教教我這一條數學哦...

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(a)
f(x) = x2 + ax + b ... (*)
代 x = 1 至 (*)，
f(1) = (1)2 + a(1) + b
f(1) = a + b + 1
代 x = 3 至 (*)，
f(3) = (3)2 + a(3) + b
f(3) = 3a + b + 9
代 x = 4 至 (*)，
f(4) = (4)2 + a(4) + b
f(4) = 4a + b + 16
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(b)
已知 f(1) = f(3),
a + b + 1 = 3a + b + 9
2a + 8 = 0
a = -4 .... (1)
已知 f(4) = 5,
4a + b + 16 = 5 【由 (a)】
4a + b + 11 = 0
4(-4) + b + 11 = 0 【由 (1)】
b - 5 = 0
b = 5 .... (2)
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(c)
若 f(x) = 2x
x2 + ax + b = 2x
x2 + (-4)x + (5) = 2x 【由 (1),(2)】
x2 - 4x + 5 = 2x
x2 - 6x + 5 = 0
(x-5)(x-1) = 0
x-5 = 0 或 x-1 = 0
x = 5 或 x = 1

f(x) = x2 + ax + b
f(1) = (1)2 + a(1) + b = 1 + a + b
f(3) = (3)2 + a(3) + b = 9 + 3a + b
f(4) = (4)2 + a(4) + b = 16 + 4a + b
if f(4)=5 ; then 16 + 4a + b = 5 ;
f(1) = f(3) gives 1 + a + b = 9 + 3a + b
1 + a + b = 9 + 3a + b gives 1 + a = 9 + 3a and -2a = 8 and a = -4
put a = -4 into 16 + 4a + b = 5 gives 16 + 4(-4) + b = 5 and this gives b = 5 ;
f(x) becomes f(x) = x2 -4x + 5
When f(x) = 2x ; it gives x2 -4x + 5 = 2x
x2 -6x + 5 = 0
(x-5)(x-1) = 0
x=5 or x=1

a)
f(1)=1+a+b
f(3)=9+3a+b
f(4)=16+4a+b

b)
f(1)=f(3)
1+a+b=9+3a+b
-8=2a
a=-4

f(4)=5
16+4a+b=5
16-16+b=5
b=5

c)
if f(x)=2x
x^2-4x+5=2x
x^2-6x+5=0
(x-5)(x-1)=0
x=5 or x=1

f(x)=x^2+ax+b



a.

f(1)=1^2+1a+b

=a+b+1

f(3)=3^2+3a+b

=3a+b+9

f(4)=4^2+4a+b

=4a+b+16



b.

∵ f(1)=f(3)

∴a+b+1=3a+b+9

a-3a+b-b=9-1

-2a=8

∴a=-4

∵f(4)=5

∴4a+b+16=5

b=5-16-4a

b=-11-4a

b=-11-4(-4)

b=-11+16

∴b=5

∴a=-4,b=5



c.

f(x)=2x

x^2+ax+b=2x

x^2-4x+5=2x

x^2-6x+5=0

x=[ 6 ± √[ (-6)^2 - 4(1)(5) ] ]/2

=[6±√(36-20) ]/2

=(6±√16)/2

=(6±4)/2

=5 or 1