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P2 2024 Oct Q4

A Level / Edexcel / P2

IAL 2024 Oct Paper · Question 4

题目

Problem

In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable.

The curve CC has equation

y=4x12+9x−12+3,x>0y=4x^{\frac12}+9x^{-\frac12}+3,\qquad x>0

(a) Find dydx\dfrac{\mathrm{d}y}{\mathrm{d}x} giving each term in simplest form.

(2)

(b) Hence find the xx coordinate of the stationary point of CC.

(2)

(c) (i) Find d2ydx2\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} giving each term in simplest form.

(ii) Hence determine the nature of the stationary point of CC, giving a reason for your answer.

(2)

(d) State the range of values of xx for which yy is decreasing.

(1)
题目中文翻译

本题必须展示计算过程的所有步骤。

完全依赖计算器技术的解答不予接受。

曲线 CC 的方程为

y=4x1/2+9x−1/2+3,x>0y=4x^{1/2}+9x^{-1/2}+3,\qquad x>0

(a) 求 dydx\dfrac{\mathrm{d}y}{\mathrm{d}x}，将每项化为最简形式。(2)

(b) 由此求 CC 的驻点的 xx 坐标。(2)

(c) (i) 求 d2ydx2\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}，将每项化为最简形式。

(ii) 由此判断 CC 的驻点的性质，并说明理由。(2)

(d) 写出 yy 递减时 xx 的取值范围。(1)

解答

(a)

解法一

思路

展开

直接用幂函数求导。9x−1/29x^{-1/2} 求导后指数会变成 −3/2-3/2。

答题过程

展开 dydx= 4⋅12x−12 +9(−12)x−32= 2x−12−92x−32.\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} =&\,4\cdot\frac12x^{-\frac12}\\ &\,\hspace{2pt}+9\bigg(-\frac12\bigg)x^{-\frac32}\\ =&\,2x^{-\frac12}-\frac92x^{-\frac32}. \end{align*}

(b)

解法一

思路

展开

stationary point 满足 dydx=0\frac{\mathrm{d}y}{\mathrm{d}x}=0。把两个负指数项同时乘以 x3/2x^{3/2} 会很干净。

答题过程

展开

At a stationary point,

2x−12−92x−32= 0.\begin{align*} 2x^{-\frac12}-\frac92x^{-\frac32}=&\,0. \end{align*}

Multiply by 2x3/22x^{3/2}:

4x−9= 0x= 94.\begin{align*} 4x-9=&\,0\\ x=&\,\frac94. \end{align*}

Therefore, the xx coordinate is 94\frac94.

(c)

解法一

思路

展开

第二导数用于判断极值性质。把 x=94x=\frac94 代入，若第二导数大于 00，就是 minimum。

答题过程

展开

From

dydx= 2x−12 −92x−32,\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} =&\,2x^{-\frac12}\\ &\,\hspace{2pt}-\frac92x^{-\frac32}, \end{align*}

we get

d2ydx2= −x−32+274x−52.\begin{align*} \frac{\mathrm{d}^2y}{\mathrm{d}x^2} =&\,-x^{-\frac32}+\frac{27}{4}x^{-\frac52}. \end{align*}

At x=94x=\frac94,

d2ydx2= −(94)−32 +274(94)−52= 1627> 0.\begin{align*} \frac{\mathrm{d}^2y}{\mathrm{d}x^2} =&\,-\bigg(\frac94\bigg)^{-\frac32}\\ &\,\hspace{2pt}+\frac{27}{4}\bigg(\frac94\bigg)^{-\frac52}\\ =&\,\frac{16}{27}\\ >&\,0. \end{align*}

Therefore, the stationary point is a minimum.

(d)

解法一

思路

展开

将第一导数合并为分式，分母在 x>0x>0 时为正，只需判断分子 4x−94x-9 的符号。

答题过程

展开

From part (a),

dydx=4x−92x3/2.\frac{\mathrm{d}y}{\mathrm{d}x} =\frac{4x-9}{2x^{3/2}}.

Since x>0x>0, the denominator is positive. Thus the derivative is negative exactly when 4x−9<04x-9<0. Therefore the decreasing interval is

0<x<94.\begin{align*} 0<x<\frac94. \end{align*}