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 $C$ has equation

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

(a) Find $\dfrac{dy}{dx}$ giving each term in simplest form.

(2)

(b) Hence find the $x$ coordinate of the stationary point of $C$ .

(2)

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

(2)

(d) State the range of values of $x$ for which $y$ is decreasing.

(1)

解答

(a)

解法一

思路

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

答题过程

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\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} =&\,4\cdot\frac12x^{-\frac12} +9\left(-\frac12\right)x^{-\frac32}\\ =&\,2x^{-\frac12}-\frac92x^{-\frac32}. \end{align*}

(b)

解法一

思路

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

答题过程

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At a stationary point,

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

Multiply by $2x^{3/2}$ :

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

Therefore, the $x$ coordinate is $\frac94$ .

(c)

解法一

思路

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第二导数用于判断极值性质。把 $x=\frac94$ 代入，若第二导数大于 $0$ ，就是 minimum。

答题过程

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From

\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x}=2x^{-\frac12}-\frac92x^{-\frac32}, \end{align*}

we get

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

At $x=\frac94$ ,

\begin{align*} \frac{\mathrm{d}^2y}{\mathrm{d}x^2} =&\,-\left(\frac94\right)^{-\frac32} +\frac{27}{4}\left(\frac94\right)^{-\frac52}\\ =&\,\frac{16}{27}\\ &>0. \end{align*}

Therefore, the stationary point is a minimum.

(d)

解法一

思路

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曲线在 minimum 之前下降，在 minimum 之后上升。由于题目给出 $x>0$ ，所以下降区间从 $0$ 到 $\frac94$ 。

答题过程

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The curve is decreasing before the stationary point. Since $x>0$ ,

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

IAL 2024 Oct Q4

题目

解答

(a)

解法一

思路

答题过程

(b)

解法一

思路

答题过程

(c)

解法一

思路

答题过程

(d)

解法一

思路

答题过程