Symmetric difference A Δ B contains elements present in exactly one of the sets. Common elements 2 and 3 are removed, leaving 1 from A and 4 from B.
If A⊆U, then A ∪ A′ equals
A U
B A′
C ∅
D A
A′ contains everything in U not in A. Union combines A and not-A, covering the whole universal set. So A ∪ A′ always equals U.
If A⊆U, then A ∩ A′ equals
A U
B A
C ∅
D A′
A and its complement have no common elements by definition. Therefore their intersection is always empty, meaning A ∩ A′ = ∅.
If A={1,2,3}, then P(A) has
A 3 subsets
B 8 subsets
C 6 subsets
D 9 subsets
A set with n elements has 2^n subsets. Here n=3, so total subsets are 2^3=8, including ∅ and the full set A.
Number of proper subsets of a 4-element set is
A 16
B 14
C 15
D 8
Total subsets of a 4-element set are 2^4=16. Proper subsets exclude the set itself, so proper subsets = 16−1 = 15.
If A has 3 elements, how many 2-element subsets exist?
A 1
B 3
C 6
D 2
Number of 2-element subsets from 3 elements is combination C(3,2)=3. They are formed by choosing any two elements from the three.
If A×B = ∅, then
A Both nonempty
B A empty only
C B empty only
D A or B empty
Cartesian product has pairs only when both sets have elements. If A×B is empty, at least one of A or B must be empty.
If A and B are finite, then |A×B| equals
A |A|/|B|
B |A|−|B|
C |A||B|
D |A|+|B|
Each element of A pairs with every element of B. So total ordered pairs are |A| choices for first element times |B| choices for second element.
If A={1,2} and B={a,b,c}, then (2,b) belongs to
A B×A
B A×B
C A∩B
D A∪B
In A×B, first component must be from A and second from B. Since 2∈A and b∈B, (2,b) is in A×B.
A relation R on A is antisymmetric if
A (a,b)&(b,a)⇒a=b
B (a,a) for all a
C (a,b),(b,c)⇒(a,c)
D (a,b)⇒(b,a)
Antisymmetric means if a relates to b and b relates to a, then a must equal b. It allows (a,a) and prevents two-way links between distinct elements.
Which relation is antisymmetric on integers?
A a is friend
B a ≠ b
C a ≤ b
D a is parallel
For ≤, if a ≤ b and b ≤ a, then a = b. That matches antisymmetric definition. Friendship is usually symmetric, not antisymmetric.
If R is symmetric, then R ∪ R⁻¹ equals
A R⁻¹
B R
C A×A
D ∅
For a symmetric relation, every (a,b) already implies (b,a). So R⁻¹ equals R. Therefore R ∪ R⁻¹ remains R.
If R is any relation, then (R⁻¹)⁻¹ equals
A R×R
B A×A
C ∅
D R
Taking inverse swaps components. Doing it twice swaps back to original order. Hence inverse of inverse of a relation always returns the same relation R.
A relation matrix (intro) represents relation using
A Venn shading
B Ordered list
C 0 and 1 entries
D Only real numbers
Relation matrix uses 1 to show an ordered pair is present and 0 to show it is absent. Rows correspond to domain elements and columns to codomain elements.
If R is reflexive on A, then number of diagonal pairs is
A |A|
B 2|A|
C 0
D |A|²
Reflexive requires (a,a) for every a in A. These are diagonal entries in matrix form. Hence there must be exactly |A| diagonal ordered pairs.
If R is equivalence, then every element belongs to
A No class
B Exactly one class
C Infinite classes
D Two classes always
Equivalence classes partition the set, meaning they cover all elements with no overlap. So each element of the set lies in one and only one equivalence class.
In quotient set A/R, elements are
A Functions only
B Ordered pairs
C Equivalence classes
D Single numbers
The quotient set A/R is the collection of all equivalence classes formed by relation R on A. It groups elements that are equivalent under R.
If a~b means a≡b (mod 3), then class of 1 is
A {…,−2,1,4,…}
B {1,2,3}
C {…,0,3,6,…}
D {…,−1,2,5,…}
Mod 3 equivalence groups numbers with same remainder on division by 3. Numbers congruent to 1 mod 3 are …, −2, 1, 4, 7, etc.
A function f: A→B is injective when
A Output constant always
B Distinct inputs distinct outputs
C Two inputs one output
D Range equals codomain
Injective (one-one) means if f(a)=f(b), then a=b. So different inputs cannot share the same output. This is crucial for inverse to exist.
A function f: A→B is surjective when
A f is constant
B Domain equals B
C Range equals B
D f is many-one
Surjective (onto) means every element of B is an output of at least one element of A. So range covers the entire codomain B.
If f is bijective, then f⁻¹ is
A Relation only
B Empty mapping
C Not a function
D A function
Bijective means one-one and onto, so each y in codomain has exactly one preimage x. Therefore inverse mapping assigns a unique x to each y, making f⁻¹ a function.
Function equality f=g requires
A Same rule only
B Same domain and outputs
C Same codomain only
D Same range only
Two functions are equal if they have the same domain and give the same output for every input. Having same formula alone is not enough if domains differ.
If f(x)=x² and g(x)=x² for x≥0, then f and g are
A Inverse functions
B Constant functions
C Not equal functions
D Always equal
Even though formula looks same, domains differ: f is on all real numbers, while g is defined only for x≥0. Different domains means different functions.
If f(x)=2x+1, then f(0) equals
A 1
B 0
C 3
D 2
Substitute x=0 into 2x+1 gives 2·0+1=1. Function evaluation means replace input by given value and simplify correctly.
For f(x)=x−3, the graph is
A Shift up 3
B Shift down 3
C Shift right 3
D Shift left 3
Subtracting 3 outside x lowers every output by 3. So y=x−3 is the line y=x shifted downward by 3 units.
For f(x)= (x+3), the graph is
A Shift right 3
B Shift down 3
C Shift up 3
D Shift left 3
Adding 3 inside the input shifts the graph left. f(x)=x+3 is line y=x moved up, but for transformation form g(x)=f(x+3) shifts left by 3.
If h(x)=f(x+2), then h is f shifted
A Down 2
B Right 2
C Up 2
D Left 2
Replacing x by x+2 moves graph left by 2 units. Points that were at x become at x−2, so the entire shape shifts left.
If h(x)=f(x)−2, then graph shifts
A Up 2
B Down 2
C Left 2
D Right 2
Subtracting 2 from function output decreases y-values by 2. This shifts the graph downward by 2 units without changing its shape.
If f(x)=|x−1|, vertex occurs at
A x=2
B x=−1
C x=1
D x=0
For |x−a|, the minimum value 0 occurs at x=a. Here a=1, so the V-shaped graph has its vertex at x=1.
If f(x)=|x|, then range is
A y ≥ 0
B y ≤ 0
C y ≠ 0
D All reals
Absolute value is never negative. It outputs distance from zero, so smallest value is 0 and all outputs are non-negative real numbers.
If f(x)=x²−4, the minimum value is
A 0
B No minimum
C −4
D 4
x² is always ≥0, so x²−4 is always ≥−4. Minimum occurs when x=0, giving f(0)=−4. This is basic parabola shifting.
If f(x)=x²−4, then range is
A y ≤ −4
B y ≥ −4
C All reals
D y ≥ 0
Since x²≥0, x²−4≥−4. The function takes all values from −4 upward. So range is [−4,∞).
A function is increasing on an interval if
A Domain is empty
B y constant always
C Larger x gives smaller y
D Larger x gives larger y
Increasing means as x increases, function values do not decrease. For x1
A function is decreasing on an interval if
A Larger x gives smaller y
B Outputs all positive
C Always one-one
D Larger x gives larger y
Decreasing means as x increases, the output goes down. For x1 f(x2) in strictly decreasing behavior.
For periodic function, f(x+T)=f(x) where T is
A Range
B Codomain
C Period
D Domain
A periodic function repeats its values after a fixed positive interval T. That smallest such positive T is called the period, e.g., sin(x) has period 2π.
If f(x) is odd, then condition is
A f(x)≥0
B f(−x)=−f(x)
C f(x+T)=f(x)
D f(−x)=f(x)
Odd function has origin symmetry. Replacing x by −x changes sign of output. Common examples are f(x)=x and f(x)=x³.
If f and g are invertible, then (f∘g) is
A Constant always
B Many-one always
C Always invertible
D Never invertible
Composition of two bijections is also a bijection. Therefore it is invertible, and its inverse is given by (f∘g)⁻¹ = g⁻¹∘f⁻¹.
If f is invertible, then f must be
A Bijective
B Many-one
C Into only
D One-one only
For inverse to be a function on codomain, each y must come from exactly one x. That requires f to be one-one and onto, i.e., bijective.
If f(x)=2x+3 and g(x)=x−1, then (f∘g)(x) equals
A 2x+3
B 2x+7
C 2x+5
D 2x+1
Apply g first: g(x)=x−1. Then f(g(x))=2(x−1)+3=2x−2+3=2x+1. Composition must follow correct order.
If f(x)=x+2 and g(x)=3x, then (g∘f)(x) equals
A x+6
B 3x+6
C 3x
D 3x+2
g∘f means g(f(x)). First f(x)=x+2. Then g(x+2)=3(x+2)=3x+6. Order matters in composition.
If f∘g = I, then g is
A Into function
B Constant function
C Inverse of f
D Equal to f
If f(g(x))=x for all x in the domain, then g undoes f. So g is the inverse of f on that set, and f is inverse of g.
If f is one-one but not onto, then inverse
A Never exists
B Exists on range
C Always constant
D Exists on codomain
One-one ensures uniqueness, so inverse mapping can be defined for outputs actually produced. But if not onto, some codomain values have no preimage, so inverse works only on range.
If f is onto but not one-one, then inverse
A Not a function
B Exists as function
C Always one-one
D Equals identity
If not one-one, two different inputs can give same output. Then inverse would need to assign one output to multiple inputs, which violates definition of a function.
Relation vs function difference: function must have
A Only diagonal pairs
B Multiple outputs allowed
C Exactly one output
D No ordered pairs
A relation can pair an input with many outputs. A function is a special relation where each input has exactly one image. This is the key distinguishing rule.
If R is an equivalence relation, then it is
A Antisymmetric transitive
B Reflexive symmetric transitive
C Symmetric only
D Reflexive only
Equivalence relation must satisfy all three properties so that “being equivalent” behaves consistently. This creates clean equivalence classes that partition the set.
If A={1,2} and B={1,2}, then A×B equals
A Two ordered pairs
B {(1,1),(2,2)}
C {(1,2),(2,1)}
D Four ordered pairs
A×B includes all pairs (a,b) with a in A and b in B. With two elements each, total pairs are 2×2=4.
If A∩B=∅, then A and B are
A Complement sets
B Universal sets
C Disjoint sets
D Equal sets
Intersection being empty means there are no common elements. Such sets are called disjoint. They can still be non-empty individually.
If A⊆B and B⊆A, then
A A and B disjoint
B A=B
C A is empty
D A is proper subset
If each set contains the other, then they have exactly the same elements. This is the standard condition for equality of sets using subset relations.
If f(x)=x² and domain is integers, then f is
A Many-one function
B Onto integers
C Constant function
D One-one function
f(2)=4 and f(−2)=4, so different integers map to the same output. Hence it is many-one and not one-one over integers.
If f(x)=x³, then f is
A Constant function
B Even function
C Odd function
D Periodic function
For x³, f(−x)=(−x)³=−x³=−f(x). This satisfies the odd function condition, giving symmetry about the origin.