maths~

1.
(a)
f(x) is divided by x - 5, the remainder is 4 :
f(5) = 4
(5a - b)² - 5 = 4
(5a - b)² = 9
5a - b = ±3
b = 5a ± 3
b = 5a + 3 or b = 5a - 3

(b)
x - 1 is a factor of f(x) :
f(1) = 0
(a - b)² - 1 = 0
(a - b)² = 1
a - b = 1 or a - b = -1
a = b + 1 or a = b - 1

Put a = b + 1 into the expression b = 5a ± 3
b = 5(b + 1) + 3 or b = 5(b + 1) - 3
b = 5b + 5 + 3 or b = 5b + 5 - 3
4b = -8 or 4b = -2
b = -2 or b = -1/2
When b = -2, a = -1
When b = -1/2, a = 1/2

Similarly, put a = b - 1 into the expression b = 5a ± 3
b = 5(b - 1) + 3 or b = 5(b - 1) - 3
b = 5b - 5 + 3 or b = 5b - 5 - 3
4b = 2 or 4b = 8
b = 1/2 or b = 2
When b = 1/2, a = -1/2
When b = 2, a = 1

Ans: a = -1 and b = -2
or: a = 1 and b = 2
or: a = 1/2 and b = -1/2
or: a = -1/2 and b = 1/2


2.
(a)
x is a factor of f(x) :
f(0) = 0
(0 + m)(0 + n)(0 - 1) + 5 = 0
mn = 5
Hence, m = 1 and n = 5
or: m = 5 and n = 1
or: m = -1 and n = -5
or: m = -5 and n = -1

(b)
Case I : f(x) = (x + 5)(x + 1)(x - 1)+ 5
f(x) = (x + 5)(x² - 1) + 5
f(x) = x³ + 5x² - x - 5 + 5
f(x) = x³ + 5x² - x

f(x) - g(x) = 0
(x³ + 5x² - x) - (x³ - 9x² + 10x + k) = 0
x³ + 5x² - x - x³ + 9x² - 10x - k = 0
14x² - 11x - k = 0

The above equation has real roots.
Then, the discriminant Δ ≥ 0
(-11)² - 4(14)(-k) ≥ 0
56k ≥ -121
k ≥ -121/56

Case II : f(x) = (x - 5)(x - 1)² + 5
f(x) = (x - 5)(x² - 2x + 1) + 5
f(x) = x³ - 7x² + 11x - 5 + 5
f(x) = x³ - 7x² + 11x

f(x) - g(x) = 0
(x³ - 7x² + 11x) - (x³ - 9x² + 10x + k) = 0
x³ - 7x² + 11x - x³ + 9x² - 10x - k = 0
2x² + x - k = 0

The above equation has real roots.
Then, the discriminant Δ ≥ 0
(1)² - 4(2)(-k) ≥ 0
8k ≥ -1
k ≥ -1/8

Ans:
When (m = 1 and n = 5) or (m = 5 and n = 1), k ≥ -121/56
When (m = -1 and n = -5) or(m = -5 and n = -1), k ≥ -1/8

麻辣(002)的第2題答案錯誤。

因為 x is a factor of f(x)，所以 f(0) = 0 而「不是」f(x) = 0

就算 f(x) = 0，你也不能設定 x 是一個整數，題目只說 m 和 m 是整數。

簡單的驗算就是你把你答案的 m 和 n 代入 f(x) 中，則 x is NOT a factor of f(x)。

1.f(x)=(ax-b)^2-x. f(5)=4.(a) express b in terms of a.f(5)=(5a-b)^2-5=4 => (5a-b)^2=9b-5a=+-3 => b=5a+-3
(b) f(1)=0,find the values of a and b.f(1)=(a-b)^2-1=0 => (a-b)^2=1b=a+-1=5a+-3 => 4a=+-1-+3=+-2a=+-1/2, b=+-1/2+-1=+-3/2......ans
2.f(x)=(x+m)(x+n)(x-1)+5, m and n are integers. f(x)=0.(a) find the values of m and n.f(x)=(x+m)(x+n)(x-1)+5=0There exist 6 sets of answers: -5,1,1; 5,1,-1x-1=-5, x+m=1, x+n=1 => x=-4, m=n=5x+m=-5, x-1=1, x+n=1 => x=2, m=-7, n=-1 or x=2, m=-1, n=-7x-1=-1, x+m=5, x+n=1 => x=0, m=n=5x+m=-1, x-1=5, x+n=1 => x=6, m=-7, n=-5 or x=6, m=-5, n=-7
(b) g(x)=x^3-9x^2+10x+k. Find the range of values of k such that f(x)-g(x)=0 has a real roots. h(x)=f(x)-g(x)=(x+m)(x+n)(x-1)+5-(x^3-9x^2+10x+k)=(m+n-1)x^2+(mn-m-n)x-mn+5+9x^2-10x-k=(m+n+8)x^2+(mn-m-n-10)x+(5-mn-k)Discriminant>=0: D=[(mn-2)-(m+n+8)]^2+4(m+n+8)(mn+k-5)=(mn-2)^2-2(mn-2)(m+n+8)+(m+n+8)^2+4(m+n+8)(mn+k-5)=(mn-2)^2+2(m+n+8)[2(mn+k-5)-mn+2]+(m+n+8)^2=(mn-2)^2+4(m+n+8)(k-4)+(m+n+8)^2=(mn-2)^2+(m+n+8)(4k-16+m+n+8)=(mn-2)^2+(m+n+8)(4k+m+n-8)=(mn)^2-4mn+4+4k(m+n+8)+(m+n)^2-64=(mn)^2+m^2+n^2-2mn-60+4k(m+n+8)=(mn)^2+(m-n)^2-60+4k(m+n+8)>=04k(m+n+8)>=60-(mn)^2-(m-n)^2b.1.m=n=5 => k>=(60-25^2-0)/4*18=-565/72b.2.m=-7, n=-1 => m+n+8=0, k=None b.3.m=-7, n=-5 => k<=(60-1225-4)/(-4)=1169/4