题目

Problem

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

Given that

\begin{align*} f'(x)=6x-\frac{(2x-1)(3x+2)}{2\sqrt{x}} \end{align*}

and the point $P(4,12)$ lies on $C$ ,

(a) find the equation of the normal to $C$ at $P$ , giving your answer in the form $y=mx+c$ where $m$ and $c$ are integers to be found,

(4)

(b) find $f(x)$ , giving each term in simplest form.

(6)

解答

展开

先代入 $x=4$ 求切线斜率，再用垂线斜率关系求法线斜率。最后用点 $P(4,12)$ 写法线方程。

展开

At $x=4$ ,

\begin{align*} f'(4) =&\,6(4)-\frac{(2(4)-1)(3(4)+2)}{2\sqrt4}\\ =&\,24-\frac{(7)(14)}{4}\\ =&\,24-\frac{98}{4}\\ =&\,24-\frac{49}{2}\\ =&\,-\frac12. \end{align*}

So the tangent gradient is $-\frac12$ .

The normal gradient is the negative reciprocal:

\begin{align*} m=2. \end{align*}

Using $P(4,12)$ ,

\begin{align*} y-12=&\,2(x-4)\\ y-12=&\,2x-8\\ y=&\,2x+4. \end{align*}

展开

先展开分子：

\begin{align*} (2x-1)(3x+2)=6x^2+x-2. \end{align*}

再除以 $2\sqrt{x}$ ，把每项写成 $x$ 的幂，之后逐项积分。最后用 $P(4,12)$ 求积分常数。

展开

First expand and split the fraction:

\begin{align*} \frac{(2x-1)(3x+2)}{2\sqrt{x}} =&\,\frac{6x^2+x-2}{2x^{1/2}}\\ =&\,3x^{3/2}+\frac12x^{1/2}-x^{-1/2}. \end{align*}

\begin{align*} f'(x) =&\,6x-\left(3x^{3/2}+\frac12x^{1/2}-x^{-1/2}\right)\\ =&\,6x-3x^{3/2}-\frac12x^{1/2}+x^{-1/2}. \end{align*}

Integrate:

\begin{align*} f(x) =&\,\int\left(6x-3x^{3/2}-\frac12x^{1/2}+x^{-1/2}\right)\,dx\\ =&\,3x^2-\frac65x^{5/2}-\frac13x^{3/2}+2x^{1/2}+c. \end{align*}

Use $P(4,12)$ :

\begin{align*} 12 =&\,3(4)^2-\frac65(4)^{5/2} -\frac13(4)^{3/2}+2(4)^{1/2}+c\\ =&\,48-\frac65(32)-\frac13(8)+4+c\\ =&\,52-\frac{192}{5}-\frac83+c. \end{align*}

Hence

\begin{align*} c =&\,12-52+\frac{192}{5}+\frac83\\ =&\,-40+\frac{192}{5}+\frac83\\ =&\,\frac{16}{15}. \end{align*}

Therefore

\begin{align*} f(x)=3x^2-\frac65x^{5/2}-\frac13x^{3/2}+2x^{1/2}+\frac{16}{15}. \end{align*}