变量 x 和 y 由方程 y=32x+5x−7y = \frac{3}{2}x + \frac{5}{x} - 7 相关联。 x 和 y 的部分对应值如下表所示。
x0.511.5234567
y3.75-0.5-1.42p-0.830.251.52.834.21

  1. (a)

    求 p 的值。

    答案 p = ...

  2. (b)

    在对面的网格上,画出 0.5≤x≤70.5 \le x \le 7 范围内 y=32x+5x−7y = \frac{3}{2}x + \frac{5}{x} - 7 的图像。

    在两个轴上均使用 2 cm 表示 1 个单位的比例。

  3. (c)

    通过绘制切线,求曲线在 x=2.5x = 2.5 处的斜率。

    答案 ...

  4. (d)
    1. (i)

      在部分 (b) 的网格上,画出 0≤x≤70 \le x \le 7 范围内的直线 2y+x=22y + x = 2。

    2. (ii)

      写出直线与曲线相交点的 x 坐标。

      答案 x = ... 或 ...

    3. (iii)

      证明直线与曲线的交点给出了方程 2x2−8x+5=02x^{2} - 8x + 5 = 0 的解。

      答案 ...

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完整解法

4(a)

1. Identify the x-value corresponding to p and substitute it into the formula.

From the table, pp is the value of yy when x=2x=2.

p=32(2)+52−7p=\frac{3}{2}(2)+\frac{5}{2}-7

2. Evaluate the expression accurately.

p=3+2.5−7p=3+2.5-7
p=−1.5p=-1.5

答案

p=−1.5p=-1.5

4(b)

1. Use the table values, including the value found in part (a), to prepare the coordinates to be plotted.

From the table and part (a), the points are:

(0.5,3.75), (1,−0.5), (1.5,−1.42), (2,−1.5), (3,−0.83), (4,0.25), (5,1.5), (6,2.83), (7,4.21).(0.5,3.75),\ (1,-0.5),\ (1.5,-1.42),\ (2,-1.5),\ (3,-0.83),\ (4,0.25),\ (5,1.5),\ (6,2.83),\ (7,4.21).

2. Set up the axes using the required scale.

Use the scale 2 cm2\text{ cm} to represent 11 unit on both axes.

The horizontal axis should cover at least 0.5≤x≤70.5\le x\le 7, and the vertical axis should cover the given yy-values, from about −1.5-1.5 to 4.214.21.

3. Plot the points accurately and join them with a smooth curve.

Plot all the coordinates accurately on the grid.

Join the points with a smooth curve, not straight line segments, for 0.5≤x≤70.5\le x\le 7. The curve decreases from (0.5,3.75)(0.5,3.75) to a minimum near x=2x=2, then increases through the remaining points.

答案

A smooth curve for y=32x+5x−7y=\dfrac{3}{2}x+\dfrac{5}{x}-7 drawn through the plotted points for 0.5≤x≤70.5\le x\le 7, using the scale 2 cm:12\text{ cm}:1 unit on both axes.

4(c)

1. Locate the point on the curve at $x=2.5$ and draw a tangent there.

On the graph drawn in part (b), locate the point on the curve where x=2.5x=2.5. Draw a straight line that just touches the curve at this point and follows the direction of the curve there.

2. Use two points on the tangent to calculate its gradient.

Choose two clear points on the tangent line and calculate
gradient=vertical changehorizontal change.\text{gradient}=\frac{\text{vertical change}}{\text{horizontal change}}.
From a correctly drawn tangent, this gives approximately
gradient≈0.7.\text{gradient}\approx 0.7.

答案

Gradient ≈0.7\approx 0.7

4(d)(i)

1. Rearrange the line equation into a form suitable for plotting.

2y+x=22y+x=2
2y=2−x2y=2-x
y=1−x2y=1-\frac{x}{2}

2. Find two points on the line for the required interval and draw the line.

For 0≤x≤70\le x\le 7,

when x=0x=0,
y=1−02=1,y=1-\frac{0}{2}=1,
so one point is (0,1)(0,1).

When x=7x=7,
y=1−72=−2.5,y=1-\frac{7}{2}=-2.5,
so another point is (7,−2.5)(7,-2.5).

Plot (0,1)(0,1) and (7,−2.5)(7,-2.5), then join them with a straight line.

答案

Draw the straight line y=1−x2y=1-\dfrac{x}{2} from (0,1)(0,1) to (7,−2.5)(7,-2.5) on the grid.

4(d)(ii)

1. Identify the points where the straight line meets the curve on the completed graph.

From the graph, locate the two intersection points of the curve y=32x+5x−7y=\dfrac{3}{2}x+\dfrac{5}{x}-7 and the line 2y+x=22y+x=2.

2. Read off the corresponding x-coordinates.

Reading the xx-coordinates of these two points gives approximately
x=0.8andx=3.2.x=0.8 \quad \text{and} \quad x=3.2.

答案

x≈0.8x \approx 0.8 or x≈3.2x \approx 3.2

4(d)(iii)

1. Use the fact that, at an intersection, the line and curve have the same y-value.

The curve is
y=32x+5x−7.y=\frac{3}{2}x+\frac{5}{x}-7.
The line is
2y+x=2,2y+x=2,
so
y=1−x2.y=1-\frac{x}{2}.
At a point of intersection,
32x+5x−7=1−x2.\frac{3}{2}x+\frac{5}{x}-7=1-\frac{x}{2}.

2. Simplify the equation to obtain the given quadratic equation.

32x+x2+5x−8=0\frac{3}{2}x+\frac{x}{2}+\frac{5}{x}-8=0
2x+5x−8=02x+\frac{5}{x}-8=0
Multiplying throughout by xx,
2x2+5−8x=02x^{2}+5-8x=0
2x2−8x+5=0.2x^{2}-8x+5=0.

答案

At the points of intersection, both equations are satisfied, so
32x+5x−7=1−x2\frac{3}{2}x+\frac{5}{x}-7=1-\frac{x}{2}
which simplifies to
2x2−8x+5=0.2x^{2}-8x+5=0.
Hence the xx-coordinates of the points of intersection give the solutions to 2x2−8x+5=02x^{2}-8x+5=0.

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