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CIE 9709 2022 June Paper 13 Q8

A Level / CIE / P1

CIE 9709 2022 June Paper 13 Paper · Question 8

题目

Problem

The diagram shows the curve with equation y=x12+4x−12y=x^{\frac12}+4x^{-\frac12}. The line y=5y=5 intersects the curve at the points A(1,5)A(1,5) and B(16,5)B(16,5).

(a) Find the equation of the tangent to the curve at the point AA.

(4)

(b) Calculate the area of the shaded region.

(4)
题目中文翻译

图中所示为方程为 y=x12+4x−12y=x^{\frac12}+4x^{-\frac12} 的曲线。直线 y=5y=5 与曲线交于点 A(1,5)A(1,5) 和 B(16,5)B(16,5)。

(a) 求曲线在点 AA 处的切线方程。

(4)

(b) 计算阴影区域的面积。

(4)
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

对两个幂分别求导,在 x=1x=1 求切线斜率。

答题过程

展开

The derivative is 12x−1/2−2x−3/2\frac12x^{-1/2}-2x^{-3/2}. At x=1x=1 it equals −3/2-3/2. The tangent through A=(1,5)A=(1,5) is

y−5=−32(x−1),y-5=-\frac32(x-1),

or y=−3x/2+13/2y=-3x/2+13/2.

(b)

解法一

思路

展开

以水平直线为上边界,根式曲线为下边界。积分区间由 A,BA,B 的横坐标决定。

答题过程

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The shaded area is

 ∫116(5−x1/2−4x−1/2) dx= [5x−23x3/2−8x1/2]116= 75−23(64−1)−8(4−1)=9.\begin{align*} &\,\int_1^{16}(5-x^{1/2}-4x^{-1/2})\,\mathrm{d}x\\ =&\,\Big[5x-\frac23x^{3/2}-8x^{1/2}\Big]_1^{16}\\ =&\,75-\frac23(64-1)-8(4-1)=9. \end{align*}

解法二

思路

展开

先求曲线下方的面积,再用直线下方的矩形面积减去它。

答题过程

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The area under the curve between x=1x=1 and x=16x=16 is

∫116(x1/2+4x−1/2) dx= [23x3/2+8x1/2]116= 23(64−1)+8(4−1)=66.\begin{align*} \int_1^{16}\Big(x^{1/2}+4x^{-1/2}\Big)\,\mathrm{d}x =&\,\Big[\frac23x^{3/2}+8x^{1/2}\Big]_1^{16}\\ =&\,\frac23(64-1)+8(4-1)=66. \end{align*}

The area under y=5y=5 is a rectangle of area 5(16−1)=755(16-1)=75. Hence the shaded area is 75−66=975-66=9.