150 practice questions covering Relations and Functions, organised into 2 topics.
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A relation $R$ is defined on the set of all straight lines in a plane by taking $L_1\,R\,L_2$ to mean that $L_1$ is parallel to $L_2$, where every line counts as parallel to itself. Statement 1: $R$ is an equivalence relation. Statement 2: The equivalence class of a line $L$ is the set of all lines perpendicular to $L$. Which is correct?
AStatement 1 is false and Statement 2 is true
BBoth statements are true
CBoth statements are false
DStatement 1 is true and Statement 2 is false
Show answer and explanation
Correct answer: D - Statement 1 is true and Statement 2 is false
With a line counted as parallel to itself, $R$ is reflexive; parallelism works both ways, so $R$ is symmetric; and two lines parallel to a common line are parallel to each other, so $R$ is transitive. Statement 1 is therefore true. The class of $L$ collects the lines related to $L$, which are the lines parallel to $L$, not the ones perpendicular to it, so Statement 2 is false.
Q2
easy
Let $A = \{1,2,3\}$ and $R = \{(1,2),(2,1),(3,3)\}$. Which statement is correct?
A$R$ is symmetric but not reflexive
B$R$ is reflexive but not symmetric
C$R$ is both reflexive and symmetric
D$R$ is neither reflexive nor symmetric
Show answer and explanation
Correct answer: A - $R$ is symmetric but not reflexive
Every pair in $R$ has its reverse in $R$: $(1,2)$ pairs with $(2,1)$, and $(3,3)$ is its own reverse, so $R$ is symmetric. Reflexivity fails because $(1,1)$ and $(2,2)$ are both missing even though $1$ and $2$ belong to $A$. So $R$ is symmetric without being reflexive.
Q3
medium
Assertion (A): The relation $R = \{(a,b) : a, b \in \mathbb{Z} \text{ and } a + b \text{ is even}\}$ is an equivalence relation on $\mathbb{Z}$. Reason (R): For every integer $a$, the number $a + a = 2a$ is even.
ABoth A and R are true, and R is the correct explanation of A
BBoth A and R are true, but R is not the correct explanation of A
CA is true, but R is false
DA is false, but R is true
Show answer and explanation
Correct answer: B - Both A and R are true, but R is not the correct explanation of A
Reflexivity holds since $2a$ is even, symmetry holds since $a + b = b + a$, and transitivity holds since $a + c = (a+b) + (b+c) - 2b$ is even whenever $a+b$ and $b+c$ are, so (A) is true. The fact stated in (R) is also true, but it settles only reflexivity; symmetry and transitivity still have to be checked separately, so (R) is not a complete explanation of (A).
Q4
medium
A set $A$ has $3$ elements and a set $B$ has $5$ elements. How many one-one functions $f : A \to B$ are there?
A$125$
B$243$
C$10$
D$60$
Show answer and explanation
Correct answer: D - $60$
Give the elements of $A$ images one at a time: the first has $5$ choices, the second must avoid that image and has $4$, and the third has $3$, so the count is $5 \times 4 \times 3 = 60$. The value $125 = 5^3$ counts all functions, repeats allowed, and $243 = 3^5$ also exchanges the roles of the two sets. The value $10$ chooses which three elements of $B$ get used but ignores the order in which the inputs take them.
Q5
medium
The real function $f$ is given by $f(x)=|x|$ for every real $x$. What is the value of $(3f)(-4)-(2f)(5)$?
A$-22$
B$-2$
C$2$
D$22$
Show answer and explanation
Correct answer: C - $2$
Here $(3f)(-4)=3|-4|=12$ and $(2f)(5)=2|5|=10$, so the difference is $12-10=2$. The value $-2$ reverses the order of subtraction. The value $22$ adds the two results. The value $-22$ drops the modulus and computes $3(-4)-2(5)$.
Q6
medium
Let $f : \mathbb{R} \to [0,\ \infty)$ be given by $f(x) = x^2$. Which statement is correct?
A$f$ is neither one-one nor onto, because $f(0) = 0$
B$f$ is onto but not one-one, because $[0,\ \infty)$ is exactly the set of squares
C$f$ is one-one but not onto, because $f$ never takes a negative value
D$f$ is both one-one and onto, because every non-negative number has a square root
Show answer and explanation
Correct answer: B - $f$ is onto but not one-one, because $[0,\ \infty)$ is exactly the set of squares
Every $y \ge 0$ is the square of $\sqrt{y}$, and the declared codomain holds exactly the non-negative reals, so each element of the codomain is an image and $f$ is onto. Injectivity fails because $f(-3) = 9 = f(3)$, so having a square root for every codomain element is not enough for a bijection. Judging onto against $\mathbb{R}$ instead of the declared codomain is what makes the negative values look like a gap. That $f(0) = 0$ breaks neither property.
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