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IAL 2026 Jan A Q5

A Level / Edexcel / P1

IAL 2026 Jan A Paper · Question 5

题目

Problem

Figure 2 shows a sketch of part of the curve with equation y=f(x)y=f(x).

Figure 2

The curve crosses the yy-axis at the point (0,8)(0,8).

The line with equation y=10y=10 is the only asymptote to the curve.

The curve has a single turning point, a minimum point at (2,5)(2,5), as shown in Figure 2.

(a) State the coordinates of the minimum point of the curve with equation

y=f(14x)\begin{align*} y=f\left(\frac14x\right) \end{align*}
(1)

(b) State the equation of the asymptote to the curve with equation y=f(x)3y=f(x)-3

(1)

The curve with equation y=f(x)y=f(x) meets the line with equation y=ky=k, where kk is a constant, at two distinct points.

(c) State the set of possible values for kk.

(2)

(d) Sketch the curve with equation y=f(x)y=-f(x). On your sketch, show clearly the coordinates of the turning point, the coordinates of the intersection with the yy-axis and the equation of the asymptote.

(3)

解答

(a)

解法一

思路

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y=f(14x)y=f\left(\frac14x\right) 表示水平方向放大 44 倍。原来的最小点 (2,5)(2,5) 变成 xx 坐标乘以 44yy 坐标不变。

答题过程

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The transformation

y=f(14x)\begin{align*} y=f\left(\frac14x\right) \end{align*}

is a stretch parallel to the xx-axis with scale factor 44.

Therefore

(2,5)(8,5).\begin{align*} (2,5)\mapsto(8,5). \end{align*}

The minimum point is

(8,5).\begin{align*} (8,5). \end{align*}

(b)

解法一

思路

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y=f(x)3y=f(x)-3 表示整体向下平移 33 个单位,所以水平渐近线也向下平移 33 个单位。

答题过程

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The original asymptote is

y=10.\begin{align*} y=10. \end{align*}

For y=f(x)3y=f(x)-3, it is translated down by 33 units, so the asymptote is

y=7.\begin{align*} y=7. \end{align*}

(c)

解法一

思路

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水平线 y=ky=k 要与曲线有两个不同交点。根据图像,最低点是 y=5y=5,渐近线是 y=10y=10。只有当水平线在最低点上方、渐近线下方时,才会切过左右两支各一次。

答题过程

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For two distinct intersections, the horizontal line y=ky=k must lie above the minimum value and below the asymptote.

Hence

5<k<10.\begin{align*} 5<k<10. \end{align*}

(d)

解法一

思路

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y=f(x)y=-f(x) 是把原图像关于 xx 轴反射。所有 yy 坐标变号,xx 坐标不变;最小点会变成最大点,水平渐近线 y=10y=10 会变成 y=10y=-10

答题过程

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Under the transformation y=f(x)y=-f(x):

(2,5)(2,5),(0,8)(0,8),y=10y=10.\begin{align*} (2,5)&\mapsto(2,-5),\\ (0,8)&\mapsto(0,-8),\\ y=10&\mapsto y=-10. \end{align*}

So the sketch should be the reflection of the original curve in the xx-axis, with:

turning point =(2,5),intersection with the y-axis =(0,8),asymptote :y=10.\begin{align*} \text{turning point }=&\,(2,-5),\\ \text{intersection with the }y\text{-axis }=&\,(0,-8),\\ \text{asymptote }&:y=-10. \end{align*}