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IAL 2024 May FP1 Q5

A Level / Edexcel / FP1

IAL 2024 May Paper · Question 5

题目

Problem

The equation 5x24x+2=05x^2 - 4x + 2 = 0 has roots 1p\dfrac{1}{p} and 1q\dfrac{1}{q}

(a) Without solving the equation,

(i) show that pq=52pq = \dfrac{5}{2}

(ii) determine the value of p+qp + q

(4)

(b) Hence, without finding the values of pp and qq, determine a quadratic equation with roots

pp2+1andqq2+1\frac{p}{p^2 + 1} \quad \text{and} \quad \frac{q}{q^2 + 1}

giving your answer in the form ax2+bx+c=0ax^2 + bx + c = 0 where aa, bb and cc are integers.

(5)
题目中文翻译

方程 5x24x+2=05x^2 - 4x + 2 = 0 的两个根为 1p\dfrac{1}{p}1q\dfrac{1}{q}

(a) 在不解方程的情况下:

(i) 证明 pq=52pq = \dfrac{5}{2}

(ii) 确定 p+qp + q 的值

(b) 由此,在不求出 ppqq 的具体值的情况下,确定一个以 pp2+1qq2+1\frac{p}{p^2 + 1} \quad \text{与} \quad \frac{q}{q^2 + 1}

为根的二次方程,将答案表示为 ax2+bx+c=0ax^2 + bx + c = 0 的形式,其中 aabbcc 均为整数。

解答

(a)

解法一

思路

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直接对原方程的两个根 1/p1/p1/q1/q 使用根与系数关系。先由根之积求 pqpq,再把根之和写成 (p+q)/(pq)(p+q)/(pq),并使用前一结果求 p+qp+q

答题过程

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(i)

Using the product of the roots,

1p×1q=25.\frac1p\times\frac1q=\frac25.

Therefore,

1pq=25pq=52.\frac1{pq}=\frac25 \quad\Longrightarrow\quad \boxed{pq=\frac52}.

(ii)

Using the sum of the roots,

1p+1q=45.\frac1p+\frac1q=\frac45.

Hence, using part (i),

p+qpq=45p+q=45(52)=2.\begin{align*} \frac{p+q}{pq} =&\,\frac45\\ p+q =&\,\frac45\left(\frac52\right)\\ =&\,\boxed{2}. \end{align*}

解法二

思路

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官方评分资料还给出了变量替换法。令 x=1/ux=1/u 并整理,可得到一个以 p,qp,q 为根的新二次方程,再直接读取其根之和与根之积。

答题过程

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Let

x=1u.x=\frac1u.

Substituting into 5x24x+2=05x^2-4x+2=0 gives

5(1u)24(1u)+2=0.5\left(\frac1u\right)^2 -4\left(\frac1u\right)+2=0.

Multiplying by u2u^2,

2u24u+5=0.2u^2-4u+5=0.

Since its roots are pp and qq, the sum and product of the roots give

p+q=2,pq=52.\boxed{p+q=2, \qquad pq=\frac52}.

(b)

解法一

思路

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设两个新根为 u,vu,v,分别求其和与积。通分后,把所有对称式改写成只含 p+qp+qpqpq 的形式,再使用 (a) 的结果。最后用“x2x^2- 根之和乘 x+x+ 根之积”建立方程。

答题过程

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Let

u=pp2+1,v=qq2+1.u=\frac{p}{p^2+1}, \qquad v=\frac{q}{q^2+1}.

First,

(p2+1)(q2+1)=p2q2+p2+q2+1=(pq)2+(p+q)22pq+1=(52)2+222(52)+1=254.\begin{align*} &(p^2+1)(q^2+1)\\ =&\,p^2q^2+p^2+q^2+1\\ =&\,(pq)^2+(p+q)^2-2pq+1\\ =&\,\left(\frac52\right)^2+2^2 -2\left(\frac52\right)+1\\ =&\,\frac{25}{4}. \end{align*}

The sum of the new roots is

u+v=p(q2+1)+q(p2+1)(p2+1)(q2+1)=pq(p+q)+(p+q)25/4=(p+q)(pq+1)25/4=2(52+1)25/4=2825.\begin{align*} u+v =&\,\frac{p(q^2+1)+q(p^2+1)} {(p^2+1)(q^2+1)}\\ =&\,\frac{pq(p+q)+(p+q)}{25/4}\\ =&\,\frac{(p+q)(pq+1)}{25/4}\\ =&\,\frac{2\left(\frac52+1\right)}{25/4}\\ =&\,\frac{28}{25}. \end{align*}

Their product is

uv=pq(p2+1)(q2+1)=5/225/4=25.\begin{align*} uv =&\,\frac{pq}{(p^2+1)(q^2+1)}\\ =&\,\frac{5/2}{25/4}\\ =&\,\frac25. \end{align*}

Therefore, the required equation is

x22825x+25=0.x^2-\frac{28}{25}x+\frac25=0.

Multiplying by 25 gives

25x228x+10=0.\boxed{25x^2-28x+10=0}.