题目

Problem

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

Figure 4 shows a sketch of the curve $C$ with equation

\begin{align*} y=2x^3+\frac12x^2-2x+5 \end{align*}

Figure 4

The line $l$ is the normal to $C$ at the point $P$ where $x=0$ .

The line $l$ also intersects $C$ at points $Q$ and $R$ as shown in Figure 4.

(a) Find, using algebra, the $x$ coordinate of point $Q$ .

(6)

The point $T$ lies on $C$ .

Given that

the tangent to $C$ at $T$ is parallel to $l$
the $x$ coordinate of $T$ is positive

(b) find, using algebra, the exact $x$ coordinate of $T$ .

(4)

解答

(a)

解法一

思路

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先求 $x=0$ 时曲线的切线斜率，再取负倒数得到法线 $l$ 的斜率。写出 $l$ 的方程后，与三次曲线联立求交点。 $x=0$ 是已知点 $P$ ，正根对应 $R$ ，负根对应 $Q$ 。

答题过程

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Differentiate:

\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} =&\,6x^2+x-2. \end{align*}

At $x=0$ ,

\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x}=-2. \end{align*}

So the gradient of the normal is

\begin{align*} \frac12. \end{align*}

When $x=0$ ,

\begin{align*} y=5, \end{align*}

so $P=(0,5)$ and the normal is

\begin{align*} y=\frac12x+5. \end{align*}

At intersections with $C$ ,

\begin{align*} 2x^3+\frac12x^2-2x+5 =&\,\frac12x+5. \end{align*}

Rearrange:

\begin{align*} 2x^3+\frac12x^2-\frac52x=&\,0. \end{align*}

Multiply by $2$ :

\begin{align*} 4x^3+x^2-5x=&\,0\\ x(4x^2+x-5)=&\,0\\ x(4x+5)(x-1)=&\,0. \end{align*}

Thus the intersections have

\begin{align*} x=0,\qquad x=-\frac54,\qquad x=1. \end{align*}

Point $Q$ is the left-hand intersection, so

\begin{align*} x_Q=-\frac54. \end{align*}

(b)

解法一

思路

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若 $T$ 处切线平行于 $l$ ，则 $T$ 处切线斜率也等于 $\frac12$ 。因此令导函数等于 $\frac12$ ，再取正的 $x$ 解。

答题过程

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For the tangent at $T$ to be parallel to $l$ ,

\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x}=\frac12. \end{align*}

\begin{align*} 6x^2+x-2=&\,\frac12\\ 12x^2+2x-4=&\,1\\ 12x^2+2x-5=&\,0. \end{align*}

Using the quadratic formula,

\begin{align*} x=&\,\frac{-2\pm\sqrt{2^2-4(12)(-5)}}{2(12)}\\ =&\,\frac{-2\pm\sqrt{244}}{24}\\ =&\,\frac{-2\pm2\sqrt{61}}{24}\\ =&\,\frac{-1\pm\sqrt{61}}{12}. \end{align*}

The $x$ coordinate of $T$ is positive, so

\begin{align*} x=\frac{-1+\sqrt{61}}{12}. \end{align*}

IAL 2025 Oct Q10

题目

解答

(a)

解法一

思路

答题过程

(b)

解法一

思路

答题过程