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P1 2025 May A Q7

A Level / Edexcel / P1

IAL 2025 May A Paper · Question 7

题目

Problem

In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable.

(a) Sketch the curve CC with equation

y=1x+6.\begin{align*} y=\frac1{x+6}. \end{align*}

State on your sketch

  • the equation of the vertical asymptote
  • the coordinates of the point of intersection of CC with the yy-axis.
(3)

The straight line ll has equation y=mx−4y=mx-4, where mm is a constant.

Given that ll cuts CC at least once,

(b) (i) show that

9m2+13m+4≥0\begin{align*} 9m^2+13m+4\geq0 \end{align*}

(ii) find the range of values of mm.

(6)
题目中文翻译

本题必须写出完整的解题步骤。 不接受完全依赖计算器技术所得的解答。

(a) 画出曲线 CC 的草图，其方程为

y=1x+6.\begin{align*} y=\frac1{x+6}. \end{align*}

在草图上标出

  • 竖直渐近线的方程
  • CC 与 yy 轴交点的坐标。

直线 ll 的方程为 y=mx−4y=mx-4，其中 mm 为常数。

已知 ll 与 CC 至少相交一次。

(b) (i) 证明

9m2+13m+4≥0.\begin{align*} 9m^2+13m+4\geq0. \end{align*}

(ii) 求 mm 的取值范围。

解答

(a)

解法一

思路

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y=1x+6y=\frac1{x+6} 是 y=1xy=\frac1x 向左平移 66 个单位。竖直渐近线从 x=0x=0 变成 x=−6x=-6，水平渐近线仍是 y=0y=0。

与 yy 轴交点令 x=0x=0。

答题过程

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The curve

y=1x+6\begin{align*} y=\frac1{x+6} \end{align*}

has vertical asymptote

x=−6.\begin{align*} x=-6. \end{align*}

The yy-intercept is found by putting x=0x=0:

y= 10+6= 16.\begin{align*} y=&\,\frac1{0+6}\\ =&\,\frac16. \end{align*}

So the curve intersects the yy-axis at

(0,16).\begin{align*} \left(0,\frac16\right). \end{align*}

The sketch should have one branch above the xx-axis for x>−6x>-6, and one branch below the xx-axis for x<−6x<-6.

A completed sketch is:

(b)(i)

解法一

思路

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交点满足直线和曲线的 yy 相等。整理后会得到一个关于 xx 的二次方程。若直线至少与曲线相交一次，这个二次方程必须有实根，所以判别式 Δ≥0\Delta\geq0。

答题过程

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At intersections,

mx−4= 1x+6.\begin{align*} mx-4=&\,\frac1{x+6}. \end{align*}

Multiply by x+6x+6:

(mx−4)(x+6)= 1.\begin{align*} (mx-4)(x+6)=&\,1. \end{align*}

Expand and collect terms:

mx2+6mx−4x−24= 1mx2+(6m−4)x−25= 0.\begin{align*} mx^2+6mx-4x-24=&\,1\\ mx^2+(6m-4)x-25=&\,0. \end{align*}

For the line to cut the curve at least once, this quadratic in xx must have real roots. Therefore

Δ≥0(6m−4)2−4(m)(−25)≥0(6m−4)2+100m≥0.\begin{align*} \Delta&\geq0\\ (6m-4)^2-4(m)(-25)&\geq0\\ (6m-4)^2+100m&\geq0. \end{align*}

Now expand:

36m2−48m+16+100m≥036m2+52m+16≥09m2+13m+4≥0.\begin{align*} 36m^2-48m+16+100m&\geq0\\ 36m^2+52m+16&\geq0\\ 9m^2+13m+4&\geq0. \end{align*}

This is the required result.

(b)(ii)

解法一

思路

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解二次不等式 9m2+13m+4≥09m^2+13m+4\geq0。先因式分解，找临界值，再根据开口向上选择外侧区间。

答题过程

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Solve

9m2+13m+4≥0.\begin{align*} 9m^2+13m+4&\geq0. \end{align*}

Factorise:

9m2+13m+4= (9m+4)(m+1).\begin{align*} 9m^2+13m+4=&\,(9m+4)(m+1). \end{align*}

The critical values are

m=−1,m=−49.\begin{align*} m=-1,\qquad m=-\frac49. \end{align*}

Since the quadratic has positive leading coefficient, it is non-negative outside the roots. Therefore

m≤−1orm≥−49.\begin{align*} m\leq-1\quad\text{or}\quad m\geq-\frac49. \end{align*}