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IAL 2022 Oct Q1

A Level / Edexcel / P1

IAL 2022 Oct Paper · Question 1

题目

Problem

The curve CC has equation

y=x34−x2+17x,x>0.\begin{align*} y=\frac{x^3}{4}-x^2+\frac{17}{x}, \qquad x>0. \end{align*}

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

(3)

The point R(2,132)R\left(2,\dfrac{13}{2}\right) lies on CC.

(b) Find the equation of the tangent to CC at the point RR. Write your answer in the form ax+by+c=0ax+by+c=0, where aa, bb and cc are integers to be found.

(3)

解答

(a)

解法一

思路

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先把 17x\frac{17}{x} 写成 17x−117x^{-1}，再逐项求导。这样负指数项也可以直接用幂函数求导法。

答题过程

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Rewrite the curve as

y=14x3−x2+17x−1.\begin{align*} y=\frac14x^3-x^2+17x^{-1}. \end{align*}

Differentiate term by term:

dydx= 34x2−2x−17x−2.\begin{align*} \frac{dy}{dx} =&\,\frac34x^2-2x-17x^{-2}. \end{align*}

(b)

解法一

思路

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切线斜率是点 RR 处的 dydx\frac{dy}{dx}。求出斜率后，用点斜式写直线，再整理成整数系数的一般式。

答题过程

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At x=2x=2,

dydx= 34(2)2−2(2)−17(2)−2= 3−4−174= −214.\begin{align*} \frac{dy}{dx} =&\,\frac34(2)^2-2(2)-17(2)^{-2}\\ =&\,3-4-\frac{17}{4}\\ =&\,-\frac{21}{4}. \end{align*}

The tangent passes through R(2,132)R\left(2,\frac{13}{2}\right), so

y−132= −214(x−2).\begin{align*} y-\frac{13}{2} =&\,-\frac{21}{4}(x-2). \end{align*}

Multiply by 44:

4y−26= −21x+4221x+4y−68= 0.\begin{align*} 4y-26=&\,-21x+42\\ 21x+4y-68=&\,0. \end{align*}