For finite sets A and B, the correct counting formula for |A ∪ B| is
A |A|+|B|
B |A|+|B|−|A∩B| only
C |A|−|B|
D |A|+|B|−|A∩B|
When adding sizes of A and B, common elements are counted twice. Subtracting |A∩B| removes the extra count, giving the correct size of the union.
For any finite sets A and B, |A − B| always equals
A |A|+|B|
B |B|−|A∩B|
C |A|−|A∩B|
D |A∩B|
Set A splits into two disjoint parts: (A−B) and (A∩B). Their sizes add to |A|, so |A−B| = |A| − |A∩B|.
The set operation A − B can be rewritten as
A A ∩ B′
B A ∪ B
C A′ ∩ B
D A′ ∪ B′
A−B means elements in A but not in B. “Not in B” is B′, and “in A” is intersection with A, so A−B = A∩B′.
If A ⊆ B in the same universal set, then which is always true
A A′ ⊆ B′
B A ⊆ B′
C B′ ⊆ A′
D B ⊆ A′
If A is inside B, then elements outside B are certainly outside A as well. So the complement of B must be a subset of the complement of A.
If A and B are disjoint finite sets, then the value of |A ∪ B| is
A |A|−|B|
B |A|+|B|
C |A∩B|
D |A|+|B|−1
Disjoint sets have no overlap, so no element is double-counted in the union. Therefore the size of A∪B is just the sum of sizes.
The expression (A ∩ B) ∪ (A ∩ C) simplifies to
A A ∩ (B ∪ C)
B A ∪ (B ∩ C)
C (A ∪ B) ∩ C
D (A ∩ B) ∩ C
This is the distributive law of sets. Intersection with A distributes over union: A∩B or A∩C together equals A∩(B∪C).
The complement of (A − B) in universal set U is
A A′ ∩ B
B A ∪ B′
C A′ ∪ B
D A′ ∩ B′
A−B equals A∩B′. Taking complement gives (A∩B′)′ = A′ ∪ (B′)′ = A′ ∪ B by De Morgan’s law.
If A is a proper subset of B, then which must be true
A A = B
B B − A is empty
C A − B nonempty
D B − A nonempty
Proper subset means A⊂B but A≠B, so at least one element is in B but not in A. That element lies in B−A, making it nonempty.
When A ⊆ B, the correct relation between power sets is
A P(A) ⊆ P(B)
B P(B) ⊆ P(A)
C P(A) = ∅
D P(B) = ∅
Every subset of A is automatically a subset of B if A is inside B. So each element of P(A) is also an element of P(B).
If a set A has n elements, then |P(P(A))| equals
A 2n
B 2^n
C 2^(2^n)
D (2^n)^2
P(A) has 2^n elements. The power set of that set has 2^(2^n) subsets. This grows extremely fast even for small n.
The identity A × (B ∪ C) equals
A (A×B) ∩ (A×C)
B (A×B) ∪ (A×C)
C (A∪B) × C
D A × (B∩C)
Pairing A with elements from (B∪C) means pairing A with B plus pairing A with C. Hence the Cartesian product distributes over union.
The identity (A ∩ B) × C equals
A (A×C) ∪ (B×C)
B (A×B) × C
C (A×C) ∩ (B×C)
D A × (B×C)
A pair (x,z) belongs to both A×C and B×C exactly when x is in both A and B, and z is in C. That matches (A∩B)×C.
If nonempty sets A and B satisfy A×B = B×A, then
A A = B
B A ⊂ B
C B ⊂ A
D A ∩ B = ∅
With nonempty sets, equality of products forces every a in A to also be in B and every b in B to also be in A. Hence A⊆B and B⊆A.
If |A|=m and |B|=n, number of relations from A to B is
A m+n
B mn
C 2^(mn)
D (mn)^2
A relation is any subset of A×B. Since A×B has mn ordered pairs, the number of subsets is 2^(mn), giving total possible relations.
If a set A has n elements, number of relations on A is
A 2^n
B n^2
C n!
D 2^(n^2)
A relation on A is a subset of A×A. Since A×A has n^2 ordered pairs, the number of subsets (relations) is 2^(n^2).
If |A|=m and |B|=n, then a relation matrix for A→B has size
A m×n
B n×m
C m+n
D m−n
Rows correspond to elements of A (domain side) and columns correspond to elements of B (codomain side). So the matrix has m rows and n columns.
For a set A with n elements, maximum possible ordered pairs in a relation on A is
A n
B 2n
C n^2
D 2^n
A relation on A is any subset of A×A. The largest subset is the whole A×A itself, which contains n^2 ordered pairs.
A relation matrix represents a symmetric relation when the matrix is
A Diagonal only
B Symmetric about diagonal
C All ones
D All zeros
Symmetry means (a,b) implies (b,a). In a matrix, that means entry (i,j) equals entry (j,i) for all i and j, making it symmetric.
Equivalence relations are important because their equivalence classes
A Form a partition
B Always overlap
C Are always empty
D Are always singleton
Equivalence classes cover the whole set without overlap. Every element belongs to exactly one class, so these classes partition the set into clear, non-overlapping groups.
If R ⊆ A×B and S ⊆ B×C, then S∘R is a subset of
A A×A
B B×B
C A×C
D C×A
In composition, an element starts in A, moves to B using R, then to C using S. So the composed pairs connect A directly to C.
For relations R and S, the inverse of a composition satisfies
A (S∘R)⁻¹ = S⁻¹∘R⁻¹
B (S∘R)⁻¹ = S∘R
C (S∘R)⁻¹ = R∘S
D (S∘R)⁻¹ = R⁻¹∘S⁻¹
Inverse reverses direction. Composition applies R then S, so the inverse must undo S first and then undo R. That becomes R⁻¹∘S⁻¹.
A relation from A to B becomes a function only when each a in A has
A At least two images
B No ordered pairs
C Exactly one image
D Only diagonal pair
A function assigns each input a unique output. If any input maps to two different outputs, it is not a function. If an input maps to none, it fails as function too.
If |A|=m and |B|=n, number of functions A→B equals
A m^n
B n^m
C mn
D n+m
Each of m elements of A can be sent to any of n elements of B independently. That gives n choices repeated m times, so total functions are n^m.
If m≤n, the number of one-one functions A→B is
A nPm
B n^m
C nCm
D mPn
For injective functions, choose distinct images in B for m elements of A. That is permutation count nPm = n!/(n−m)!, assigning different outputs to different inputs.
If |A|=|B|=n, number of bijections A→B is
A n
B n^2
C n!
D 2^n
A bijection pairs every element of A with a unique element of B. The number of such one-to-one matchings equals the number of permutations of n elements, which is n!.
For f: R→R defined by f(x)=x²+1, the function is
A One-one only
B Onto R
C Constant function
D Not onto R
Since x²≥0, x²+1≥1. So outputs are never less than 1. Many real numbers like 0 or −5 are not achieved, so it is not onto R.
For f: R→R defined by f(x)=x³+1, the function is
A Onto R
B Not onto R
C Not one-one
D Constant function
For any real y, solve y=x³+1 ⇒ x=∛(y−1), which is real. So every real value is hit, making the function onto R and also one-one.
If f(x)=(x−1)/2, then the inverse function is
A (x−1)/2
B 2x−1
C 2x+1
D x/2−1
Set y=(x−1)/2. Solve: 2y=x−1 ⇒ x=2y+1. Replacing y with x gives f⁻¹(x)=2x+1. Linear with nonzero slope is invertible.
For f(x)=(x+1)² on real numbers, a correct restriction to make it invertible is
A x ≤ 1
B x ≥ −1
C x ≥ 1
D x ≤ 0
Squaring creates two inputs giving same output. Restricting to one side of the vertex makes it one-one. For (x+1)², vertex is at x=−1, so x≥−1 works.
If a function f is even, then which statement always holds
A f(0)=1
B f(−x)=−f(x)
C f(x)=f(|x|)
D f(x+1)=f(x)
Even means f(−x)=f(x). So the value depends only on distance from zero, i.e., |x|. Therefore f(x) equals f(|x|) for all x in domain.
If a function f is odd and defined at 0, then f(0) must be
A 0
B 1
C −1
D Undefined
Odd means f(−x)=−f(x). Put x=0 gives f(0)=−f(0). Only value equal to its negative is 0, so f(0)=0.
If T is a period of f(x), then which is also a period
A T/2
B −T/3 only
C 2T
D 0
If f(x+T)=f(x), then repeating twice gives f(x+2T)=f(x). So 2T is always a period. The smallest positive period may still be T.
The range of f(x)=|x−2|+3 over all real x is
A y > 3
B y ≤ 3
C All real y
D y ≥ 3
|x−2| is never negative, minimum 0 at x=2. Adding 3 shifts outputs up by 3. So smallest output is 3, and all larger values occur.
The range of g(x)=1/(x²+1) over real x is
A 0
B y≥1
C y<0 only
D All real y
Since x²+1 ≥ 1, the fraction is at most 1 (at x=0). Denominator grows without bound, so values approach 0 but never reach 0.
The domain of h(x)=√(x−5) over real numbers is
A x ≤ 5
B x ≠ 5
C x ≥ 5
D All real x
Square root requires the inside to be non-negative. So x−5 ≥ 0, giving x ≥ 5. This ensures h(x) is real-valued.
The relation “a divides b” on natural numbers is
A Symmetric
B Transitive
C Not reflexive
D Not antisymmetric
If a divides b and b divides c, then c=bk and b=al, so c=a(lk). Hence a divides c, so the relation is transitive.
The relation “a=b” on any set is
A Only symmetric
B Only transitive
C Equivalence relation
D Only reflexive
Equality is reflexive (a=a), symmetric (a=b ⇒ b=a), and transitive (a=b and b=c ⇒ a=c). Thus it satisfies all three conditions.
The relation “a
A Not symmetric
B Symmetric relation
C Reflexive relation
D Equivalence relation
If a
For functions with suitable domains, composition satisfies
A f∘g = g∘f
B f∘f = I always
C (f∘g)∘h = f∘(g∘h)
D (f∘g)=f+g
Composition is associative: grouping does not change the final result as long as each composition is defined. But changing order generally changes result, so it is not commutative.
If f is one-one and f∘g₁ = f∘g₂, then
A f = g₁
B g₁ is onto
C f is constant
D g₁ = g₂
One-one allows left cancellation. If applying f to outputs of g₁ and g₂ gives same result for every input, then outputs of g₁ and g₂ must match, so g₁=g₂.
If g is onto and f₁∘g = f₂∘g, then
A g is one-one
B f₁ = f₂
C f₁ is constant
D f₂ is identity
Onto allows right cancellation. Since every element of codomain of g is reached by some input, equality after composing with g forces f₁ and f₂ to agree everywhere.
The inverse of the identity function I(x)=x is
A I(x)=x
B −x
C 1/x
D x²
Identity maps each element to itself. Undoing it requires no change, so the inverse is also the identity function. Composition with identity keeps any function unchanged.
For a bijection f, the expression (f⁻¹)⁻¹ equals
A f⁻¹
B I only
C f
D ∅
Taking inverse swaps input and output roles. Doing that twice returns the original mapping. This works when inverse exists as a function, i.e., when f is bijective.
For f(x)=ax+b on real numbers, f is invertible when
A a = 0
B b = 0
C a = b
D a ≠ 0
If a=0, f becomes constant and many inputs give same output, so no inverse function exists. When a≠0, f is one-one and onto, so it is invertible.
If f(x)=1/(x−2), then f⁻¹(x) equals
A 1/(x−2)
B 2 − 1/x
C 2 + 1/x
D (x−2)/x
Solve y=1/(x−2). Then x−2=1/y, so x=2+1/y. Replace y by x to write the inverse: f⁻¹(x)=2+1/x (x≠0).
The domain of (f∘g)(x) is the set of x such that
A g(x) in dom f
B x in dom f
C x in dom g
D f(x) in dom g
Composition f∘g means apply g first, then f. So x must be allowed in g, and the output g(x) must be an allowed input for f.
If a function f is many-one, then its inverse relation
A Is always a function
B Is not a function
C Equals identity
D Becomes constant
Many-one means different inputs can share the same output. Then reversing the mapping gives one output pointing to multiple inputs, so inverse cannot assign a unique value.
A relation written as ordered pairs is a function exactly when
A Same second repeats
B All pairs diagonal
C Same first unique
D No pair exists
In a function, each first component (input) must appear with exactly one second component (output). If one input appears with two different outputs, it fails the function rule.
The function f(x)=⌊x⌋ with domain R and codomain Z is
A Onto Z
B One-one on R
C Constant function
D Not well-defined
For every integer k, choose x=k. Then ⌊k⌋=k, so every integer is achieved. It is not one-one because many real numbers share the same floor value.
If f: R→R is defined by f(x)=|x|, then f is
A Onto R
B Bijective on R
C Not onto R
D Constant on R
|x| never produces negative outputs, so values like −3 are not reached. Hence it is not onto R. It becomes onto if codomain is restricted to y≥0.