函数 $\mathrm{f}$ 定义为

其中 $x\in\mathbb{R}$，$p$ 为常数。

(a) 求使方程 $\mathrm{f}(x)=0$ 有重根的 $p$ 的精确值。

函数 $\mathrm{g}$ 定义为

其中 $x\ge a$，$p$ 和 $a$ 为正常数。

(b) (i) 证明使反函数 $\mathrm{g}^{-1}$ 存在的 $a$ 的最小值为 $\frac{1}{p}$。

(ii) 已知 $p=2$ 且 $\mathrm{g}^{-1}$ 确实存在，求 $\mathrm{g}^{-1}(x)$ 的表达式。

Answer all questions. You should use a calculator where appropriate.

You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.

Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.

请回答所有题目。适当时应使用计算器。必须清楚展示所有必要的计算过程；仅给出计算器答案而没有过程，不予评分。除非题目另有精度要求，非精确数值答案保留 3 位有效数字，以度为单位的角度保留 1 位小数。

二次方程有重根时判别式为零。先注意 $p=0$ 时原式为一次方程。

For $p\ne0$, the discriminant is $(-2)^2-4p(p+4)$. Setting it to zero gives

Thus $p=-2\pm\sqrt5$. Both are nonzero.

配方后，重根意味着平方项以外的顶点值为零。此时方程只剩一个平方等于零。

If $p=0$, the equation is linear and has no repeated root. For $p\ne0$, completing the square gives

There is a repeated root precisely when $p+4-1/p=0$, since the equation then reduces to $p(x-1/p)^2=0$. Multiplying by the nonzero $p$ gives $p^2+4p-1=0$, so

Both values are nonzero and therefore give a repeated root.

用配方找顶点。定义域的左端若在顶点左边，顶点两侧会出现相同的函数值。

Completing the square gives

For $p>0$ this is strictly increasing on $x\ge1/p$. If $a<1/p$, the domain contains distinct points equally far to either side of $1/p$, with equal outputs. Thus the least possible $a$ is $1/p$.

一阶导数可以找出函数由减转增的位置。定义域从这个位置或其右侧开始，才能保留单调的一支。

Since $p>0$,

The derivative is negative for $x<1/p$ and positive for $x>1/p$. Thus $g$ decreases to its minimum at $x=1/p$ and then increases. On $x\ge1/p$ it is strictly increasing and one-to-one. If $a<1/p$, the domain includes parts of both sides of the minimum, giving repeated output values near that minimum. Hence $g^{-1}$ exists exactly when $a\ge1/p$, and the least possible $a$ is $1/p$.

代入 $p=2$ 后配方。反函数存在时 $a\ge1/2$，所以反解时取非负平方根。

With $p=2$, write $y=2(x-1/2)^2+11/2$. Since $x\ge a\ge1/2$, solving for $x$ gives

Hence