题目
Problem
Figure 2 shows a sketch of the curve C with equation y=x2−5x+13.
Figure 2
The point M is the minimum point of C.
The straight line l passes through the origin O and intersects C at the points M and N as shown.
Find, showing your working,
(a) the coordinates of M,
(3)
(b) the coordinates of N.
(5)
Figure 3 shows the curve C and the line l. The finite region R, shown shaded in Figure 3, is bounded by C, l and the y-axis.
Figure 3
(c) Use inequalities to define the region R.
(2)
解答
(a)
解法一
思路
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用配方法找二次函数的最小点。括号平方项最小时等于 0。
答题过程
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y===x2−5x+13(x−25)2−425+13(x−25)2+427.
Therefore
M=(25,427).
(b)
解法一
思路
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直线 l 经过原点和 M,所以先求斜率,再与曲线联立。一个交点是 M,另一个就是 N。
答题过程
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The gradient of l is
5/227/4=1027.
Thus
l:y=1027x.
Intersect with C:
x2−5x+13=10x2−50x+130=10x2−77x+130=(2x−5)(5x−26)=1027x27x00.
The point M has x=25, so for N,
x=526.
Then
y=1027⋅526=25351.
Therefore
N=(526,25351).
(c)
解法一
思路
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区域 R 在 y 轴右侧、在直线 l 上方、在曲线 C 下方。右端到 M 为止,所以 x 从 0 到 25。
答题过程
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The region is defined by
0≤x≤25,1027x≤y≤x2−5x+13.