How Relations and Functions Appears in the Board Exam

This chapter opens the paper's syllabus and shows up every year, in a very stable pattern:

The marking scheme rewards structure: name the property, take arbitrary elements, run the algebra (or exhibit the counterexample), conclude. Each property proved or disproved cleanly earns its own credit.

Below are 12 board-style written questions with complete solutions, organised by marks, followed by a 15-question MCQ quiz.

2-Mark Questions

Question 1 — Ordering of the reals

Show that the relation $R = \{(a, b) : a \leq b\}$ in $\mathbb{R}$ is reflexive and transitive but not symmetric.

Step 1 — reflexive: $a \leq a$ holds for every real $a$.

Step 2 — transitive: $a \leq b$ and $b \leq c$ give $a \leq c$.

Step 3 — not symmetric: $2 \leq 3$ but $3 \not\leq 2$ — one counterexample suffices.

Answer: reflexive and transitive, not symmetric.

Question 2 — Doubling on the naturals

Show that $f : \mathbb{N} \to \mathbb{N}$ given by $f(x) = 2x$ is one-one but not onto.

Step 1 — one-one: $f(x_1) = f(x_2) \Rightarrow 2x_1 = 2x_2 \Rightarrow x_1 = x_2$.

Step 2 — not onto: $1$ in the co-domain would require $2x = 1$, i.e. $x = \frac{1}{2} \notin \mathbb{N}$.

Answer: one-one, not onto — the range is only the even naturals.

Question 3 — Two quick compositions

If $f(x) = \vert x \vert$ and $g(x) = 5x - 2$ on $\mathbb{R}$, find $g \circ f$ and $f \circ g$.

Step 1 — $g \circ f$: $\ (g \circ f)(x) = g(\vert x \vert) = 5\vert x \vert - 2$.

Step 2 — $f \circ g$: $\ (f \circ g)(x) = \vert 5x - 2 \vert$.

Answer: $5\vert x \vert - 2$ and $\vert 5x - 2 \vert$ — different functions (at $x = 0$: values $-2$ and $2$), so the order of composition matters.

Question 4 — Symmetric only

Show that the relation $R = \{(1, 2), (2, 1)\}$ in the set $\{1, 2, 3\}$ is symmetric but neither reflexive nor transitive.

Step 1 — symmetric: the reverse of each pair present — $(2,1)$ for $(1,2)$ and vice versa — is present ✓.

Step 2 — not reflexive: $(1, 1) \notin R$.

Step 3 — not transitive: $(1,2) \in R$ and $(2,1) \in R$, but $(1,1) \notin R$.

Answer: symmetric only. Note the transitivity failure uses a chain that returns to its start — a pattern worth remembering.

3-Mark Questions

Question 5 — All three properties fail

Show that the relation $R = \{(a, b) : a \leq b^2\}$ in $\mathbb{R}$ is neither reflexive nor symmetric nor transitive.

Step 1 — not reflexive: for $a = \frac{1}{2}$, $\ \frac{1}{2} \leq \frac{1}{4}$ is false.

Step 2 — not symmetric: $(1, 2) \in R$ since $1 \leq 4$, but $(2, 1) \notin R$ since $2 \not\leq 1$.

Step 3 — not transitive: $(3, 2) \in R$ ($3 \leq 4$) and $(2, 1.5) \in R$ ($2 \leq 2.25$), but $(3, 1.5) \notin R$ ($3 \not\leq 2.25$).

Answer: none of the three properties holds. Each disproof is one concrete counterexample — the fastest full-marks route.

Question 6 — The greatest integer function

Prove that $f : \mathbb{R} \to \mathbb{R}$, given by $f(x) = [x]$, is neither one-one nor onto.

Step 1 — not one-one: $f(1.2) = 1 = f(1.9)$ while $1.2 \neq 1.9$.

Step 2 — not onto: $[x]$ is always an integer, so the co-domain element $\frac{1}{2}$ satisfies $f(x) = \frac{1}{2}$ for no $x$.

Step 3 — conclude: with a collision exhibited and a missed target exhibited, $f$ is neither injective nor surjective.

Answer: neither one-one nor onto.

Question 7 — A shifted parabola

Discuss whether $f : \mathbb{R} \to \mathbb{R}$, $f(x) = 1 + x^2$, is one-one and whether it is onto.

Step 1 — not one-one: $f(-1) = 2 = f(1)$, and $-1 \neq 1$.

Step 2 — not onto: $1 + x^2 \geq 1$ for every real $x$, so no $y < 1$ — for instance $y = 0$ — is attained.

Step 3 — conclude: the range is $[1, \infty)$, a proper subset of the co-domain.

Answer: neither one-one nor onto.

Question 8 — Compositions in both orders

If $f(x) = x^2$ and $g(x) = 2x + 1$ on $\mathbb{R}$, find $g \circ f$ and $f \circ g$, and show they are not equal.

Step 1 — $g \circ f$: $\ g(x^2) = 2x^2 + 1$.

Step 2 — $f \circ g$: $\ f(2x + 1) = (2x + 1)^2 = 4x^2 + 4x + 1$.

Step 3 — compare at $x = 1$: $\ (g \circ f)(1) = 3$ but $(f \circ g)(1) = 9$.

Answer: $g \circ f \neq f \circ g$ — one point of disagreement completes the proof.

5-Mark Questions

Question 9 — Equivalence relation with classes exhibited

Show that the relation $R = \{(a, b) : \vert a - b \vert$ is even$\}$ in $A = \{1, 2, 3, 4, 5\}$ is an equivalence relation, and show that all elements of $\{1, 3, 5\}$ are related to each other, all elements of $\{2, 4\}$ are related to each other, but no element of $\{1, 3, 5\}$ is related to any element of $\{2, 4\}$.

Step 1 — reflexive: $\vert a - a \vert = 0$, which is even.

Step 2 — symmetric: $\vert a - b \vert = \vert b - a \vert$, so evenness transfers.

Step 3 — transitive: if $a - b$ and $b - c$ are even, then $a - c = (a - b) + (b - c)$ is a sum of even numbers, hence even.

Step 4 — the classes: two odds differ by an even number, so $1, 3, 5$ are mutually related; two evens likewise for $2, 4$; an odd and an even differ by an odd number, so no cross-pair lies in $R$.

Answer: $R$ is an equivalence relation partitioning $A$ into $\{1, 3, 5\}$ and $\{2, 4\}$.

Question 10 — Similarity of triangles

Show that the relation $R = \{(T_1, T_2) : T_1$ is similar to $T_2\}$ on the set of all triangles is an equivalence relation. Consider the right triangles $T_1$ with sides 3, 4, 5; $\ T_2$ with sides 5, 12, 13; and $T_3$ with sides 6, 8, 10. Which among them are related?

Step 1 — reflexive: every triangle is similar to itself (ratio 1).

Step 2 — symmetric: if $T_1 \sim T_2$ with ratio $k$, then $T_2 \sim T_1$ with ratio $\frac{1}{k}$.

Step 3 — transitive: ratios compose — similarity with ratios $k_1$ then $k_2$ gives similarity with ratio $k_1 k_2$.

Step 4 — test the triples: $\frac{6}{3} = \frac{8}{4} = \frac{10}{5} = 2$, so $T_1 \sim T_3$; but $\frac{5}{3} \neq \frac{12}{4}$, so $T_2$ is similar to neither.

Answer: equivalence relation; only $T_1$ and $T_3$ are related.

Question 11 — The rational bijection

Let $A = \mathbb{R} - \{3\}$ and $B = \mathbb{R} - \{1\}$. Show that the function $f : A \to B$ defined by $f(x) = \dfrac{x - 2}{x - 3}$ is one-one and onto.

Step 1 — one-one: assume $\dfrac{x_1 - 2}{x_1 - 3} = \dfrac{x_2 - 2}{x_2 - 3}$. Cross-multiplying: $(x_1 - 2)(x_2 - 3) = (x_2 - 2)(x_1 - 3)$.

Step 2 — expand and cancel: $x_1 x_2 - 3x_1 - 2x_2 + 6 = x_1 x_2 - 3x_2 - 2x_1 + 6$, so $-3x_1 - 2x_2 = -3x_2 - 2x_1$, giving $x_1 = x_2$.

Step 3 — onto: for $y \in B$ solve $y = \dfrac{x - 2}{x - 3}$: $\ x(y - 1) = 3y - 2$, so $x = \dfrac{3y - 2}{y - 1}$, defined since $y \neq 1$.

Step 4 — the preimage is legal: $\dfrac{3y - 2}{y - 1} = 3$ would need $3y - 2 = 3y - 3$, impossible — so $x \in A$, and substituting back gives $f(x) = y$.

Answer: $f$ is one-one and onto (hence invertible, with $f^{-1}(y) = \frac{3y - 2}{y - 1}$).

Question 12 — Invertibility with the inverse computed

Let $f : [0, \infty) \to [4, \infty)$ be given by $f(x) = x^2 + 4$. Show that $f$ is invertible and find $f^{-1}$.

Step 1 — one-one: $x_1^2 + 4 = x_2^2 + 4 \Rightarrow x_1^2 = x_2^2 \Rightarrow x_1 = x_2$, since both are nonnegative.

Step 2 — onto: given $y \geq 4$, set $x = \sqrt{y - 4}$, which is a legal input ($y - 4 \geq 0$), and $f(x) = (y - 4) + 4 = y$.

Step 3 — conclude invertibility by the theorem (bijective $\iff$ invertible), with candidate $g(y) = \sqrt{y - 4}$.

Step 4 — verify both compositions: $$g(f(x)) = \sqrt{x^2 + 4 - 4} = \sqrt{x^2} = x \ (x \geq 0), \qquad f(g(y)) = (\sqrt{y - 4})^2 + 4 = y$$

Answer: $f$ is invertible with $f^{-1}(y) = \sqrt{y - 4}$. The restricted domain $[0, \infty)$ is what rescues one-one-ness from the parabola.

Presentation tip: every 5-mark solution above follows the same skeleton — properties one at a time, arbitrary elements, a clean counterexample where needed, and the final answer in words. Board examiners award each block separately.