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IAL 2022 Oct Q7

A Level / Edexcel / P3

IAL 2022 Oct Paper · Question 7

题目

Problem

Figure 3 shows a sketch of the graph of C1C_1 with equation

y=5−∣3x−22∣y=5-\lvert 3x-22\rvert

(a) Write down the coordinates of

(i) the vertex of C1C_1

(ii) the intersection of C1C_1 with the yy-axis.

(2)

(b) Find the xx coordinates of the intersections of C1C_1 with the xx-axis.

(2)

Diagram 1, shown on page 21, is a copy of Figure 3.

(c) On Diagram 1, sketch the curve C2C_2 with equation

y=19x2−9y=\frac19x^2-9

Identify clearly the coordinates of any points of intersection of C2C_2 with the coordinate axes.

(3)

(d) Find the coordinates of the points of intersection of C1C_1 and C2C_2.

Solutions relying entirely on calculator technology are not acceptable.

(5)
题目中文翻译

图 3 给出了图像 C1C_1 的示意图，其方程为

y=5−∣3x−22∣y=5-\lvert 3x-22\rvert

(a) 写出

(i) C1C_1 的顶点坐标；

(ii) C1C_1 与 yy 轴交点的坐标。

(b) 求 C1C_1 与 xx 轴交点的 xx 坐标。

第 21 页上的 Diagram 1 是 Figure 3 的副本。

(c) 在 Diagram 1 上画出曲线 C2C_2

y=19x2−9y=\frac19x^2-9

并清楚标出 C2C_2 与坐标轴交点的坐标。

(d) 求 C1C_1 与 C2C_2 交点的坐标。

不接受完全依赖计算器技术的解法。

解答

(a)(i)

The vertex occurs when the expression inside the modulus is zero:

3x−22=03x-22=0

So

x=223x=\frac{22}{3}

At this point,

y=5−0=5y=5-0=5

Therefore the vertex is

(223,5)\boxed{\left(\frac{22}{3},5\right)}

(a)(ii)

At the yy-axis,

x=0x=0

So

y=5−∣3(0)−22∣y=5-|3(0)-22|

Hence

y=5−22=−17y=5-22=-17

Therefore the intersection with the yy-axis is

(0,−17)\boxed{(0,-17)}

(b)

For intersections with the xx-axis,

y=0y=0

So

5−∣3x−22∣=05-|3x-22|=0

Hence

∣3x−22∣=5|3x-22|=5

Therefore

3x−22=53x-22=5

or

3x−22=−53x-22=-5

So

x=9x=9

or

x=173x=\frac{17}{3}

Thus the xx coordinates are

x=173, 9\boxed{x=\frac{17}{3},\ 9}

(c)

The curve

y=19x2−9y=\frac19x^2-9

is an upward-opening parabola.

Its yy-intercept is found by setting x=0x=0:

y=−9y=-9

so the yy-intercept is

(0,−9)(0,-9)

Its xx-intercepts are found by setting y=0y=0:

19x2−9=0\frac19x^2-9=0

So

x2=81x^2=81

and hence

x=±9x=\pm9

Thus the xx-intercepts are

(−9,0)and(9,0)(-9,0) \qquad\text{and}\qquad (9,0)

(d)

One intersection is already visible from part (b) and part (c):

(9,0)(9,0)

For the other intersection, use the left branch of C1C_1.

When

x<223x<\frac{22}{3}

we have

∣3x−22∣=22−3x|3x-22|=22-3x

so

y=5−(22−3x)=3x−17y=5-(22-3x)=3x-17

Set this equal to the equation of C2C_2:

3x−17=19x2−93x-17=\frac19x^2-9

Multiply by 99:

27x−153=x2−8127x-153=x^2-81

So

x2−27x+72=0x^2-27x+72=0

Factorise:

(x−3)(x−24)=0(x-3)(x-24)=0

On this branch, the valid solution is

x=3x=3

Then

y=3(3)−17=−8y=3(3)-17=-8

Therefore the points of intersection are

(3,−8) and (9,0)\boxed{(3,-8)\ \text{and}\ (9,0)}