In set-builder form, the set of even natural numbers is
A {2,4,6,…}
B {x|x even}
C {1,3,5,…}
D {x|x prime}
Set-builder form describes a set using a rule or property. “{x|x even}” means all values x that satisfy “x is even,” usually with a stated universe like natural numbers.
The roster form of {x∈N | x<4} is
A {1,2,3}
B {0,1,2,3}
C {2,3,4}
D {1,3,5}
Natural numbers are typically {1,2,3,…}. The condition x<4 gives elements 1, 2, and 3. Roster form lists them directly inside braces.
If A={1,2}, then 1 belongs to A means
A 1 ⊆ A
B A ∈ 1
C 1 = A
D 1 ∈ A
Symbol “∈” means “is an element of.” Since 1 is listed inside set A, we write 1 ∈ A. Subset symbol ⊆ is used only between sets.
If A={1,2}, then {1} is
A Element of A
B Equal to A
C Subset of A
D Complement of A
{1} is a set containing the element 1. Every element of {1} is in A, so {1} ⊆ A. It is not an element of A because A contains numbers, not sets.
In Venn diagrams, A ∪ B is shaded as
A Only overlap part
B Outside both circles
C Only A side
D Both circles total
Union includes everything in A or in B, including the common overlap. In a Venn diagram, you shade the entire region of both circles together.
In Venn diagrams, A ∩ B is shaded as
A Only overlap region
B Both circles total
C Outside union region
D Only B side
Intersection means elements common to both sets. So in a Venn diagram, only the shared overlapping area of circles A and B is shaded.
For any set A, A ∪ ∅ equals
A ∅
B U
C A′
D A
Union with the empty set adds no new elements because ∅ has none. So A ∪ ∅ remains A. This is a basic identity property of sets.
For any set A, A ∩ ∅ equals
A A
B ∅
C U
D A′
Intersection keeps only common elements. Since ∅ has no elements, nothing can be common with it, so A ∩ ∅ is always the empty set.
Idempotent law in sets states
A A∪A = A
B A∪U = U
C A∩U = A
D A−A = U
Idempotent laws say combining a set with itself changes nothing: A∪A=A and A∩A=A. Repeating the same set does not create new elements.
Absorption law gives A ∪ (A ∩ B) equals
A B
B A′
C U
D A
Absorption laws simplify expressions: A ∪ (A ∩ B) = A and A ∩ (A ∪ B) = A. The set A already covers all elements in A∩B.
If U={1,2,3,4} and A={1,3}, then A′ is
A {1,3}
B {1,2}
C {2,4}
D {3,4}
Complement A′ means elements in universal set U that are not in A. From U remove 1 and 3, leaving {2,4}.
If A ⊆ B, then A ∩ B equals
A A
B ∅
C B
D U
When A is a subset of B, every element of A is already in B. So common elements of A and B are exactly the elements of A, giving A ∩ B = A.
If A ⊆ B, then A ∪ B equals
A A
B B
C ∅
D A′
If A is inside B, union adds nothing extra beyond B. So A ∪ B = B. This is a standard property used to simplify set expressions.
If A={1,2,3} and B={3,4}, then A−B is
A {3}
B {4}
C {1,2,4}
D {1,2}
A−B keeps elements in A that are not in B. Since 3 is common, it is removed. The remaining elements from A are 1 and 2.
If A={a,b} and B={1}, then A×B is
A {(a,1),(b,1)}
B {(1,a),(1,b)}
C {(a,b),(1,1)}
D {(a,1,b)}
Cartesian product A×B forms ordered pairs (x,y) with x∈A and y∈B. With B having only 1, pair each element of A with 1.
If A has 0 elements, then A×B is
A Always non-empty
B Equals B
C Equals A
D Always empty
If A is empty, there is no first element available to form an ordered pair (a,b). Hence no ordered pairs exist and A×B becomes the empty set.
If A={1,2} then A×A has how many pairs?
A 2
B 4
C 6
D 8
|A×A| = |A|·|A|. Here |A|=2, so total ordered pairs are 2×2=4: (1,1),(1,2),(2,1),(2,2).
A relation is written as a set of
A Ordered pairs
B Single numbers
C Unordered pairs
D Subsets only
A relation is a subset of a Cartesian product, so it consists of ordered pairs (a,b). Each ordered pair represents that a is related to b.
For relation R={(1,2),(2,3)}, domain is
A {2,3}
B {1,3}
C {1,2}
D {3}
Domain is the set of all first components in ordered pairs. Here first elements are 1 and 2, so domain is {1,2}.
For relation R={(1,2),(2,3)}, range is
A {1,2}
B {1,3}
C {1,2,3}
D {2,3}
Range is the set of all second components that actually appear. Here the second elements are 2 and 3, so range is {2,3}.
A relation R on A is symmetric if
A (a,b)⇒(b,a)
B (a,a) always
C (a,b)⇒(a,c)
D (a,b),(b,c)⇒(a,c)
Symmetry means reversing any related pair keeps it in the relation. If a is related to b, then b must be related to a for every pair.
A relation R on A is reflexive if
A (a,b) always
B (a,b)⇒(b,a)
C No pair repeats
D (a,a) always
Reflexive requires every element to relate to itself. So for each a in A, the pair (a,a) must be present in the relation.
A relation can be both reflexive and symmetric but not
A Many-one
B Onto
C Transitive
D Constant
Reflexive and symmetric do not automatically force transitivity. A relation may have (a,b) and (b,c) without (a,c), so it fails transitive condition.
Inverse of R={(1,3),(2,4)} is
A {(3,1),(4,2)}
B {(1,3),(2,4)}
C {(1,2),(3,4)}
D {(3,2),(4,1)}
Inverse relation swaps each ordered pair. So (1,3) becomes (3,1) and (2,4) becomes (4,2). Direction of relation is reversed.
A relation is a function only if
A One input per output
B Range equals domain
C Domain equals U
D One output per input
A function assigns each input exactly one output. It can have different inputs sharing the same output, but a single input cannot map to two outputs.
Which relation is NOT a function?
A {(1,2),(2,3)}
B {(2,5),(3,5)}
C {(1,2),(1,3)}
D {(0,1),(4,1)}
Input 1 maps to two different outputs 2 and 3, violating “one output per input.” Therefore it is not a function from the domain containing 1.
For f: A→B, the range is a subset of
A A
B B
C A×B
D P(A)
Outputs of a function lie in the codomain B. The actual set of outputs is called range, so it must be contained in B.
If f(x)=3 for all x, then f is
A Identity function
B Onto function
C One-one function
D Constant function
A constant function gives the same output for every input. Here output is always 3, so it is constant. It is generally not one-one unless domain has one element.
If f(x)=x on real numbers, then f is
A Identity function
B Constant function
C Modulus function
D Step function
Identity function maps each element to itself, so f(x)=x. It is both one-one and onto when domain and codomain are the same set.
If f(x)=x² on R, then f is
A One-one function
B Not one-one
C Onto R
D Invertible on R
f(2)=4 and f(−2)=4, so different inputs give the same output. Hence it is not one-one on all real numbers and cannot have an inverse on R.
If f(x)=x² on x≥0, then f becomes
A Still many-one
B Not defined
C Not a function
D One-one function
Restricting domain to x≥0 removes the symmetry causing duplicates. Now each output y has exactly one input x=√y, so it becomes one-one.
A function f: R→R, f(x)=x³ is
A One-one onto
B Many-one onto
C One-one into
D Constant into
x³ is strictly increasing on R, so it is one-one. Also every real y has solution x=∛y, so it is onto R. Hence it is bijective.
A function is “into” when
A Range equals codomain
B Domain is empty
C Range proper subset
D Inverse always exists
Into function does not hit every element of the codomain. Some outputs in codomain are never achieved, so the range is a proper subset of the codomain.
If g maps every element of A to different elements of B, g is
A Onto function
B One-one function
C Constant function
D Into function
One-one means distinct inputs give distinct outputs. If no two elements of A share the same image in B, the mapping is injective.
If every element of B is hit, function is
A One-one
B Many-one
C Piecewise
D Onto
Onto (surjective) means range equals codomain. Every element of the target set B has at least one preimage in A.
A piecewise function is defined by
A Different rules intervals
B One rule only
C Only polynomials
D Only rationals
Piecewise functions use different formulas on different parts of the domain, like one rule for x<0 and another for x≥0. This helps model real situations.
Greatest integer ⌊2.9⌋ equals
A 3
B 1
C 2
D 0
The greatest integer function ⌊x⌋ gives the largest integer less than or equal to x. For 2.9, integers ≤2.9 are …,2, so answer is 2.
Greatest integer ⌊−1.2⌋ equals
A −1
B 0
C 1
D −2
For negative numbers, floor goes to more negative integer. Integers less than or equal to −1.2 include −2, −3, … The greatest among them is −2.
Modulus |−7| equals
A −7
B 7
C 0
D 1
Modulus gives distance from zero, always non-negative. So |−7| = 7. It changes sign of negative inputs but keeps positive inputs unchanged.
For f(x)=1/x, domain is
A Real except 0
B All real numbers
C Only positive reals
D Only integers
1/x is undefined at x=0 because division by zero is not allowed. Therefore domain is all real numbers except 0.
For f(x)=1/x, range is
A All real numbers
B Only positive reals
C Real except 0
D Only integers
Output y=1/x can never be 0 for any real x. For any nonzero y, choosing x=1/y gives that y. So range is R excluding 0.
If f(x)=x+5, then f⁻¹(x) is
A x+5
B 5−x
C x/5
D x−5
To find inverse, set y=x+5 and solve for x: x=y−5. Replacing y with x gives f⁻¹(x)=x−5. This works since function is one-one and onto.
If f(x)=2x, then f⁻¹(x) is
A x/2
B 2/x
C 2x
D x−2
Let y=2x, then x=y/2. So inverse is f⁻¹(x)=x/2. Linear functions with nonzero slope are invertible on real numbers.
If (f∘g)(x)=f(g(x)), then (g∘f)(x) is
A f(g(x))
B f(x)g(x)
C f(x)+g(x)
D g(f(x))
Composition order matters. g∘f means apply f first, then g. So (g∘f)(x)=g(f(x)), which is generally different from f(g(x)).
Composition of functions is
A Always commutative
B Generally not commutative
C Always distributive
D Always symmetric
Usually f∘g ≠ g∘f because changing order changes the result. Only in special cases do two functions commute under composition.
Identity element for composition is
A Identity function
B Zero function
C Constant function
D Modulus function
Identity function I(x)=x acts as a neutral element: f∘I = f and I∘f = f. This is similar to number 1 in multiplication.
Associative property means
A f∘g = g∘f
B f+g = g+f
C f∘f = I
D (f∘g)∘h = f∘(g∘h)
Composition is associative: the grouping does not change the final function, as long as domains fit. But the order still matters, so it is not commutative.
If f is one-one, cancellation says
A f(a)=b ⇒ a=b
B f(a)=f(b) ⇒ a≠b
C f(a)=f(b) ⇒ a=b
D f(a)=a always
Injective (one-one) functions preserve distinctness. If two outputs match, the inputs must have been the same. This property is used to prove uniqueness and inverses.
If f is onto, then for every y in codomain
A At least one x exists
B Exactly one x exists
C No x exists
D x must be 0
Surjective means every codomain element is achieved by the function. So for each y in codomain, there exists at least one x in domain such that f(x)=y.
If f(x)=|x|, then f is
A Odd function
B Periodic function
C Constant function
D Even function
For even functions, f(−x)=f(x). Since |−x|=|x| for all real x, modulus function is even and its graph is symmetric about the y-axis