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IAL 2022 Oct Q7

A Level / Edexcel / P2

IAL 2022 Oct Paper · Question 7

题目

Problem

The curve CC has equation

y=12x4−84x3+182x2−210x21xx>0y = \frac{12x^4 - 84x^3 + 182x^2 - 210x}{21\sqrt{x}} \qquad x > 0

(a) Write the equation of CC in the form

y=ax72+bx52+cx32+dx12y = ax^{\frac{7}{2}} + bx^{\frac{5}{2}} + cx^{\frac{3}{2}} + dx^{\frac{1}{2}}

where aa, bb, cc and dd are fully simplified constants.

(3)

The curve CC has three turning points.

Using calculus,

(b) show that the xx coordinates of the three turning points satisfy the equation

2x3−10x2+13x−5=02x^3 - 10x^2 + 13x - 5 = 0

(3)

Given that the xx coordinate of one of the turning points is 11

(c) find, using algebra, the exact xx coordinates of the other two turning points.

(Solutions based entirely on calculator technology are not acceptable.)

(3)

解答

(a)

解法一

思路

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本题要求我们将分式形式的曲线方程化简为多项式指数形式。 分母为 21x=21x1/221\sqrt{x} = 21x^{1/2}。我们将分子上的每一项都除以分母：

y=12x421x1/2−84x321x1/2+182x221x1/2−210x21x1/2y = \frac{12x^4}{21x^{1/2}} - \frac{84x^3}{21x^{1/2}} + \frac{182x^2}{21x^{1/2}} - \frac{210x}{21x^{1/2}}

利用指数法则 xmxn=xm−n\dfrac{x^m}{x^n} = x^{m-n} 化简每一项的指数，并约分系数：

  • 第一项：1221=47\dfrac{12}{21} = \dfrac{4}{7}，指数 4−12=724 - \dfrac{1}{2} = \dfrac{7}{2}。
  • 第二项：8421=4\dfrac{84}{21} = 4，指数 3−12=523 - \dfrac{1}{2} = \dfrac{5}{2}。
  • 第三项：18221=263\dfrac{182}{21} = \dfrac{26}{3}（因为 182182 和 2121 都能被 7 整除，182÷7=26182 \div 7 = 26），指数 2−12=322 - \dfrac{1}{2} = \dfrac{3}{2}。
  • 第四项：21021=10\dfrac{210}{21} = 10，指数 1−12=121 - \dfrac{1}{2} = \dfrac{1}{2}。

答题过程

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Express the denominator as 21x1221x^{\frac{1}{2}}. Divide each term in the numerator by the denominator:

y=  12x4−84x3+182x2−210x21x12=  12x421x12−84x321x12+182x221x12−210x21x12=  47x4−12−4x3−12+263x2−12−10x1−12=  47x72−4x52+263x32−10x12\begin{align*} y =&\,\, \frac{12x^4 - 84x^3 + 182x^2 - 210x}{21x^{\frac{1}{2}}} \\[2mm] =&\,\, \frac{12x^4}{21x^{\frac{1}{2}}} - \frac{84x^3}{21x^{\frac{1}{2}}} + \frac{182x^2}{21x^{\frac{1}{2}}} - \frac{210x}{21x^{\frac{1}{2}}} \\[2mm] =&\,\, \frac{4}{7}x^{4 - \frac{1}{2}} - 4x^{3 - \frac{1}{2}} + \frac{26}{3}x^{2 - \frac{1}{2}} - 10x^{1 - \frac{1}{2}} \\[2mm] =&\,\, \frac{4}{7}x^{\frac{7}{2}} - 4x^{\frac{5}{2}} + \frac{26}{3}x^{\frac{3}{2}} - 10x^{\frac{1}{2}} \end{align*}

Thus, the constants are:

a=47,b=−4,c=263,d=−10a = \frac{4}{7}, \quad b = -4, \quad c = \frac{26}{3}, \quad d = -10

(b)

解法一

思路

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求拐点/驻点（Turning points）需要用到微积分，当导数 dydx=0\dfrac{\mathrm{d}y}{\mathrm{d}x} = 0 时成立。 对 (a) 中求得的 yy 的表达式进行求导：

y=47x72−4x52+263x32−10x12y = \frac{4}{7}x^{\frac{7}{2}} - 4x^{\frac{5}{2}} + \frac{26}{3}x^{\frac{3}{2}} - 10x^{\frac{1}{2}}

运用求导的幂法则（ddxxn=nxn−1\frac{\mathrm{d}}{\mathrm{d}x} x^n = n x^{n-1}）：

dydx=47(72)x52−4(52)x32+263(32)x12−10(12)x−12\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{4}{7}\left(\frac{7}{2}\right)x^{\frac{5}{2}} - 4\left(\frac{5}{2}\right)x^{\frac{3}{2}} + \frac{26}{3}\left(\frac{3}{2}\right)x^{\frac{1}{2}} - 10\left(\frac{1}{2}\right)x^{-\frac{1}{2}}

化简得到：

dydx=2x52−10x32+13x12−5x−12\frac{\mathrm{d}y}{\mathrm{d}x} = 2x^{\frac{5}{2}} - 10x^{\frac{3}{2}} + 13x^{\frac{1}{2}} - 5x^{-\frac{1}{2}}

令 dydx=0\dfrac{\mathrm{d}y}{\mathrm{d}x} = 0：

2x52−10x32+13x12−5x−12=02x^{\frac{5}{2}} - 10x^{\frac{3}{2}} + 13x^{\frac{1}{2}} - 5x^{-\frac{1}{2}} = 0

因为 x>0x > 0，所以 x1/2≠0x^{1/2} \neq 0。两边同时乘以 x1/2x^{1/2} 消除负指数项：

2x3−10x2+13x−5=02x^3 - 10x^2 + 13x - 5 = 0

即可证出。

答题过程

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Differentiating yy with respect to xx:

dydx=  ddx(47x72−4x52+263x32−10x12)=  47(72)x52−4(52)x32+263(32)x12−10(12)x−12=  2x52−10x32+13x12−5x−12\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} =&\,\, \frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{4}{7}x^{\frac{7}{2}} - 4x^{\frac{5}{2}} + \frac{26}{3}x^{\frac{3}{2}} - 10x^{\frac{1}{2}}\right) \\[2mm] =&\,\, \frac{4}{7}\left(\frac{7}{2}\right)x^{\frac{5}{2}} - 4\left(\frac{5}{2}\right)x^{\frac{3}{2}} + \frac{26}{3}\left(\frac{3}{2}\right)x^{\frac{1}{2}} - 10\left(\frac{1}{2}\right)x^{-\frac{1}{2}} \\[2mm] =&\,\, 2x^{\frac{5}{2}} - 10x^{\frac{3}{2}} + 13x^{\frac{1}{2}} - 5x^{-\frac{1}{2}} \end{align*}

At turning points, the gradient dydx=0\frac{\mathrm{d}y}{\mathrm{d}x} = 0:

2x52−10x32+13x12−5x−12=02x^{\frac{5}{2}} - 10x^{\frac{3}{2}} + 13x^{\frac{1}{2}} - 5x^{-\frac{1}{2}} = 0

Since x>0x > 0, we can multiply both sides by x12x^{\frac{1}{2}}:

(2x52−10x32+13x12−5x−12)×x12=  0×x122x3−10x2+13x−5=  0(proven)\begin{align*} \left(2x^{\frac{5}{2}} - 10x^{\frac{3}{2}} + 13x^{\frac{1}{2}} - 5x^{-\frac{1}{2}}\right) \times x^{\frac{1}{2}} =&\,\, 0 \times x^{\frac{1}{2}} \\[2mm] 2x^3 - 10x^2 + 13x - 5 =&\,\, 0 \quad \text{(proven)} \end{align*}

(c)

解法一

思路

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已知三次方程的一个根是 x=1x = 1。这意味着 (x−1)(x - 1) 是多项式 2x3−10x2+13x−52x^3 - 10x^2 + 13x - 5 的一个因式。 我们需要进行多项式除法（Long Division）或因式拆分，求出另一个二次因式：

2x3−10x2+13x−5=(x−1)(2x2+px+q)2x^3 - 10x^2 + 13x - 5 = (x - 1)(2x^2 + px + q)

通过展开对比系数或直接进行多项式除法：

  • 设二次项为 2x22x^2。
  • 展开后常数项：−1×q=−5  ⟹  q=5-1 \times q = -5 \implies q = 5。
  • 一次项系数：q−p=13  ⟹  5−p=13  ⟹  p=−8q - p = 13 \implies 5 - p = 13 \implies p = -8。 所以二次因式为 2x2−8x+52x^2 - 8x + 5。

然后，令 2x2−8x+5=02x^2 - 8x + 5 = 0，通过求根公式解出剩下的两个 xx 精确值：

x=−(−8)±(−8)2−4(2)(5)2(2)=8±64−404=8±244=8±264=4±62x = \frac{-(-8) \pm \sqrt{(-8)^2 - 4(2)(5)}}{2(2)} = \frac{8 \pm \sqrt{64 - 40}}{4} = \frac{8 \pm \sqrt{24}}{4} = \frac{8 \pm 2\sqrt{6}}{4} = \frac{4 \pm \sqrt{6}}{2}

由于题目要求“exact coordinates”（精确值），不能使用近似的小数，答案应当保留根式。

答题过程

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Given that x=1x = 1 is a root of the cubic equation, (x−1)(x - 1) is a factor of the expression.

Using algebraic division or factorisation:

2x3−10x2+13x−5=(x−1)(2x2−8x+5)2x^3 - 10x^2 + 13x - 5 = (x - 1)(2x^2 - 8x + 5)

Setting the remaining quadratic factor to 00 to find the other two roots:

2x2−8x+5=02x^2 - 8x + 5 = 0

Using the quadratic formula:

x=  −(−8)±(−8)2−4(2)(5)2(2)=  8±64−404=  8±244=  8±264=  4±62(or 2±62)\begin{align*} x =&\,\, \frac{-(-8) \pm \sqrt{(-8)^2 - 4(2)(5)}}{2(2)} \\[2mm] =&\,\, \frac{8 \pm \sqrt{64 - 40}}{4} \\[2mm] =&\,\, \frac{8 \pm \sqrt{24}}{4} \\[2mm] =&\,\, \frac{8 \pm 2\sqrt{6}}{4} \\[2mm] =&\,\, \frac{4 \pm \sqrt{6}}{2} \quad \left(\text{or } 2 \pm \frac{\sqrt{6}}{2}\right) \end{align*}

Therefore, the exact xx coordinates of the other two turning points are:

x=4−62,x=4+62x = \frac{4 - \sqrt{6}}{2}, \quad x = \frac{4 + \sqrt{6}}{2}