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IAL 2025 Jan Q2

A Level / Edexcel / FP2

IAL 2025 Jan Paper · Question 2

题目

Problem

(a) Use algebra to determine the exact xx coordinates of the points of intersection of the curves with equations

y=2xx2+1andy=1x+4\begin{align*} y=\frac{2x}{x^2+1} \qquad \text{and} \qquad y=\frac{1}{x+4} \end{align*}
(3)

Hence

(b) determine the values of xx for which

2xx2+1<1x+4\begin{align*} \frac{2x}{x^2+1}<\frac{1}{x+4} \end{align*}
(2)

(c) state the values of xx for which

2xx2+1<1∣x+4∣\begin{align*} \frac{2x}{x^2+1}<\frac{1}{|x+4|} \end{align*}
(2)

解答

(a)

解法一

思路

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交点的 xx 坐标由两个 yy 表达式相等得到。由于 x2+1x^2+1 恒正，x+4x+4 在交点处不能为 00，所以可以交叉相乘并解二次方程。

答题过程

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At the points of intersection,

2xx2+1=1x+4\begin{align*} \frac{2x}{x^2+1} ={}& \frac1{x+4} \end{align*}

Cross-multiply:

2x(x+4)=x2+12x2+8x=x2+1x2+8x−1=0\begin{align*} 2x(x+4) ={}& x^2+1\\[2mm] 2x^2+8x ={}& x^2+1\\[2mm] x^2+8x-1 ={}& 0 \end{align*}

Using the quadratic formula,

x=−8±82−4(1)(−1)2=−8±682=−4±17\begin{align*} x ={}& \frac{-8\pm\sqrt{8^2-4(1)(-1)}}{2}\\[3mm] ={}& \frac{-8\pm\sqrt{68}}{2}\\[3mm] ={}& -4\pm\sqrt{17} \end{align*}

Therefore the exact xx coordinates are

x=−4−17,x=−4+17\begin{align*} \boxed{x=-4-\sqrt{17},\quad x=-4+\sqrt{17}} \end{align*}

解法二

思路

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将两式移到等号同一边，通过通分化为单一分式，即 2xx2+1−1x+4=0\dfrac{2x}{x^2+1} - \dfrac{1}{x+4} = 0。通分后，由于分母不为零，令分子等于 00 求解二次方程即可得到交点的 xx 坐标。此法也为后续 (b) 部分求解不等式直接提供了代数工具。

答题过程

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At the points of intersection, write the equation by moving all terms to one side:

2xx2+1−1x+4=  0\begin{align*} \frac{2x}{x^2+1} - \frac{1}{x+4} =&\,\, 0 \end{align*}

Combine the fractions over a common denominator:

2x(x+4)−(x2+1)(x2+1)(x+4)=  02x2+8x−x2−1(x2+1)(x+4)=  0x2+8x−1(x2+1)(x+4)=  0\begin{align*} \frac{2x(x+4) - (x^2+1)}{(x^2+1)(x+4)} =&\,\, 0\\[4mm] \frac{2x^2+8x-x^2-1}{(x^2+1)(x+4)} =&\,\, 0\\[4mm] \frac{x^2+8x-1}{(x^2+1)(x+4)} =&\,\, 0 \end{align*}

Since the denominator cannot be zero at the points of intersection, set the numerator to zero:

x2+8x−1=  0\begin{align*} x^2+8x-1 =&\,\, 0 \end{align*}

Solve using the quadratic formula:

x=  −8±82−4(1)(−1)2=  −8±682=  −4±17\begin{align*} x =&\,\, \frac{-8\pm\sqrt{8^2-4(1)(-1)}}{2}\\[4mm] =&\,\, \frac{-8\pm\sqrt{68}}{2}\\[4mm] =&\,\, -4\pm\sqrt{17} \end{align*}

Therefore the exact xx coordinates are:

x=−4−17,x=−4+17\begin{align*} \boxed{x=-4-\sqrt{17},\quad x=-4+\sqrt{17}} \end{align*}

(b)

解法一

思路

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用 (a) 的两个交点，再加上 x=−4x=-4 这个竖直渐近线，把数轴分段。由于 x2+1>0x^2+1>0，不等式的符号变化主要来自交点和 x+4x+4。

答题过程

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Move everything to one side:

2xx2+1−1x+4<02x(x+4)−(x2+1)(x2+1)(x+4)<0x2+8x−1(x2+1)(x+4)<0\begin{align*} \frac{2x}{x^2+1}-\frac1{x+4} <& 0\\[3mm] \frac{2x(x+4)-(x^2+1)} {(x^2+1)(x+4)} <& 0\\[3mm] \frac{x^2+8x-1} {(x^2+1)(x+4)} <& 0 \end{align*}

Since

x2+8x−1=(x+4−17)(x+4+17),\begin{align*} x^2+8x-1 ={}& \left(x+4-\sqrt{17}\right) \left(x+4+\sqrt{17}\right), \end{align*}

the critical values are

−4−17,−4,−4+17\begin{align*} -4-\sqrt{17},\qquad -4,\qquad -4+\sqrt{17} \end{align*}

A sign table gives the expression negative on

x<−4−17or−4<x<−4+17\begin{align*} \boxed{ x<-4-\sqrt{17} \quad \text{or} \quad -4<x<-4+\sqrt{17} } \end{align*}

(c)

解法一

思路

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绝对值会把 1x+4\frac1{x+4} 的负半支反到正值。分两段看：当 x>−4x>-4 时，题目和 (b) 一样；当 x<−4x<-4 时，右边变成 −1x+4-\frac1{x+4}，此时右边为正，而左边在 x<0x<0 时为负，所以全部成立。

答题过程

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For x>−4x>-4,

∣x+4∣=x+4\begin{align*} |x+4|=x+4 \end{align*}

so the inequality is the same as in part (b). This gives

−4<x<−4+17\begin{align*} -4<x<-4+\sqrt{17} \end{align*}

For x<−4x<-4,

∣x+4∣=−(x+4)\begin{align*} |x+4|=-(x+4) \end{align*}

so

1∣x+4∣>0\begin{align*} \frac1{|x+4|}>0 \end{align*}

But for x<−4x<-4,

2xx2+1<0\begin{align*} \frac{2x}{x^2+1}<0 \end{align*}

Therefore the inequality is true for all x<−4x<-4. Since x=−4x=-4 is not in the domain, the solution is

x<−4or−4<x<−4+17\begin{align*} \boxed{ x<-4 \quad \text{or} \quad -4<x<-4+\sqrt{17} } \end{align*}