Online Test — Relations and Functions
10 Questions • 20 min • Chapter MCQ
20:00
Question 1 of 10
medium
The relation $R = \{(a,b): |a-b|$ is divisible by $4\}$ on $A = \{1,2,3,4,...,10\}$ is:
Reflexive only
Symmetric only
An equivalence relation
Transitive only
Explanation: $|a-a|=0$ divisible by 4 (reflexive); $|a-b|=|b-a|$ (symmetric); if $4|a-b|$ and $4|b-c|$ then $4|(a-c)$ (transitive). So it is an equivalence relation.
Question 2 of 10
easy
A function $f: A \to B$ is onto (surjective) if:
$f(x_1)=f(x_2) \Rightarrow x_1=x_2$
Range of $f$ equals $B$ (the codomain)
$f$ has an inverse
$|A| \leq |B|$
Explanation: A function is onto/surjective if every element of the codomain $B$ has at least one preimage in $A$, i.e. Range = Codomain.
Question 3 of 10
easy
If $f(x) = 3x - 2$, then $f^{-1}(x)$ is:
$\frac{x+2}{3}$
$3x+2$
$\frac{x-2}{3}$
$\frac{2-x}{3}$
Explanation: Set $y=3x-2$, then $x=\frac{y+2}{3}$. So $f^{-1}(x)=\frac{x+2}{3}$.
Question 4 of 10
medium
The number of equivalence classes of the relation $R$ on $\mathbb{Z}$ defined by $aRb \iff 3|(a-b)$ is:
1
2
3
Infinite
Explanation: The equivalence classes are $[0]=[3]=\{...,-3,0,3,...\}$, $[1]=\{...,-2,1,4,...\}$, $[2]=\{...,-1,2,5,...\}$ — exactly 3 classes corresponding to remainders mod 3.
Question 5 of 10
easy
If $f: \mathbb{R} \to \mathbb{R}$ defined by $f(x) = x^3$ is bijective, then $f^{-1}(8)$ is:
2
$\pm2$
$8^{1/3}$
4
Explanation: $f(x)=x^3$ is bijective on $\mathbb{R}$. $f^{-1}(y)=y^{1/3}$. So $f^{-1}(8)=2$.
Question 6 of 10
easy
If $f(x) = 2x + 3$ and $g(x) = x^2$, then $(f \circ g)(2)$ is:
7
11
49
10
Explanation: $(f \circ g)(2) = f(g(2)) = f(4) = 2(4)+3 = 11$.
Question 7 of 10
medium
Which of the following relations on $\{1,2,3\}$ is reflexive but NOT symmetric?
$\{(1,1),(2,2),(3,3),(1,2),(2,1)\}$
$\{(1,1),(2,2),(3,3),(1,2)\}$
$\{(1,2),(2,1)\}$
$\{(1,1),(2,2),(3,3)\}$
Explanation: Option B has all diagonal pairs (reflexive) but $(1,2)$ without $(2,1)$ (not symmetric).
Question 8 of 10
medium
The identity element for the binary operation $a * b = a + b - 5$ on $\mathbb{Z}$ is:
0
5
$-5$
1
Explanation: Need $a*e=a$: $a+e-5=a \Rightarrow e=5$. Check: $a*5=a+5-5=a$. ✓
Question 9 of 10
hard
A function $f: \{1,2,3\} \to \{1,2,3,4,5\}$ can be one-one. How many one-one functions exist?
60
125
24
120
Explanation: Number of one-one functions from a 3-element set to a 5-element set = $P(5,3) = 5 \times 4 \times 3 = 60$.
Question 10 of 10
hard
If $f$ and $g$ are bijections from $A$ to $B$ and $B$ to $C$ respectively, then $(g \circ f)^{-1}$ equals:
$g^{-1} \circ f^{-1}$
$f^{-1} \circ g^{-1}$
$f \circ g$
$(f \circ g)^{-1}$
Explanation: For bijections, $(g \circ f)^{-1} = f^{-1} \circ g^{-1}$. The order reverses when taking inverse of a composition.