If A and B are finite with |A|=7, |B|=5 and |A∩B|=3, then |A∪B| is
A 10
B 11
C 9
D 12
Use inclusion–exclusion: |A∪B|=|A|+|B|−|A∩B|=7+5−3=9. This avoids double counting common elements.
If |A∪B|=14, |A|=9, and |B|=8, then |A∩B| is
A 1
B 3
C 2
D 4
Inclusion–exclusion gives 14=9+8−|A∩B|. So |A∩B|=17−14=3. This counts overlap correctly in finite sets.
If |A|=12 and |A−B|=7, then |A∩B| equals
A 3
B 4
C 7
D 5
A splits into disjoint parts (A−B) and (A∩B). So |A|=|A−B|+|A∩B|, hence 12=7+|A∩B| giving 5.
If U has 20 elements and |A|=12, then |A′| equals
A 6
B 10
C 8
D 12
Complement size is |A′|=|U|−|A|. With U=20 and A=12, A′ has 8 elements. Complement always depends on universal set.
If |A|=10, |B|=10 and |A∪B|=10, then A and B are
A Disjoint sets
B Equal sets
C Complement sets
D Proper subsets
If union size equals each set’s size, there is no extra element in either set. That can happen only when A and B contain exactly the same elements.
If A⊂B and |A|=|B| for finite sets, then
A Contradiction
B Always possible
C A is empty
D B is empty
Proper subset means B has at least one element not in A, so |B| must be greater than |A| for finite sets. Equal sizes contradict A⊂B.
If |A×B|=35 and |A|=5, then |B| is
A 5
B 6
C 7
D 8
For finite sets, |A×B|=|A||B|. So 35=5·|B| giving |B|=7. Each element of A pairs with all of B.
If A has 6 elements, number of relations on A that are reflexive is
A 2^36
B 2^6
C 36
D 2^30
Relations on A have 6²=36 possible pairs. Reflexive forces 6 diagonal pairs included, leaving 30 pairs optional. So number of reflexive relations is 2^30.
If A has 5 elements, number of symmetric relations on A is
A 2^25
B 2^15
C 2^10
D 5!
For symmetry, choose any subset of diagonal pairs (5 choices) and for each unordered pair {i,j}, choose include both or neither. There are C(5,2)=10 such pairs. Total choices 2^(5+10)=2^15.
If A has 4 elements, number of antisymmetric relations on A is
A 3^6·2^4
B 2^16
C 2^12
D 3^4
For each unordered pair {i,j} (there are C(4,2)=6), antisymmetry allows: include (i,j) only, include (j,i) only, or include neither → 3 choices. Diagonal 4 pairs are free → 2^4. Multiply gives 3^6·2^4.
For relation R on A, transitive means
A (a,b)⇒(b,a)
B (a,a) always
C Only diagonal pairs
D (a,b)&(b,c)⇒(a,c)
Transitivity connects chains: if a relates to b and b relates to c, then a must relate to c. This condition is central in equivalence and order relations.
If R is an equivalence relation on a set with 9 elements, possible number of equivalence classes can be
A 0
B 10
C 2
D 12
Number of equivalence classes must be between 1 and 9 because classes partition the set. Having 2 classes is possible (e.g., sizes 4 and 5). 0 or 10 is impossible.
If equivalence classes have sizes 3,3,3 on a 9-element set, then number of classes is
A 2
B 3
C 6
D 9
The set is partitioned into classes. If each class has 3 elements and total is 9, then there must be exactly 9/3=3 equivalence classes.
If R and S are relations on A, then R∘S means
A Apply R then S
B Union of relations
C Intersection only
D Apply S then R
(R∘S)(a,c) holds when there exists b with (a,b) in S and (b,c) in R. So S is applied first, then R.
If R is a relation on A, then R∘I equals
A R
B I
C R⁻¹
D ∅
Identity relation I contains (a,a) for all a. Composing any relation with I preserves pairs because the “middle” element stays the same, so R∘I = R and I∘R = R.
If f: A→B is one-one with |A|=8, then minimum |B| is
A 6
B 7
C 8
D 9
Injective requires distinct outputs for distinct inputs, so codomain must have at least as many elements as domain. Therefore |B| must be ≥8; minimum is 8.
If f: A→B is onto with |A|=6, then maximum |B| is
A 5
B 6
C 7
D 12
Onto means every element of B is hit by some element of A. So B cannot have more elements than A. Hence maximum |B| is 6.
Number of onto functions from a 3-element set to a 2-element set is
A 2
B 4
C 8
D 6
Total functions are 2^3=8. Not onto means all map to one element: 2 such constant functions. So onto functions = 8−2=6.
Number of one-one functions from a 4-element set to a 6-element set is
A 6^4
B 6P4
C 6C4
D 4!
Injective functions choose distinct images for 4 inputs from 6 outputs with order mattering. That count is 6P4 = 6·5·4·3 = 360.
If f(x)=x² and g(x)=√x with x≥0, then (g∘f)(x) equals
A x
B x²
C |x|
D √x
f(x)=x² outputs nonnegative. Then g(f(x))=√(x²)=|x|. Square root returns the nonnegative root, giving absolute value.
For f(x)=|x|, to get inverse f⁻¹(x)=x (principal), restrict domain to
A x≤0
B x≠0
C All real x
D x≥0
|x| is many-one on all reals because x and −x give same output. Restricting to x≥0 makes it one-one and onto [0,∞), so inverse exists.
If f(x)=x³−2, then f⁻¹(6) equals
A 0
B 2
C ∛8
D 4
Solve x³−2=6 ⇒ x³=8 ⇒ x=2. Since cubic is one-one on reals, the inverse exists and gives unique real output.
If f(x)=(2x−1)/(x+3), then x=−3 is excluded because
A Denominator zero
B Not one-one
C Range mismatch
D Not onto
Rational functions are undefined when denominator becomes zero. Here x+3=0 at x=−3, causing division by zero, so −3 must be removed from domain.
For f(x)=1/(x−2), the range excludes
A 1
B 2
C 0
D −2
1/(x−2) can never be 0 because numerator is 1. For any nonzero y, x=2+1/y gives that y. So only 0 is missing from range.
For f(x)= (x−1)² with domain R, f is
A One-one
B Onto R
C Constant
D Many-one
Squaring makes outputs repeat: f(0)=1 and f(2)=1. Hence it is not one-one on all reals. Also range is y≥0, so it is not onto R.
If f(x)=x² and domain is [0,∞), then f⁻¹(y) equals
A −√y
B √y
C ±√y
D 1/√y
With domain restricted to nonnegative x, x²=y implies x=√y only. The negative root is not allowed, so inverse is the principal square root.
If f and g are bijections, then (f∘g)⁻¹ equals
A f⁻¹∘g⁻¹
B (f∘g)
C g⁻¹∘f⁻¹
D g∘f
Composition applies g first then f. To undo, reverse order: undo f by f⁻¹, then undo g by g⁻¹. Thus (f∘g)⁻¹ = g⁻¹∘f⁻¹.
If f is one-one, then f(A∩B) equals
A f(A)∪f(B)
B f(A)−f(B)
C Always empty
D f(A)∩f(B)
For injective functions, images preserve intersection: any y in f(A)∩f(B) must come from some a in A and b in B with same image, forcing a=b, so it comes from A∩B.
For any function f, which statement is always true
A f(A∪B)=f(A)∩f(B)
B f(A∩B)=f(A)∩f(B)
C f(A∪B)=f(A)∪f(B)
D f(A−B)=f(A)−f(B)
Image of a union equals union of images for any function: outputs from A∪B come from A or B. But images of intersection may be larger than intersection of images unless f is one-one.
If f: A→B is onto, then for any S⊆B, f(f⁻¹(S)) equals
A A
B S
C B
D ∅
f⁻¹(S) collects all inputs mapping into S. Applying f returns exactly those outputs in S. Onto ensures no element of S is “missed” due to missing preimages.
For any function f and subset T⊆A, f⁻¹(f(T)) is always
A Superset of T
B Equal to T
C Disjoint from T
D Empty set
Every element of T maps into f(T), so it surely lies in the preimage of f(T). But extra elements outside T can share the same image, making preimage larger unless f is one-one.
If f is bijection, then f⁻¹(f(T)) equals
A A
B B
C ∅
D T
In a bijection, different inputs have different outputs, so no outside element can share the same image as an element of T. Therefore preimage of f(T) is exactly T.
If f is periodic with period 5, then f(x+15)=f(x) because
A 15 is prime
B 15 is smallest
C 15 is multiple
D 15 is not period
If 5 is a period, then any integer multiple of 5 is also a period. Since 15=3×5, shifting by 15 repeats the same values.
If f is both even and odd, then f(x) must be
A Constant 1
B Constant 0
C Always positive
D Always increasing
Even gives f(−x)=f(x) and odd gives f(−x)=−f(x). Together they imply f(x)=−f(x), so f(x)=0 for all x where defined.
A function f: R→R is defined by f(x)=⌊x⌋. The function is
A One-one
B Bijective
C Identity
D Not one-one
Many real inputs share same floor value. For example, all x in [2,3) map to 2. So it cannot be one-one, though it is onto integers if codomain is Z.
A relation R on A is an equivalence relation if and only if
A Reflexive and antisymmetric
B Symmetric and antisymmetric
C Reflexive symmetric transitive
D Only transitive
These three properties ensure “equivalent” behaves like equality in grouping: every element matches itself, swapping order holds, and chaining equivalence remains within same class.
If R is an equivalence relation, then quotient set A/R is
A Set of ordered pairs
B Set of classes
C Set of functions
D Set of integers
A/R contains the equivalence classes formed under R. Each class is a subset of A, and together they cover A without overlap, giving a partition.
If A has 6 elements, maximum number of equivalence classes possible is
A 6
B 1
C 2
D 3
Maximum number of classes occurs when each class is a singleton. That gives 6 classes for 6 elements. Minimum would be 1 class when all elements are equivalent.
If f: A→B is bijective, then |A| and |B| are
A Always different
B Unrelated sizes
C One must be zero
D Always equal
Bijection pairs each element of A with exactly one element of B and uses all of B. Hence there must be same number of elements in domain and codomain.
If f is one-one and A is finite, then |f(A)| equals
A |A|−1
B |A|
C |A|+1
D Depends only on B
Injective maps distinct inputs to distinct outputs, so no collisions occur. Therefore the image set f(A) has exactly as many elements as A when A is finite.
If f is onto, then for any y in codomain
A Exactly one preimage
B No preimage
C At least one preimage
D Only two preimages
Surjective means every codomain element is hit. It allows one or more preimages. Exactly one preimage would require one-one as well, i.e., bijection.
For function f(x)=x³, the inverse mapping is
A x²
B |x|
C ⌊x⌋
D ∛x
If y=x³, then x=∛y. Cubic is strictly increasing on reals, so it is one-one and onto. Hence inverse function exists and is cube root.
The graph of a one-one function fails the horizontal line test when
A No line intersects
B Some line intersects twice
C Only vertical lines
D Line y=x only
One-one means each output corresponds to at most one input. If a horizontal line cuts the graph at two points, the same y-value has two x-values, so function is not one-one.
If f(x)=2x+1 and g(x)=x², then (f∘g)(x) equals
A 2x²+1
B (2x+1)²
C 2x+2
D x²+1
Apply g first: g(x)=x². Then apply f: f(g(x))=2(x²)+1=2x²+1. Composition depends on correct order.
If g(x)=2x+1 and f(x)=x², then (f∘g)(x) equals
A 2x²+1
B x²+1
C (2x+1)²
D 4x²+1
Here g is applied first: g(x)=2x+1. Then f squares it: f(g(x))=(2x+1)². Swapping order changes result.
If f∘g = g∘f for all x, then f and g are
A Always inverses
B Always bijections
C Always constants
D Commuting functions
When f∘g equals g∘f, the functions commute under composition. This is a special property and not true for most function pairs.
If f(x)=x+1 on integers, then f is
A Not onto
B Onto integers
C Many-one
D Constant
For any integer y, choose x=y−1, then f(x)=y. So every integer is hit, making it onto Z. It is also one-one, so it is bijective on integers.
If f: R→R is defined by f(x)=e^x, then f is
A Onto R
B Odd function
C Into R
D Periodic
e^x is always positive, so its range is (0,∞), which is a proper subset of R. Hence it is into, not onto. It is also one-one and increasing.
If f(x)=sin x as R→R, then f is
A One-one onto
B One-one into
C Bijective
D Many-one into
sin x repeats values periodically, so many different x give same output, making it many-one. Its range is [−1,1], not all real numbers, so it is into R.
If A has 3 elements, number of equivalence relations on A is
A 10
B 15
C 5
D 27
Equivalence relations correspond to partitions of the set. For 3 elements, number of partitions (Bell number B3) is 5: one class, three singletons, and three “pair+single” cases.