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IAL 2019 Oct Q3

A Level / Edexcel / P1

IAL 2019 Oct Paper · Question 3

题目

Problem

Figure 2 shows a sketch of the curve CC with equation y=x25x+13y=x^2-5x+13.

Figure 2

The point MM is the minimum point of CC.

The straight line ll passes through the origin OO and intersects CC at the points MM and NN as shown.

Find, showing your working,

(a) the coordinates of MM,

(3)

(b) the coordinates of NN.

(5)

Figure 3 shows the curve CC and the line ll. The finite region RR, shown shaded in Figure 3, is bounded by CC, ll and the yy-axis.

Figure 3

(c) Use inequalities to define the region RR.

(2)

解答

(a)

解法一

思路

展开

用配方法找二次函数的最小点。括号平方项最小时等于 00

答题过程

展开 y=x25x+13=(x52)2254+13=(x52)2+274.\begin{align*} y=&\,x^2-5x+13\\ =&\,\left(x-\frac52\right)^2-\frac{25}{4}+13\\ =&\,\left(x-\frac52\right)^2+\frac{27}{4}. \end{align*}

Therefore

M=(52,274).\begin{align*} M=\left(\frac52,\frac{27}{4}\right). \end{align*}

(b)

解法一

思路

展开

直线 ll 经过原点和 MM,所以先求斜率,再与曲线联立。一个交点是 MM,另一个就是 NN

答题过程

展开

The gradient of ll is

27/45/2=2710.\begin{align*} \frac{27/4}{5/2} =&\,\frac{27}{10}. \end{align*}

Thus

l:y=2710x.\begin{align*} l:\quad y=\frac{27}{10}x. \end{align*}

Intersect with CC:

x25x+13=2710x10x250x+130=27x10x277x+130=0(2x5)(5x26)=0.\begin{align*} x^2-5x+13=&\,\frac{27}{10}x\\ 10x^2-50x+130=&\,27x\\ 10x^2-77x+130=&\,0\\ (2x-5)(5x-26)=&\,0. \end{align*}

The point MM has x=52x=\frac52, so for NN,

x=265.\begin{align*} x=\frac{26}{5}. \end{align*}

Then

y=2710265=35125.\begin{align*} y=\frac{27}{10}\cdot\frac{26}{5} =\frac{351}{25}. \end{align*}

Therefore

N=(265,35125).\begin{align*} N=\left(\frac{26}{5},\frac{351}{25}\right). \end{align*}

(c)

解法一

思路

展开

区域 RRyy 轴右侧、在直线 ll 上方、在曲线 CC 下方。右端到 MM 为止,所以 xx0052\frac52

答题过程

展开

The region is defined by

0x52,2710xyx25x+13.\begin{align*} 0\leq x\leq\frac52, \qquad \frac{27}{10}x\leq y\leq x^2-5x+13. \end{align*}