题目

Problem

The curve $C$ has equation $y=f(x)$ where $x>0$ .

Given that

\begin{align*} f'(x)=\frac{4x^2+10-7x^{\frac12}}{4x^{\frac12}} \end{align*}

and the point $P(4,-1)$ lies on $C$ ,

(a) (i) find the value of the gradient of $C$ at $P$

(ii) Hence find the equation of the normal to $C$ at $P$ , giving your answer in the form $ax+by+c=0$ where $a$ , $b$ and $c$ are integers to be found.

(4)

(b) Find $f(x)$ .

(6)

解答

(a)(i)

解法一

思路

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曲线在 $P$ 点的梯度就是 $f'(4)$ 。直接把 $x=4$ 代入即可。

答题过程

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At $P$ , $x=4$ . Therefore

\begin{align*} f'(4) =&\,\frac{4(4)^2+10-7(4)^{\frac12}}{4(4)^{\frac12}}\\ =&\,\frac{64+10-14}{8}\\ =&\,\frac{60}{8}\\ =&\,\frac{15}{2}. \end{align*}

So the gradient of $C$ at $P$ is

\begin{align*} \frac{15}{2}. \end{align*}

(a)(ii)

解法一

思路

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法线斜率是切线斜率的负倒数。切线斜率为 $\frac{15}{2}$ ，所以法线斜率为 $-\frac{2}{15}$ 。再用点 $P(4,-1)$ 写直线方程。

答题过程

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The normal gradient is

\begin{align*} -\frac{2}{15}. \end{align*}

Using $P(4,-1)$ ,

\begin{align*} y+1=&\,-\frac{2}{15}(x-4)\\ 15y+15=&\,-2x+8\\ 2x+15y+7=&\,0. \end{align*}

(b)

解法一

思路

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先把 $f'(x)$ 拆成幂函数，再积分。最后用 $P(4,-1)$ 求积分常数。

答题过程

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Rewrite $f'(x)$ :

\begin{align*} f'(x) =&\,\frac{4x^2}{4x^{\frac12}} +\frac{10}{4x^{\frac12}} -\frac{7x^{\frac12}}{4x^{\frac12}}\\ =&\,x^{\frac32}+\frac52x^{-\frac12}-\frac74. \end{align*}

Integrate:

\begin{align*} f(x) =&\,\int\left(x^{\frac32}+\frac52x^{-\frac12}-\frac74\right)\,dx\\ =&\,\frac25x^{\frac52}+5x^{\frac12}-\frac74x+c. \end{align*}

Use $P(4,-1)$ :

\begin{align*} -1 =&\,\frac25(4)^{\frac52}+5(4)^{\frac12}-\frac74(4)+c\\ =&\,\frac25(32)+10-7+c\\ =&\,\frac{64}{5}+3+c\\ =&\,\frac{79}{5}+c. \end{align*}

\begin{align*} c=-\frac{84}{5}. \end{align*}

Therefore

\begin{align*} f(x)=\frac25x^{\frac52}+5x^{\frac12}-\frac74x-\frac{84}{5}. \end{align*}

IAL 2023 Oct Q7

题目

解答

(a)(i)

解法一

思路

答题过程

(a)(ii)

解法一

思路

答题过程

(b)

解法一

思路

答题过程