Skip to content
CalcGospel 國際數學圖譜
返回

CIE 9709 2023 November Paper 12 Q9

A Level / CIE / P1

CIE 9709 2023 Nov Paper 12 Paper · Question 9

题目

Problem

The diagram shows curves with equations y=2x12+13x−12y=2x^{\frac{1}{2}}+13x^{-\frac{1}{2}} and y=3x−12+12y=3x^{-\frac{1}{2}}+12. The curves intersect at points AA and BB.

(a) Find the coordinates of AA and BB.

(4)

(b) Hence find the area of the shaded region.

(5)
题目中文翻译

图中两条曲线的方程分别为 y=2x12+13x−12y=2x^{\frac{1}{2}}+13x^{-\frac{1}{2}} 和 y=3x−12+12y=3x^{-\frac{1}{2}}+12。两条曲线相交于点 AA 和 BB。

(a) 求 AA 和 BB 的坐标。

(4)

(b) 由此求阴影区域的面积。

(5)
原卷通用作答要求

Answer all questions. You should use a calculator where appropriate.

You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.

Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.

请回答所有题目。适当时应使用计算器。必须清楚展示所有必要的计算过程;仅给出计算器答案而没有过程,不予评分。除非题目另有精度要求,非精确数值答案保留 3 位有效数字,以度为单位的角度保留 1 位小数。

解答

(a)

解法一

思路

展开

令两式相等并乘以正的 x\sqrt x。设 u=xu=\sqrt x 后便得到二次方程。

答题过程

展开

The intersection equation gives 2x+10=12x2x+10=12\sqrt x on x>0x>0. With u=xu=\sqrt x this is

u2−6u+5=(u−1)(u−5)=0.u^2-6u+5=(u-1)(u-5)=0.

Thus x=1x=1 or 2525. The heights from y=12+3/xy=12+3/\sqrt x are 1515 and 63/563/5. From left to right, A=(1,15)A=(1,15) and B=(25,63/5)B=(25,63/5).

解法二

思路

展开

也可以在根式单独一边后两边平方。所得候选根仍要回代根式方程。

答题过程

展开

From x+5=6xx+5=6\sqrt x, squaring gives

x2−26x+25=(x−1)(x−25)=0.x^2-26x+25=(x-1)(x-25)=0.

The roots 11 and 2525 give 1+5=6(1)1+5=6(1) and 25+5=6(5)25+5=6(5), so both work. The heights 12+3/x12+3/\sqrt x give A=(1,15)A=(1,15) and B=(25,63/5)B=(25,63/5).

(b)

解法一

思路

展开

图中上方是 12+3/x12+3/\sqrt x。先相减,再逐项积分并代入 (a) 的横坐标。

答题过程

展开

The shaded area is

 ∫125(12−2x1/2−10x−1/2) dx= [12x−43x3/2−20x1/2]125= 1003−(−283)= 1283.\begin{align*} &\,\int_1^{25}\bigl(12-2x^{1/2}-10x^{-1/2}\bigr)\,\mathrm{d}x\\ =&\,\Big[12x-\frac43x^{3/2}-20x^{1/2}\Big]_1^{25}\\ =&\,\frac{100}{3}-\Big(-\frac{28}{3}\Big)\\ =&\,\frac{128}{3}. \end{align*}