题目

Problem

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

A curve $C$ has equation $y=f(x)$ , $x>0$ . A point $P$ lies on $C$ .

Given that

$f'(x)=4x^2+\dfrac{6}{x^2}-4$
the equation of the tangent to $C$ at $P$ is $y=10x-6\sqrt3$

(a) (i) verify that $\sqrt3$ is a possible $x$ coordinate of $P$ ,

(ii) find, using algebra, the other possible $x$ coordinate of $P$ .

(6)

Given that the $x$ coordinate of $P$ is $\sqrt3$

(b) find an equation of the normal to $C$ at $P$ .

(2)

(4)

解答

(a)

解法一

思路

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切线斜率是 $10$ ，所以可能的 $x$ 坐标满足 $f'(x)=10$ 。先代入 $x=\sqrt3$ 验证，再解方程找另一个正的 $x$ 值。

答题过程

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The gradient of the tangent is $10$ .

For $x=\sqrt3$ ,

\begin{align*} f'(\sqrt3) =&\, 4(\sqrt3)^2+\frac{6}{(\sqrt3)^2}-4\\[3mm] =&\, 12+2-4\\[3mm] =&\, 10 \end{align*}

This is equal to the gradient of the tangent, so $\sqrt3$ is a possible $x$ coordinate of $P$ .

For the other possible coordinate, solve

\begin{align*} 4x^2+\frac{6}{x^2}-4=10 \end{align*}

Then

\begin{align*} 4x^2+\frac{6}{x^2} =&\, 14\\[3mm] 4x^4+6 =&\, 14x^2\\[3mm] 2x^4-7x^2+3 =&\, 0 \end{align*}

Factorise:

\begin{align*} (2x^2-1)(x^2-3)=0 \end{align*}

\begin{align*} x^2=\frac12 \quad\text{or}\quad x^2=3 \end{align*}

Since $x>0$ , the possible $x$ coordinates are

\begin{align*} x=\frac{1}{\sqrt2} \quad\text{or}\quad x=\sqrt3 \end{align*}

Therefore the other possible $x$ coordinate is

\begin{align*} \frac{\sqrt2}{2} \end{align*}

(b)

解法一

思路

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切线斜率是 $10$ ，所以法线斜率是 $-\dfrac1{10}$ 。点 $P$ 的 $y$ 坐标可由切线方程代入 $x=\sqrt3$ 得到。

答题过程

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When $x=\sqrt3$ , using the tangent equation,

\begin{align*} y =&\, 10\sqrt3-6\sqrt3\\[3mm] =&\, 4\sqrt3 \end{align*}

\begin{align*} P=(\sqrt3,4\sqrt3) \end{align*}

The gradient of the normal is

\begin{align*} -\frac1{10} \end{align*}

Therefore the normal is

\begin{align*} y-4\sqrt3 = -\frac1{10}(x-\sqrt3) \end{align*}

(c)

解法一

思路

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对 $f'(x)$ 积分得到 $f(x)$ ，再用点 $P(\sqrt3,4\sqrt3)$ 求积分常数。

答题过程

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Since

\begin{align*} f'(x)=4x^2+6x^{-2}-4, \end{align*}

integrating gives

\begin{align*} f(x) =&\, \frac43x^3-6x^{-1}-4x+C\\[3mm] =&\, \frac43x^3-\frac6x-4x+C \end{align*}

Using $P(\sqrt3,4\sqrt3)$ ,

\begin{align*} 4\sqrt3 =&\, \frac43(\sqrt3)^3 -\frac6{\sqrt3} -4\sqrt3+C\\[3mm] =&\, 4\sqrt3-2\sqrt3-4\sqrt3+C\\[3mm] =&\, -2\sqrt3+C \end{align*}

\begin{align*} C=6\sqrt3 \end{align*}

Therefore

\begin{align*} f(x) = \frac43x^3-\frac6x-4x+6\sqrt3 \end{align*}

IAL 2026 Jan Q8

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