Class 12 Mathematics · 65 questions · 0 with answers
Let A = {0, 1, 2, 3} and define a relation R on A as follows: R = {(0, 0), (0, 1), (0, 3), (1, 0), (1, 1), (2, 2), (3, 0), (3, 3)}. Is R reflexive? symmetric? transitive? Solution R is reflexive and symmetric, but not transitive since for (1, 0) ∈ R and (0, 3) ∈ R whereas (1, 3) ∉ R.
For the set A = {1, 2, 3}, define a relation R in the set A as follows: R = {(1, 1), (2, 2), (3, 3), (1, 3)}. Write the ordered pairs to be added to R to make it the smallest equivalence relation. Solution (3, 1) is the single ordered pair which needs to be added to R to make it the smallest equivalence relation.
Let R be the equivalence relation in the set Z of integers given by R = {(a, b) : 2 divides a – b}. Write the equivalence class [0]. Solution [0] = {0, ± 2, ± 4, ± 6,...}
Let the function f : R → R be defined by f (x) = 4x – 1, ∀ x ∈ R. Then, show that f is one-one. Solution For any two elements x1, x2 ∈ R such that f (x1) = f (x2), we have 4x1 – 1 = 4x2 – 1 ⇒ 4x1 = 4x2, i.e., x1 = x2 Hence f is one-one.
If f = {(5, 2), (6, 3)}, g = {(2, 5), (3, 6)}, write f o g. Solution f o g = {(2, 2), (3, 3)}
Let f : R → R be the function defined by f (x) = 4x – 3 ∀ x ∈ R. Then write f –1.
Is the binary operation * defined on Z (set of integer) by m * n = m – n + mn ∀ m, n ∈ Z commutative? Solution No. Since for 1, 2 ∈ Z, 1 * 2 = 1 – 2 + 1.2 = 1 while 2 * 1 = 2 – 1 + 2.1 = 3 so that 1 * 2 ≠ 2 * 1.
If f = {(5, 2), (6, 3)} and g = {(2, 5), (3, 6)}, write the range of f and g. Solution The range of f = {2, 3} and the range of g = {5, 6}.
If A = {1, 2, 3} and f, g are relations corresponding to the subset of A × A indicated against them, which of f, g is a function? Why? f = {(1, 3), (2, 3), (3, 2)} g = {(1, 2), (1, 3), (3, 1)} Solution f is a function since each element of A in the first place in the ordered pairs is related to only one element of A in the second place while g is not a function because 1 is related to more than one element of A, namely, 2 and 3.
If A = {a, b, c, d} and f = {a, b), (b, d), (c, a), (d, c)}, show that f is one- one from A onto A. Find f –1. Solution f is one-one since each element of A is assigned to distinct element of the set A. Also, f is onto since f (A) = A. Moreover, f –1 = {(b, a), (d, b), (a, c), (c, d)}.
In the set N of natural numbers, define the binary operation * by m * n = g.c.d (m, n), m, n ∈ N. Is the operation * commutative and associative? Solution The operation is clearly commutative since m * n = g.c.d (m, n) = g.c.d (n, m) = n * m ∀ m, n ∈ N. It is also associative because for l, m, n ∈ N, we have l * (m * n) = g. c. d (l, g.c.d (m, n)) = g.c.d. (g. c. d (l, m), n) = (l * m) * n. RELATIONS AND FUNCTIONS 5
In the set of natural numbers N, define a relation R as follows: ∀ n, m ∈ N, nRm if on division by 5 each of the integers n and m leaves the remainder less than 5, i.e. one of the numbers 0, 1, 2, 3 and 4. Show that R is equivalence relation. Also, obtain the pairwise disjoint subsets determined by R. Solution R is reflexive since for each a ∈ N, aRa. R is symmetric since if aRb, then bRa for a, b ∈ N. Also, R is transitive since for a, b, c ∈ N, if aRb and bRc, then aRc. Hence R is an equivalence relation in N which will partition the set N into the pairwise disjoint subsets. The equivalent classes are as mentioned below: A0 = {5, 10, 15, 20 ...} A1 = {1, 6, 11, 16, 21 ...} A2 = {2, 7, 12, 17, 22, ...} A3 = {3, 8, 13, 18, 23, ...} A4 = {4, 9, 14, 19, 24, ...} It is evident that the above five sets are pairwise disjoint and A0 ∪ A1 ∪ A2 ∪ A3 ∪ A4 = ∪ Ai = N . i =0
Show that the function f : R → R defined by f (x) = , ∀ x∈R , is x +1 neither one-one nor onto. Solution For x1, x2 ∈ R, consider f (x1) = f (x2) x1 x ⇒ = 22 x1 +1 x2 +1 ⇒ x1 x22 + x1 = x2 x12 + x2 ⇒ x1 x2 (x2 – x1) = x2 – x1 ⇒ x1 = x2 or x1 x2 = 1 We note that there are point, x1 and x2 with x1 ≠ x2 and f (x1) = f (x2), for instance, if 1 2 2 1 we take x1 = 2 and x2 = , then we have f (x1) = and f (x2) = but 2 ≠ . Hence 2 5 5 2 f is not one-one. Also, f is not onto for if so then for 1∈R ∃ x ∈ R such that f (x) = 1
Let f, g : R → R be two functions defined as f (x) = x + x and g (x) = x – x ∀ x ∈ R. Then, find f o g and g o f. Solution Here f (x) = x + x which can be redefined as 2 x if x ≥ 0 f (x) = 0 if x < 0 Similarly, the function g defined by g (x) = x – x may be redefined as 0 if x ≥ 0 g (x) = –2 x if x < 0 Therefore, g o f gets defined as : For x ≥ 0, (g o f ) (x) = g (f (x) = g (2x) = 0 and for x < 0, (g o f ) (x) = g (f (x) = g (0) = 0. Consequently, we have (g o f ) (x) = 0, ∀ x ∈ R. Similarly, f o g gets defined as: For x ≥ 0, (f o g ) (x) = f (g (x) = f (0) = 0, and for x < 0, (f o g ) (x) = f (g(x)) = f (–2 x) = – 4x. 0, x > 0 i.e. ( f o g ) ( x) = −4 x, x < 0
Let R be the set of real numbers and f : R → R be the function defined by f (x) = 4x + 5. Show that f is invertible and find f –1. Solution Here the function f : R → R is defined as f (x) = 4x + 5 = y (say). Then y −5 4x = y – 5 or x= . RELATIONS AND FUNCTIONS 7 This leads to a function g : R → R defined as y −5 g (y) = . Therefore, ( g o f ) (x) = g(f (x) = g (4x + 5) 4x + 5 − 5 = = x or g o f = IR Similarly (f o g) (y) = f (g(y)) y −5 = f 4 y −5 = 4 +5 = y 4 or f o g = IR . Hence f is invertible and f –1 = g which is given by x −5 f –1 (x) =
Let * be a binary operation defined on Q. Find which of the following binary operations are associative
Let R be a relation on the set N of natural numbers defined by nRm if n divides m. Then R is
Let L denote the set of all straight lines in a plane. Let a relation R be defined by lRm if and only if l is perpendicular to m ∀ l, m ∈ L. Then R is
Let N be the set of natural numbers and the function f : N → N be defined by f (n) = 2n + 3 ∀ n ∈ N. Then f is
Set A has 3 elements and the set B has 4 elements. Then the number of RELATIONS AND FUNCTIONS 9 injective mappings that can be defined from A to B is
Let f : R → R be defined by f (x) = sin x and g : R → R be defined by g (x) = x2, then f o g is
Let f : R → R be defined by f (x) = 3x – 4. Then f –1 (x) is given by x+4 x
Let f : R → R be defined by f (x) = x2 + 1. Then, pre-images of 17 and – 3, respectively, are
For real numbers x and y, define xRy if and only if x – y + 2 is an irrational number. Then the relation R is
Consider the set A = {1, 2, 3} and R be the smallest equivalence relation on A, then R = ________
The domain of the function f : R → R defined by f (x) = x 2 – 3x + 2 is ________. Solution Here x2 – 3x + 2 ≥ 0 ⇒ (x – 1) (x – 2) ≥ 0 ⇒ x ≤ 1 or x ≥ 2 Hence the domain of f = (– ∞, 1] ∪ [2, ∞)
Consider the set A containing n elements. Then, the total number of injective functions from A onto itself is ________. Solution n!
Let Z be the set of integers and R be the relation defined in Z such that aRb if a – b is divisible by 3. Then R partitions the set Z into ________ pairwise disjoint subsets. Solution Three.
Let R be the set of real numbers and * be the binary operation defined on R as a * b = a + b – ab ∀ a, b ∈ R. Then, the identity element with respect to the binary operation * is _______. Solution 0 is the identity element with respect to the binary operation *. State True or False for the statements in each of the Examples 30 to 34.
Consider the set A = {1, 2, 3} and the relation R = {(1, 2), (1, 3)}. R is a transitive relation. Solution True.
Let A be a finite set. Then, each injective function from A into itself is not surjective. Solution False.
For sets A, B and C, let f : A → B, g : B → C be functions such that g o f is injective. Then both f and g are injective functions. Solution False.
For sets A, B and C, let f : A → B, g : B → C be functions such that g o f is surjective. Then g is surjective Solution True. RELATIONS AND FUNCTIONS 11
Let N be the set of natural numbers. Then, the binary operation * in N defined as a * b = a + b, ∀ a, b ∈ N has identity element. Solution False.
MATHEMATICS Solution Given that f (x) = 4x – 3 = y (say), then 4x = y + 3 y +3 ⇒ x= y +3 x+3 Hence f –1 (y) = ⇒ f –1 (x) = 4 4
MATHEMATICS which gives =1 . But there is no such x in the domain R, since the equation x +1 x2 – x + 1 = 0 does not give any real value of x.
MATHEMATICS Thus (a * b) * c ≠ a * (b * c) and hence * is not associative. (ii) * is associative since Q is associative with respect to multiplication. (iii) * is not associative for if we take a = 2, b = 3 and c = 4, then (a * b) * c = (2 * 3) * 4 = (2 – 3 + 6) * 4 = 5 * 4 = 5 – 4 + 20 = 21, and a * (b * c) = 2 * (3 * 4) = 2 * (3 – 4 + 12) = 2 * 11 = 2 – 11 + 22 = 13 Thus (a * b) * c ≠ a * (b * c) and hence * is not associative. (iv) * is not associative for if we take a = 1, b = 2 and c = 3, then (a * b) * c = (1 * 2) * 3 = 4 * 3 = 4 × 9 = 36 and a * (b * c) = 1 * (2 * 3) = 1 * 18 = 1 × 182 = 324. Thus (a * b) * c ≠ a * (b * c) and hence * is not associative.
MATHEMATICS Solution R = {(1, 1), (2, 2), (3, 3)}.
Let A = {a, b, c} and the relation R be defined on A as follows: R = {(a, a), (b, c), (a, b)}. Then, write minimum number of ordered pairs to be added in R to make R reflexive and transitive.
Let D be the domain of the real valued function f defined by f (x) = 25 − x 2 . Then, write D.
Let f , g : R → R be defined by f (x) = 2x + 1 and g (x) = x2 – 2, ∀ x ∈ R, respectively. Then, find g o f.
Let f : R → R be the function defined by f (x) = 2x – 3 ∀ x ∈ R. write f –1.
If A = {a, b, c, d} and the function f = {(a, b), (b, d), (c, a), (d, c)}, write f –1.
If f : R → R is defined by f (x) = x2 – 3x + 2, write f (f (x)).
Is g = {(1, 1), (2, 3), (3, 5), (4, 7)} a function? If g is described by g (x) = αx + β, then what value should be assigned to α and β.
Are the following set of ordered pairs functions? If so, examine whether the mapping is injective or surjective.
If the mappings f and g are given by f = {(1, 2), (3, 5), (4, 1)} and g = {(2, 3), (5, 1), (1, 3)}, write f o g.
Let C be the set of complex numbers. Prove that the mapping f : C → R given by f (z) = |z|, ∀ z ∈ C, is neither one-one nor onto.
Let the function f : R → R be defined by f (x) = cosx, ∀ x ∈ R. Show that f is neither one-one nor onto.
Let X = {1, 2, 3}and Y = {4, 5}. Find whether the following subsets of X ×Y are functions from X to Y or not.
If functions f : A → B and g : B → A satisfy g o f = IA, then show that f is one- one and g is onto. 12 MATHEMATICS
Let f : R → R be the function defined by f (x) = 2 – cos x ∀ x ∈ R.Then, find the range of f.
Let n be a fixed positive integer. Define a relation R in Z as follows: ∀ a, b ∈ Z, aRb if and only if a – b is divisible by n . Show that R is an equivalance relation.
If A = {1, 2, 3, 4 }, define relations on A which have properties of being:
Let R be relation defined on the set of natural number N as follows: R = {(x, y): x ∈ N, y ∈ N, 2x + y = 41}. Find the domain and range of the relation R. Also verify whether R is reflexive, symmetric and transitive.
Given A = {2, 3, 4}, B = {2, 5, 6, 7}. Construct an example of each of the following:
Give an example of a map
Let A = R – {3}, B = R – {1}. Let f : A → B be defined by f (x) = x – 3 ∀ x ∈ A . Then show that f is bijective.
Let A = [–1, 1]. Then, discuss whether the following functions defined on A are one-one, onto or bijective:
Each of the following defines a relation on N:
Let A = {1, 2, 3, ... 9} and R be the relation in A ×A defined by (a, b) R (c, d) if a + d = b + c for (a, b), (c, d) in A ×A. Prove that R is an equivalence relation and also obtain the equivalent class [(2, 5)].
Using the definition, prove that the function f : A → B is invertible if and only if f is both one-one and onto.
Functions f , g : R → R are defined, respectively, by f (x) = x2 + 3x + 1, g (x) = 2x – 3, find
Let * be the binary operation defined on Q. Find which of the following binary operations are commutative
Let * be binary operation defined on R by a * b = 1 + ab, ∀ a, b ∈ R. Then the operation * is
to 47 (M.C.Q.). 28. Let T be the set of all triangles in the Euclidean plane, and let a relation R on T be defined as aRb if a is congruent to b ∀ a, b ∈ T. Then R is
Consider the non-empty set consisting of children in a family and a relation R defined as aRb if a is brother of b. Then R is
The maximum number of equivalence relations on the set A = {1, 2, 3} are
If a relation R on the set {1, 2, 3} be defined by R = {(1, 2)}, then R is
Let us define a relation R in R as aRb if a ≥ b. Then R is
Let A = {1, 2, 3} and consider the relation R = {1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1,3)}. Then R is
The identity element for the binary operation * defined on Q ~ {0} as a, b ∈ Q ~ {0} is 2 ∀ a*b=
If the set A contains 5 elements and the set B contains 6 elements, then the number of one-one and onto mappings from A to B is
Let A = {1, 2, 3, ...n} and B = {a, b}. Then the number of surjections from A into B is
Let f : R → R be defined by f (x) = x ∈ R. Then f is x ∀
Let f : R → R be defined by f (x) = 3x2 – 5 and g : R → R by g (x) = . x +1 Then g o f is 3 x 2 −5 3 x 2 −5
Which of the following functions from Z into Z are bijections?
Let f : R → R be the functions defined by f (x) = x3 + 5. Then f –1 (x) is 1 1
Let f : A → B and g : B → C be the bijective functions. Then (g o f)–1 is
Let f : R − → R be defined by f (x) = 5 x − 3 . Then 5 –1
Let f : [0, 1] → [0, 1] be defined by f (x) = 1 − x, if x isirrational 16 MATHEMATICS Then (f o f) x is
Let f : [2, ∞) → R be the function defined by f (x) = x2 – 4x + 5, then the range
Let f : N → R be the function defined by f (x) = and g : Q → R be another function defined by g (x) = x + 2. Then (g o f) is
Let f : R → R be defined by 2x: x > 3 f ( x) = x 2 :1< x ≤ 3 3 x : x ≤1 Then f (– 1) + f (2) + f (4) is
Let f : R → R be given by f (x) = tan x. Then f –1 (1) is
Let the relation R be defined in N by aRb if 2a + 3b = 30. Then R = ______.
Let the relation R be defined on the set A = {1, 2, 3, 4, 5} by R = {(a, b) : |a2 – b2| < 8. Then R is given by _______.
Let f = {(1, 2), (3, 5), (4, 1) and g = {(2, 3), (5, 1), (1, 3)}. Then g o f = ______ and f o g = ______. RELATIONS AND FUNCTIONS 17
Let f : R → R be defined by f ( x ) = . Then ( f o f o f ) (x) = _______ 1 + x2
If f (x) = (4 – (x–7)3}, then f –1(x) = _______. State True or False for the statements in each of the Exercises 53 to 63.
Let R = {(3, 1), (1, 3), (3, 3)} be a relation defined on the set A = {1, 2, 3}. Then R is symmetric, transitive but not reflexive.
Let f : R → R be the function defined by f (x) = sin (3x+2) ∀ x ∈ R. Then f is invertible.
Every relation which is symmetric and transitive is also reflexive.
An integer m is said to be related to another integer n if m is a integral multiple of n. This relation in Z is reflexive, symmetric and transitive.
Let A = {0, 1} and N be the set of natural numbers. Then the mapping f : N → A defined by f (2n–1) = 0, f (2n) = 1, ∀ n ∈ N, is onto. 58.The relation R on the set A = {1, 2, 3} defined as R = {{1, 1), (1, 2), (2, 1), (3, 3)} is reflexive, symmetric and transitive.
The composition of functions is commutative.
The composition of functions is associative.
Every function is invertible.
A binary operation on a set has always the identity element.