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IAL 2022 Oct Q2

A Level / Edexcel / P2

IAL 2022 Oct Paper · Question 2

题目

Problem

A curve CC has equation y=f(x)y = \mathrm{f}(x) where

f(x)=(2−kx)5\mathrm{f}(x) = (2 - kx)^5

and kk is a constant.

Given that when f(x)\mathrm{f}(x) is divided by (4x−5)(4x - 5) the remainder is 24332\dfrac{243}{32}

(a) show that k=25k = \dfrac{2}{5}

(2)

(b) Find the first three terms, in ascending powers of xx, of the binomial expansion of

(2−25x)5\left(2 - \frac{2}{5}x\right)^5

giving each term in simplest form.

(3)

Using the solution to part (b) and making your method clear,

(c) find the gradient of CC at the point where x=0x = 0

(2)

解答

(a)

解法一

思路

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根据余数定理（Remainder Theorem），当多项式 f(x)\mathrm{f}(x) 除以一次因式 (ax−b)(ax - b) 时，所得的余数等于 f(ba)\mathrm{f}\left(\dfrac{b}{a}\right)。 在本题中，除式为 4x−54x - 5，令其为零得到 x=54x = \dfrac{5}{4}。 因此余数可以表示为 f(54)\mathrm{f}\left(\dfrac{5}{4}\right)，我们将其代入原方程并等于已知余数 24332\dfrac{243}{32}：

(2−k⋅54)5=24332\left(2 - k \cdot \frac{5}{4}\right)^5 = \frac{243}{32}

由于 243=35243 = 3^5 且 32=2532 = 2^5，上式右边即为 (32)5\left(\dfrac{3}{2}\right)^5。两边同时开五次方根即可解出 kk。

答题过程

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According to the Remainder Theorem, when f(x)\mathrm{f}(x) is divided by (4x−5)(4x - 5), the remainder is f(54)\mathrm{f}\left(\dfrac{5}{4}\right).

Given that the remainder is 24332\dfrac{243}{32}:

f(54)=  24332(2−k⋅54)5=  24332\begin{align*} \mathrm{f}\left(\frac{5}{4}\right) =&\,\, \frac{243}{32} \\[2mm] \left(2 - k \cdot \frac{5}{4}\right)^5 =&\,\, \frac{243}{32} \end{align*}

Taking the 5th5\text{th} root of both sides:

2−54k=  2433252−54k=  3254k=  2−3254k=  12k=  12×45k=  25(proven)\begin{align*} 2 - \frac{5}{4}k =&\,\, \sqrt[5]{\frac{243}{32}} \\[2mm] 2 - \frac{5}{4}k =&\,\, \frac{3}{2} \\[2mm] \frac{5}{4}k =&\,\, 2 - \frac{3}{2} \\[2mm] \frac{5}{4}k =&\,\, \frac{1}{2} \\[2mm] k =&\,\, \frac{1}{2} \times \frac{4}{5} \\[2mm] k =&\,\, \frac{2}{5} \quad \text{(proven)} \end{align*}

(b)

解法一

思路

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使用二项式定理（Binomial Theorem）展开式公式：

(a+b)n=an+(n1)an−1b+(n2)an−2b2+⋯(a + b)^n = a^n + \binom{n}{1}a^{n-1}b + \binom{n}{2}a^{n-2}b^2 + \cdots

这里 a=2a = 2，b=−25xb = -\dfrac{2}{5}x，n=5n = 5。我们需要求前三项（即含 x0x^0, x1x^1 和 x2x^2 的项），并将其系数化简到最简形式。

答题过程

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Using the binomial expansion formula for (2−25x)5\left(2 - \dfrac{2}{5}x\right)^5:

(2−25x)5=  25+(51)(2)4(−25x)+(52)(2)3(−25x)2+⋯=  32+5×16×(−25x)+10×8×(425x2)+⋯=  32−32x+645x2+⋯\begin{align*} \left(2 - \frac{2}{5}x\right)^5 =&\,\, 2^5 + \binom{5}{1}(2)^4\left(-\frac{2}{5}x\right) + \binom{5}{2}(2)^3\left(-\frac{2}{5}x\right)^2 + \cdots \\[2mm] =&\,\, 32 + 5 \times 16 \times \left(-\frac{2}{5}x\right) + 10 \times 8 \times \left(\frac{4}{25}x^2\right) + \cdots \\[2mm] =&\,\, 32 - 32x + \frac{64}{5}x^2 + \cdots \end{align*}

Thus, the first three terms in ascending powers of xx are:

32−32x+12.8x2(or 32−32x+645x2)32 - 32x + 12.8x^2 \quad \left(\text{or } 32 - 32x + \frac{64}{5}x^2\right)

(c)

解法一：利用 (b) 的二项式展开式求导

思路

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曲线 CC 在某一点处的斜率（Gradient）等于其导数 f′(x)\mathrm{f}'(x) 在该点的值。 题目要求**“利用 (b) 的结论”**，因此我们需要将 (b) 中求得的展开式对 xx 求导：

f(x)≈32−32x+645x2\mathrm{f}(x) \approx 32 - 32x + \frac{64}{5}x^2

求导后代入 x=0x=0，可得 f′(0)=−32\mathrm{f}'(0) = -32。

答题过程

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From part (b), the expansion of C:y=f(x)C: y = \mathrm{f}(x) for the first three terms is:

y≈32−32x+645x2y \approx 32 - 32x + \frac{64}{5}x^2

Differentiating with respect to xx:

dydx=  ddx(32−32x+645x2+⋯ )=  −32+1285x+⋯\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} =&\,\, \frac{\mathrm{d}}{\mathrm{d}x}\left(32 - 32x + \frac{64}{5}x^2 + \cdots\right) \\[2mm] =&\,\, -32 + \frac{128}{5}x + \cdots \end{align*}

At the point where x=0x = 0:

dydx∣x=0=  −32+0=  −32\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x}\bigg|_{x=0} =&\,\, -32 + 0 \\[2mm] =&\,\, -32 \end{align*}

Therefore, the gradient of CC at x=0x = 0 is −32-32.

解法二：直接求导法（链式法则验算）

思路

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作为备用解法（或验算方法），我们也可以直接对原函数 (2−25x)5(2 - \frac{2}{5}x)^5 运用链式法则（Chain Rule）求导并代入 x=0x = 0。

答题过程

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We can differentiate the original function y=(2−25x)5y = \left(2 - \dfrac{2}{5}x\right)^5 using the chain rule:

dydx=  5(2−25x)4×(−25)=  −2(2−25x)4\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} =&\,\, 5\left(2 - \frac{2}{5}x\right)^4 \times \left(-\frac{2}{5}\right) \\[2mm] =&\,\, -2\left(2 - \frac{2}{5}x\right)^4 \end{align*}

At the point where x=0x = 0:

dydx∣x=0=  −2(2−0)4=  −2(16)=  −32\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x}\bigg|_{x=0} =&\,\, -2(2 - 0)^4 \\[2mm] =&\,\, -2(16) \\[2mm] =&\,\, -32 \end{align*}