Skip to content
CalcGospel 國際數學圖譜
返回

IAL 2021 Jan Q1

A Level / Edexcel / P1

IAL 2021 Jan Paper · Question 1

题目

Problem

A curve has equation

y=2x35x232x+7,x>0.\begin{align*} y=2x^3-5x^2-\frac{3}{2x}+7, \qquad x>0. \end{align*}

(a) Find, in simplest form, dydx\dfrac{dy}{dx}.

(3)

The point PP lies on the curve and has xx coordinate 12\dfrac12.

(b) Find an equation of the normal to the curve at PP, writing your answer in the form ax+by+c=0ax+by+c=0, where aa, bb and cc are integers to be found.

(5)

解答

(a)

解法一

思路

展开

把分式写成负指数后逐项求导。

答题过程

展开 y=2x35x232x1+7,dydx=6x210x+32x2=6x210x+32x2.\begin{align*} y=&\,2x^3-5x^2-\frac32x^{-1}+7,\\ \frac{dy}{dx} =&\,6x^2-10x+\frac32x^{-2}\\ =&\,6x^2-10x+\frac{3}{2x^2}. \end{align*}

(b)

解法一

思路

展开

先求点 PP 的坐标,再求切线斜率。法线斜率是切线斜率的负倒数。

答题过程

展开

When x=12x=\frac12,

y=2(12)35(12)232(12)+7=14543+7=3.\begin{align*} y =&\,2\left(\frac12\right)^3-5\left(\frac12\right)^2 -\frac{3}{2\left(\frac12\right)}+7\\ =&\,\frac14-\frac54-3+7\\ =&\,3. \end{align*}

So P=(12,3)P=\left(\frac12,3\right).

At x=12x=\frac12,

dydx=6(12)210(12)+32(12)2=325+6=52.\begin{align*} \frac{dy}{dx} =&\,6\left(\frac12\right)^2-10\left(\frac12\right) +\frac{3}{2\left(\frac12\right)^2}\\ =&\,\frac32-5+6\\ =&\,\frac52. \end{align*}

So the normal gradient is 25-\frac25. Hence

y3=25(x12)5y15=2x+12x+5y16=0.\begin{align*} y-3=&\,-\frac25\left(x-\frac12\right)\\ 5y-15=&\,-2x+1\\ 2x+5y-16=&\,0. \end{align*}