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IAL 2023 Jan Q5

A Level / Edexcel / P2

IAL 2023 Jan Paper · Question 5

题目

Problem

f(x)=x3+(p+3)x2−x+q\mathrm{f}(x) = x^3 + (p + 3)x^2 - x + q

where pp and qq are constants and p>0p > 0.

Given that (x−3)(x - 3) is a factor of f(x)\mathrm{f}(x),

(a) show that 9p+q=−519p + q = -51.

(2)

Given also that when f(x)\mathrm{f}(x) is divided by (x+p)(x + p) the remainder is 99,

(b) show that 3p2+p+q−9=03p^2 + p + q - 9 = 0.

(2)

(c) Hence find the value of pp and the value of qq.

(3)

(d) Hence find a quadratic expression g(x)\mathrm{g}(x) such that f(x)=(x−3)g(x)\mathrm{f}(x) = (x - 3)\mathrm{g}(x).

(2)

解答

(a)

解法一

思路

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根據因式定理（Factor Theorem），如果 (x−3)(x - 3) 是 f(x)\mathrm{f}(x) 的因式，那麼 f(3)=0\mathrm{f}(3) = 0。 將 x=3x = 3 代入函數表達式，令其等於 00，化簡即可得到目標關係式。

答题过程

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Since (x−3)(x - 3) is a factor of f(x)\mathrm{f}(x), by the factor theorem, we have f(3)=0\mathrm{f}(3) = 0:

f(3)=  0(3)3+(p+3)(3)2−3+q=  027+9(p+3)−3+q=  024+9p+27+q=  09p+q+51=  0.\begin{align*} \mathrm{f}(3) =&\,\, 0 \\[2mm] (3)^3 + (p + 3)(3)^2 - 3 + q =&\,\, 0 \\[2mm] 27 + 9(p + 3) - 3 + q =&\,\, 0 \\[2mm] 24 + 9p + 27 + q =&\,\, 0 \\[2mm] 9p + q + 51 =&\,\, 0. \end{align*}

Therefore,

9p+q=  −51.\begin{align*} 9p + q =&\,\, -51. \end{align*}

(b)

解法一

思路

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根據餘式定理（Remainder Theorem），如果 f(x)\mathrm{f}(x) 除以 (x+p)(x + p) 的餘數是 99，那麼 f(−p)=9\mathrm{f}(-p) = 9。 將 x=−px = -p 代入函數表達式，令其等於 99，進行代數化簡即可。

答题过程

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Since f(x)\mathrm{f}(x) divided by (x+p)(x + p) leaves a remainder of 99, by the remainder theorem, we have f(−p)=9\mathrm{f}(-p) = 9:

f(−p)=  9(−p)3+(p+3)(−p)2−(−p)+q=  9−p3+(p+3)p2+p+q=  9−p3+p3+3p2+p+q=  93p2+p+q=  9.\begin{align*} \mathrm{f}(-p) =&\,\, 9 \\[2mm] (-p)^3 + (p + 3)(-p)^2 - (-p) + q =&\,\, 9 \\[2mm] -p^3 + (p + 3)p^2 + p + q =&\,\, 9 \\[2mm] -p^3 + p^3 + 3p^2 + p + q =&\,\, 9 \\[2mm] 3p^2 + p + q =&\,\, 9. \end{align*}

Therefore,

3p2+p+q−9=  0.\begin{align*} 3p^2 + p + q - 9 =&\,\, 0. \end{align*}

(c)

解法一

思路

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  1. 消元：由 (a) 小題可得 q=−51−9pq = -51 - 9p。
  2. 代入求解：將其代入 (b) 小題的方程中，得到一個關於 pp 的一元二次方程。
  3. 求解並篩選：解出 pp 的兩個值。根據已知條件 p>0p > 0，捨去負值，從而確定 pp。再代入求出 qq。

答题过程

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From part (a), we can express qq in terms of pp:

q=  −51−9p.\begin{align*} q =&\,\, -51 - 9p. \end{align*}

Substitute this expression for qq into the equation from part (b):

3p2+p+(−51−9p)−9=  03p2−8p−60=  0(3p+10)(p−6)=  0.\begin{align*} 3p^2 + p + (-51 - 9p) - 9 =&\,\, 0 \\[2mm] 3p^2 - 8p - 60 =&\,\, 0 \\[2mm] (3p + 10)(p - 6) =&\,\, 0. \end{align*}

This gives:

p=−103orp=6.\begin{align*} p = -\frac{10}{3} \quad \text{or} \quad p = 6. \end{align*}

Since p>0p > 0, we choose:

p=  6.\begin{align*} p =&\,\, 6. \end{align*}

Substituting p=6p = 6 back into the expression for qq:

q=  −51−9(6)=  −51−54=  −105.\begin{align*} q =&\,\, -51 - 9(6) \\[2mm] =&\,\, -51 - 54 \\[2mm] =&\,\, -105. \end{align*}

Thus, the values are:

p=6,q=−105.\begin{align*} p = 6, \quad q = -105. \end{align*}

(d)

解法一

思路

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將 p=6p = 6 和 q=−105q = -105 代入原函數，得到 f(x)=x3+9x2−x−105\mathrm{f}(x) = x^3 + 9x^2 - x - 105。 利用多項式除法（polynomial division）或待定係數法將 f(x)\mathrm{f}(x) 除以 (x−3)(x - 3)，所得的商式即為二次多項式 g(x)\mathrm{g}(x)。

答题过程

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Substituting p=6p = 6 and q=−105q = -105 into f(x)\mathrm{f}(x):

f(x)=  x3+9x2−x−105.\begin{align*} \mathrm{f}(x) =&\,\, x^3 + 9x^2 - x - 105. \end{align*}

We divide f(x)\mathrm{f}(x) by (x−3)(x - 3) using algebraic long division or division by inspection:

x3+9x2−x−105=  x2(x−3)+3x2+9x2−x−105=  x2(x−3)+12x2−x−105=  x2(x−3)+12x(x−3)+36x−x−105=  x2(x−3)+12x(x−3)+35x−105=  x2(x−3)+12x(x−3)+35(x−3)=  (x−3)(x2+12x+35).\begin{align*} x^3 + 9x^2 - x - 105 =&\,\, x^2(x - 3) + 3x^2 + 9x^2 - x - 105 \\[2mm] =&\,\, x^2(x - 3) + 12x^2 - x - 105 \\[2mm] =&\,\, x^2(x - 3) + 12x(x - 3) + 36x - x - 105 \\[2mm] =&\,\, x^2(x - 3) + 12x(x - 3) + 35x - 105 \\[2mm] =&\,\, x^2(x - 3) + 12x(x - 3) + 35(x - 3) \\[2mm] =&\,\, (x - 3)(x^2 + 12x + 35). \end{align*}

Comparing this to f(x)=(x−3)g(x)\mathrm{f}(x) = (x - 3)\mathrm{g}(x), the quadratic expression is:

g(x)=  x2+12x+35.\begin{align*} \mathrm{g}(x) =&\,\, x^2 + 12x + 35. \end{align*}