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IAL 2020 Jan Q9

A Level / Edexcel / P1

IAL 2020 Jan Paper · Question 9

题目

Problem

In this question you must show all stages of your working.

Solutions relying on calculator technology are not acceptable.

A curve has equation

y=4x2+92x,x>0.\begin{align*} y=\frac{4x^2+9}{2\sqrt{x}},\qquad x>0. \end{align*}

Find the xx coordinate of the point on the curve at which dydx=0\dfrac{dy}{dx}=0.

(6)

解答

解法一

思路

展开

先把分式拆成两项,并写成指数形式,再求导。因为 x>0x>0,最后只保留正的 xx 值。

答题过程

展开 y=4x22x1/2+92x1/2=2x3/2+92x1/2.\begin{align*} y =&\,\frac{4x^2}{2x^{1/2}}+\frac{9}{2x^{1/2}}\\ =&\,2x^{3/2}+\frac92x^{-1/2}. \end{align*}

Differentiate:

dydx=3x1/294x3/2.\begin{align*} \frac{dy}{dx} =&\,3x^{1/2}-\frac94x^{-3/2}. \end{align*}

Set dydx=0\dfrac{dy}{dx}=0:

3x1/294x3/2=03x1/2=94x3/2.\begin{align*} 3x^{1/2}-\frac94x^{-3/2}=&\,0\\ 3x^{1/2}=&\,\frac94x^{-3/2}. \end{align*}

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

3x2=94x2=34.\begin{align*} 3x^2=&\,\frac94\\ x^2=&\,\frac34. \end{align*}

Since x>0x>0,

x=32.\begin{align*} x=\frac{\sqrt3}{2}. \end{align*}