对于方程 py=xy+1py = xy + 1(其中 pp 为未知常数),可以用一条直线表示,其形式为 Y=mX+cY = mX + c,其中 XX 和 YY 是 xx 和/或 yy 的函数,而 mm 和 cc 是常数。利用下表,在其中填入 YY、XX、mm 和 cc 的表达式。 变量 xx 和 yy 满足关系式 xay=bx^{a}y = b,其中 aa 和 bb 是常数。下表显示了 xx 和 yy 的一些对应值。 本页提供了用于绘制 lg⁡y\lg y 对 lg⁡x\lg x 图像的坐标纸。 绘制 lg⁡y\lg y 对 lg⁡x\lg x 的图并画出直线图。 利用你的图像估计 aa 和 bb 的值。

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完整解法

9(a)

1. Rearrange the given equation to isolate a term involving only $y$ or to separate the variables into a linear form $Y = mX + c$.

Given py=xy+1py = xy + 1.
Divide the entire equation by yy:
p=x+1yp = x + \frac{1}{y}

2. Express the equation in the form $Y = mX + c$ and identify the components.

Rearranging the terms gives:
1y=−x+p\frac{1}{y} = -x + p
Comparing this with Y=mX+cY = mX + c, we can identify:
Y=1yY = \frac{1}{y}
X=xX = x
m=−1m = -1
c=pc = p

答案

Y=1yY = \frac{1}{y}, X=xX = x, m=−1m = -1, c=pc = p
(Note: Other valid forms exist, such as Y=yY = y, X=xyX = xy, m=1pm = \frac{1}{p}, c=1pc = \frac{1}{p})

9(b)(i)

1. Calculate the values of $\lg x$ and $\lg y$ to 3 significant figures.

Using the given table:

xx1.311.311.631.631.971.972.542.544.034.03
yy1270127039839813113135.135.12.632.63
lg⁡x\lg x0.1170.1170.2120.2120.2940.2940.4050.4050.6050.605
lg⁡y\lg y3.1043.1042.6002.6002.1172.1171.5451.5450.4200.420

2. Plot the points and draw the straight line graph.

Plot the points (0.117,3.104)(0.117, 3.104), (0.212,2.600)(0.212, 2.600), (0.294,2.117)(0.294, 2.117), (0.405,1.545)(0.405, 1.545), and (0.605,0.420)(0.605, 0.420) on the graph paper with lg⁡y\lg y on the vertical axis and lg⁡x\lg x on the horizontal axis. Draw a straight line of best fit passing through or near these points.

答案

A straight line graph plotted with points (0.117,3.104)(0.117, 3.104), (0.212,2.600)(0.212, 2.600), (0.294,2.117)(0.294, 2.117), (0.405,1.545)(0.405, 1.545), and (0.605,0.420)(0.605, 0.420).

9(b)(ii)

1. Convert the non-linear equation to linear form

Given xay=bx^a y = b.
Taking logarithms on both sides:
lg⁡(xay)=lg⁡b\lg(x^a y) = \lg b
alg⁡x+lg⁡y=lg⁡ba \lg x + \lg y = \lg b
lg⁡y=−alg⁡x+lg⁡b\lg y = -a \lg x + \lg b
This is in the form Y=mX+cY = mX + c, where Y=lg⁡yY = \lg y, X=lg⁡xX = \lg x, gradient m=−am = -a, and YY-intercept c=lg⁡bc = \lg b.

2. Calculate the gradient to find $a$

Using the points (lg⁡1.31,lg⁡1270)(\lg 1.31, \lg 1270) and (lg⁡4.03,lg⁡2.63)(\lg 4.03, \lg 2.63), which are approximately (0.117,3.104)(0.117, 3.104) and (0.605,0.420)(0.605, 0.420):
Gradient m=0.420−3.1040.605−0.117=−2.6840.488≈−5.5m = \frac{0.420 - 3.104}{0.605 - 0.117} = \frac{-2.684}{0.488} \approx -5.5
Since m=−am = -a, we have −a=−5.5  ⟹  a=5.5-a = -5.5 \implies a = 5.5.

3. Find the $Y$-intercept to determine $b$

Using the point (0.117,3.104)(0.117, 3.104) and m=−5.5m = -5.5:
c=Y−mX=3.104−(−5.5)(0.117)=3.104+0.6435≈3.75c = Y - mX = 3.104 - (-5.5)(0.117) = 3.104 + 0.6435 \approx 3.75
Since c=lg⁡bc = \lg b, we have lg⁡b=3.75\lg b = 3.75.

4. Calculate the value of $b$

b=103.75≈5620b = 10^{3.75} \approx 5620 (accept values around 5600 depending on graph reading)

答案

a=5.5a = 5.5, b=5600b = 5600

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