(1)
Sub f(-1)=10,
10=(-1-1)(-1-4)(-a+b)
1=-a+b...............(2)
(1)+2(2),
3b=6
b=2
a=1.
2. Since b is the roots of the equation x^2+4x+k=0,
we have b^2+4b+k=0
b^2=-4b-k
Hence,
4a-b^2
=4a-(-4b-k)
=4a+4b+k
=4(a+b)+k
=k-16. (sum of roots=a+b=-4/1=-4)
3. From the question, we know that Amy can finish 1/x of the job in a day.
Similarly, we also know that Billy can finish 1/(x+6) of the job in a day.
Hence,
1/x+1/(x+6)=1/4
(x+6+x)/x(x+6)=1/4
8x+24=x(x+6)
x^2-2x-24=0
(x-6)(x+4)=0
x=-4(rej.), 6
4. Let y=2x-1,
then 4x-3=2(2x-1)-1=2y-1
Therefore f(y) is divisible by (2y-1)
and f(1/2)=0
Thus, f(x) is divisible by (2x-1).


2011-12-22 12:09:29 補充：
1. Thus, f(x)=(x-1)(x-4)(x+2).

Q3:
1/((1/x+6)+(1/x))=4
1/(2x+6/x^2+6x)=4
x^2+6x/2x+6=4
x^2+6x=8x+24
x^2-2x-24=0
(x+4)(x-6)=0
x=6 or -4 (rej.)