(a) Linear self-inverse classification M1·A1·A1·A1·R1
Let $f(x) = mx + k$. Compute:
$$ f(f(x)) \;=\; m(mx + k) + k \;=\; m^{2}\,x + (mk + k) \;=\; m^{2}\,x + k(m + 1). $$
Self-inverse $\Leftrightarrow$ $f(f(x)) = x$ for every $x \Leftrightarrow$ $m^{2} = 1$ and $k(m + 1) = 0$ (equate coefficients of $x$ and the constant).
- $m^{2} = 1 \Rightarrow m = 1$ or $m = -1$.
- If $m = 1$: $k(1 + 1) = 0 \Rightarrow k = 0$. Function: $f(x) = x$ (identity).
- If $m = -1$: $k(-1 + 1) = 0 \Rightarrow 0 = 0$, satisfied for every $k$. Functions: $f(x) = -x + k$, i.e. $f(x) = c - x$ with $c = k \in \mathbb{R}$.
Conclusion: the linear self-inverse functions on $\mathbb{R}$ are the identity together with the one-parameter family $f(x) = c - x$, $c \in \mathbb{R}$. The non-trivial ones are exactly $f(x) = c - x$.
(b) Symmetry about $y = x$ M1·A1·R1
Suppose $f$ is self-inverse: $f(f(x)) = x$. If $(p, q)$ lies on $y = f(x)$, then $q = f(p)$, and applying $f$ again gives $f(q) = f(f(p)) = p$. So $(q, p)$ also lies on $y = f(x)$. The reflection of $(p, q)$ in $y = x$ is $(q, p)$, which is also on the graph. Hence the graph is invariant under reflection in $y = x$, i.e.
symmetric about $y = x$.
Fixed points: $f(x) = x$ means $(x, x)$ is on the graph, which trivially lies on $y = x$. Conversely, any intersection of the graph with $y = x$ satisfies $y = x$ and $y = f(x)$, hence $f(x) = x$, i.e. a fixed point.
(c) Fixed points of $f(x) = c - x$ M1·A1
$f(x) = x \;\Longleftrightarrow\; c - x = x \;\Longleftrightarrow\; x = \dfrac{c}{2}$. So the unique fixed point is $x = c/2$. Check against (b): the graph $y = c - x$ is a line of slope $-1$ through $(c, 0)$ and $(0, c)$; it crosses $y = x$ at exactly $(c/2, c/2)$, consistent.
(d) Compose $g(x) = a - x$ and $h(x) = b - x$ M1·A1·R1
$(g \circ h)(x) = g(b - x) = a - (b - x) = (a - b) + x$. So $(g \circ h)(x) = x + (a - b)$, a translation.
$(h \circ g)(x) = h(a - x) = b - (a - x) = (b - a) + x$, a translation by $(b - a) = -(a - b)$.
Self-inverse? A translation $T(x) = x + t$ has $T(T(x)) = x + 2t$, which equals $x$ iff $t = 0$, i.e. iff $a = b$. So $g \circ h$ is self-inverse
only when $a = b$ (in which case it is the identity). For $a \ne b$, neither composite is self-inverse. (Composition of two self-inverse functions is not, in general, self-inverse — composition of involutions is not an involution.)
(e) Non-linear self-inverse on $\mathbb{R} \setminus \{0\}$ A1·A1
Take $f(x) = \dfrac{1}{x}$ on $\mathbb{R} \setminus \{0\}$.
$$ f(f(x)) \;=\; f\!\left(\frac{1}{x}\right) \;=\; \frac{1}{\,1/x\,} \;=\; x. \;\checkmark $$
So $f \circ f = \mathrm{id}$, $f$ is self-inverse. Another valid choice: $f(x) = \dfrac{k}{x}$ for any $k \ne 0$ (gives $f(f(x)) = \dfrac{k}{k/x} = x$).
Involutions form a wider zoo than "$c - x$". A self-inverse function on a set is an involution. On $\mathbb{R}$, the linear involutions are the identity and the half-turns $x \mapsto c - x$ (rotation by $180^{\circ}$ about $(c/2, c/2)$). On $\mathbb{R} \setminus \{0\}$, the rational involutions $x \mapsto k/x$ join the family. On $\widehat{\mathbb{R}} = \mathbb{R} \cup \{\infty\}$ (Mobius transforms), the full self-inverse family is exactly the trace-zero Mobius transforms (Q5d), a $3$-parameter family. The composition counter-example in (d) is the key: the set of involutions is not closed under composition. Composing $x \mapsto a - x$ and $x \mapsto b - x$ gives a translation, which has infinite order (not order $2$) whenever $a \ne b$. This is why the involutions of a set form a generating set for a group of order $2$ involutions and translations, not the group of involutions itself — a Paper 3-flavour structural observation.
(a) 线性自逆分类 M1·A1·A1·A1·R1
设 $f(x) = mx + k$。计算：
$$ f(f(x)) \;=\; m(mx + k) + k \;=\; m^{2}\,x + (mk + k) \;=\; m^{2}\,x + k(m + 1). $$
自逆 $\Leftrightarrow$ 对每个 $x$，$f(f(x)) = x$ $\Leftrightarrow$ $m^{2} = 1$ 且 $k(m + 1) = 0$（按 $x$ 与常数项比较系数）。
- $m^{2} = 1 \Rightarrow m = 1$ 或 $m = -1$。
- $m = 1$：$k(1 + 1) = 0 \Rightarrow k = 0$。函数 $f(x) = x$（恒等）。
- $m = -1$：$k(-1 + 1) = 0 \Rightarrow 0 = 0$，对任意 $k$ 成立。函数 $f(x) = -x + k$，即 $f(x) = c - x$（$c = k \in \mathbb{R}$）。
结论：$\mathbb{R}$ 上的线性自逆函数为恒等加上一参数族 $f(x) = c - x$（$c \in \mathbb{R}$）。非平凡者恰为 $f(x) = c - x$。
(b) 关于 $y = x$ 对称 M1·A1·R1
设 $f$ 自逆：$f(f(x)) = x$。若 $(p, q)$ 在 $y = f(x)$ 上，则 $q = f(p)$，再用一次 $f$：$f(q) = f(f(p)) = p$。故 $(q, p)$ 也在 $y = f(x)$ 上。$(p, q)$ 关于 $y = x$ 的反射就是 $(q, p)$，也在图像上。故图像在 $y = x$ 反射下不变，即
关于 $y = x$ 对称。
不动点：$f(x) = x$ 即 $(x, x)$ 在图像上，本就在 $y = x$ 上。反之，图像与 $y = x$ 的交点满足 $y = x$ 与 $y = f(x)$，故 $f(x) = x$，是不动点。
(c) $f(x) = c - x$ 的不动点 M1·A1
$f(x) = x \;\Longleftrightarrow\; c - x = x \;\Longleftrightarrow\; x = \dfrac{c}{2}$。唯一不动点 $x = c/2$。与 (b) 核对：直线 $y = c - x$ 斜率 $-1$，过 $(c, 0)$ 与 $(0, c)$，与 $y = x$ 恰交于 $(c/2, c/2)$，一致。
(d) 复合 $g(x) = a - x$ 与 $h(x) = b - x$ M1·A1·R1
$(g \circ h)(x) = g(b - x) = a - (b - x) = (a - b) + x$。故 $(g \circ h)(x) = x + (a - b)$，一个平移。
$(h \circ g)(x) = h(a - x) = b - (a - x) = (b - a) + x$，平移量 $(b - a) = -(a - b)$。
是否自逆？平移 $T(x) = x + t$ 满足 $T(T(x)) = x + 2t$，等于 $x$ 当且仅当 $t = 0$，即 $a = b$。故 $g \circ h$
仅在 $a = b$ 时自逆（此时即恒等）。$a \ne b$ 时两复合都不自逆。（两个自逆函数的复合一般不自逆——对合的复合不是对合。）
(e) $\mathbb{R} \setminus \{0\}$ 上非线性自逆 A1·A1
取 $f(x) = \dfrac{1}{x}$（$\mathbb{R} \setminus \{0\}$）。
$$ f(f(x)) \;=\; f\!\left(\frac{1}{x}\right) \;=\; \frac{1}{\,1/x\,} \;=\; x. \;\checkmark $$
故 $f \circ f = \mathrm{id}$，$f$ 自逆。另一族合法选择：$f(x) = \dfrac{k}{x}$（$k \ne 0$）；$f(f(x)) = \dfrac{k}{k/x} = x$。
对合家族远不止 "$c - x$"。自逆函数即对合。$\mathbb{R}$ 上线性对合为恒等与半旋 $x \mapsto c - x$（关于 $(c/2, c/2)$ 的 $180^{\circ}$ 旋转）。$\mathbb{R} \setminus \{0\}$ 上有理对合 $x \mapsto k/x$ 加入。$\widehat{\mathbb{R}} = \mathbb{R} \cup \{\infty\}$（Mobius 变换）上完整的自逆族恰为迹为零的 Mobius 变换（Q5d），是 $3$ 参数族。(d) 的反例是关键：对合集合不在复合下封闭。$x \mapsto a - x$ 与 $x \mapsto b - x$ 的复合为平移，$a \ne b$ 时阶无穷（非 $2$ 阶）。因此一个集合上的对合只是生成"含对合与平移的群"的生成系，并非自成一群——Paper 3 风格的结构性观察。