NCERT Solutions (Miscellaneous Exercise) - Relations and Functions JEE Notes | EduRev

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RELATIONS AND FUNCTIONS
Miscellaneous Exercise
Q.1. Let f: R → R be defined as f(x) = 10x + 7. Find the function g: R → R such that gof = fog = IR.
Ans. It is given that f: RR is defined as f(x) = 10x + 7.
One-one:
Let f(x) = f(y), where x, y ∈ R.
⇒ 10x + 7 = 10y + 7
⇒ x = y
∴ f is a one-one function.
Onto:
For y ∈ R, let y = 10x + 7.

NCERT Solutions (Miscellaneous Exercise) - Relations and Functions JEE Notes | EduRev
Therefore, for any y ∈ R, there exists
NCERT Solutions (Miscellaneous Exercise) - Relations and Functions JEE Notes | EduRev
∴ f is onto.
Therefore, f is one-one and onto.
Thus, f is an invertible function.
Let us define g: RR as f (y) = y-7/10.
Now, we have

NCERT Solutions (Miscellaneous Exercise) - Relations and Functions JEE Notes | EduRev

NCERT Solutions (Miscellaneous Exercise) - Relations and Functions JEE Notes | EduRev
NCERT Solutions (Miscellaneous Exercise) - Relations and Functions JEE Notes | EduRev
gof = IR, fog =  IR
Hence, the required function g: R → R is defined as g(y) = y - 7/10

Q.2. Let f: W → W be defined as f(n) = n − 1, if is odd and f(n) = n + 1, if n is even. Show that f is invertible. Find the inverse of f. Here, W is the set of all whole numbers.
Ans. It is given that:

NCERT Solutions (Miscellaneous Exercise) - Relations and Functions JEE Notes | EduRev
One-one:
Let f(n) = f(m).
It can be observed that if n is odd and m is even, then we will have n − 1 = m + 1.
⇒ n − m = 2
However, this is impossible.
Similarly, the possibility of n being even and m being odd can also be ignored under a similar argument.
∴ Both n and m must be either odd or even.
Now, if both n and m are odd, then we have:
f(n) = f(m) ⇒ n − 1 = m − 1 ⇒ n = m
Again, if both n and m are even, then we have:
f(n) = f(m) ⇒ n + 1 = m + 1 ⇒ n = m
∴ f is one-one.
It is clear that any odd number 2r + 1 in co-domain is the image of 2r in domain and any even number 2r in co-domain is the image of 2r + 1 in domain N.
∴ f is onto.
Hence, f is an invertible function.
Let us define g: W → W as:

NCERT Solutions (Miscellaneous Exercise) - Relations and Functions JEE Notes | EduRev
Now, when n is odd:
gof(n) = g(f(n)) = g(n - 1) = n - 1 + 1 = n
And, when n is even:
gof(n) = g(f(n)) = g(n + 1) = n + 1 - 1 = n
Similarly, when m is odd:
fog(m) = f(g(m)) = g(m - 1) = m - 1 + 1 = m
When m is even:
fog(m) = f(g(m)) = g(m + 1) = m + 1 - 1 = m
∴ gof = Iw and fog = Iw
Thus, f is invertible and the inverse of f is given by f—1 = g, which is the same as f.
Hence, the inverse of f is f itself.

Q.3. If f: R → R is defined by f(x) = x2 − 3x + 2, find f(f(x)).
Ans. f (f(x)) = f(x2 - 3x + 2)
= (x2 - 3x + 2)2 - 3(x2 - 3x + 2) + 2
= (x4 + 9 x 2 + 4 - 6 x3 - 12x + 4x2) + (- 3x2 + 9x - 6) + 2
= x4 - 6x3 + 10x2 - 3x

Q.4. Show that function f: R → {x ∈ R: −1 < x < 1} defined by f(x) = x/1+|x|, x ∈ R is one – one and onto function.
Ans. It is given that f: R → {x ∈ R: −1 < x < 1} is defined as
NCERT Solutions (Miscellaneous Exercise) - Relations and Functions JEE Notes | EduRev
For one – one
Suppose f(x) = f(y), where x, y ∈ R.
NCERT Solutions (Miscellaneous Exercise) - Relations and Functions JEE Notes | EduRev
It can be observed that if x is positive and y is negative,
Then, we have
NCERT Solutions (Miscellaneous Exercise) - Relations and Functions JEE Notes | EduRev
Since x is positive and y is negative:
x > y ⇒ x − y > 0
But, 2xy is negative.
Then, 2xy ≠ x - y.
Thus, the case of x being positive and y being negative can be ruled out.
Under a similar argument, x being negative and y being positive can also be ruled out.
∴ x and y have to be either positive or negative.
When x and y are both positive, we have
NCERT Solutions (Miscellaneous Exercise) - Relations and Functions JEE Notes | EduRev
When x and y are both negative, we have
NCERT Solutions (Miscellaneous Exercise) - Relations and Functions JEE Notes | EduRev
∴ f is one – one.
For onto
Now, let y ∈ R such that −1 < y < 1.
NCERT Solutions (Miscellaneous Exercise) - Relations and Functions JEE Notes | EduRev
∴ f is onto.
Hence, f is one – one and onto.

Q.5. Show that the function f: R → R given by f(x) = x3 is injective.
Ans.  f : R → R is given as f(x) = x3.
For one – one
Suppose f(x) = f(y), where x, y ∈ R.
⇒ x= y3   ...........(1)
Now, we need to show that x = y.
Suppose x ≠ y, their cubes will also not be equal.
⇒ x3 ≠ y3
However, this will be a contradiction to (1).
∴ x = y
Hence, f is injective.

Q.6. Give examples of two functions f: N → Z and g: Z → Z such that g o f is injective but g is not injective. (Hint: Consider f(x) = x and g(x) = |x|)
Ans. Define f: NZ as f(x) = x and g: ZZ as g(x) = |x|.
We first show that g is not injective.
It can be observed that:
g(-1) = |-1| = 1
g(1) = |1| = 1
∴ g(−1) = g(1), but −1 ≠ 1.
∴ g is not injective.
Now, gof: NZ is defined as gof (x) = g(f(x)) = g(x) = |x|.
Let x, y ∈ N such that gof(x) = gof(y).
⇒ |x| = |y|
Since x and y ∈ N, both are positive.
∴ |x| = |y| ⇒x = y
Hence, gof is injective

Q.7. Given examples of two functions f: N → N and ??: N → N such that ?? of is onto but f is not onto.
NCERT Solutions (Miscellaneous Exercise) - Relations and Functions JEE Notes | EduRev
Ans. Define f: N → N by f(x) = x + 1
NCERT Solutions (Miscellaneous Exercise) - Relations and Functions JEE Notes | EduRev

We first show that g is not onto.
For this, consider element 1 in co-domain N. It is clear that this element is not an image of any of the elements in domain N.
∴ f is not onto.
Now, gof: NN is defined by,

NCERT Solutions (Miscellaneous Exercise) - Relations and Functions JEE Notes | EduRev
Then, it is clear that for y ∈ N, there exists x = y ∈ N such that gof(x) = y.
Hence, gof is onto.

Q.8. Given a non-empty set X, consider P(X) which is the set of all subsets of X. Define the relation R in P(X) as follows:
For subsets A, B in P(X), ARB if and only if A ⊂ B. Is R an equivalence relation on P(X)? Justify you answer.
Ans. Since every set is a subset of itself, ARA for all A ∈ P(X).
∴ R is reflexive.
Let ARB ⇒ A ⊂ B.
This cannot be implied to B ⊂ A.
For instance, if A = {1, 2} and B = {1, 2, 3}, then it cannot be implied that B is related to A.
∴ R is not symmetric.
Further, if ARB and BRC, then A ⊂ B and B ⊂ C.
⇒ A ⊂ C
⇒ ARC
∴ R is transitive.
Hence, R is not an equivalence relation as it is not symmetric.

Q.9. Given a non-empty set X, consider the binary operation *: P(X) × P(X) → P(X) given by A * B = A ∩ B ∀ A, B in P(X) is the power set of X. Show that X is the identity element for this operation and X is the only invertible element in P(X) with respect to the operation*.
Ans. It is given the binary operation *:
P(X) × P(X) → P(X) given by A * B = A ∩ B ∀ A, B in P(X)
We know that A ∩ X = A = X ∩ A for all A ∈ P(X)
⇒ A * X = A = X * A for all A ∈ P(X)
Thus, X is the identity element for the given binary operation *.
Now, an element A ∈ P(X) is invertible if there exists B ∈ P(X) such that
A * B = X = B * A [As X is the identity element]
or A ∩ B = X = B ∩ A
This case is possible only when A = X = B.
Thus, X is the only invertible element in P(X) with respect to the given operation*.
Hence, the given result is proved.

Q.10. Find the number of all onto functions from the set {1, 2, 3, … , n) to itself.  
Ans. Onto functions from the set {1, 2, 3, … , n} to itself is simply a permutation on n symbols 1, 2, …, n. Thus, the total number of onto maps from {1, 2, … , n} to itself is the same as the total number of permutations on n symbols 1, 2, …, n, which is n. 9

Q.11. Let S = {a, b, c} and T = {1, 2, 3}. Find F−1 of the following functions F from S to T, if it exists.
(i) F = {(a, 3), (b, 2), (c, 1)}
(ii) F = {(a, 2), (b, 1), (c, 1)}
Ans. S = {a, b, c}, T = {1, 2, 3}
(i) F : S → T is defined as F = {(a, 3), (b, 2), (c, 1)}
⇒ F (a) = 3, F (b) = 2, F(c) = 1
Therefore, F−1: T → S is given by F−1 = {(3, a), (2, b), (1, c)}.
(ii) F : S → T is defined as F = {(a, 2), (b, 1), (c, 1)}
Since F (b) = F (c) = 1, F is not one – one.
Hence, F is not invertible i.e., F−1 does not exist.

Q.12. Consider the binary operations*: R ×R → and o: R × R → R defined as a*b = |a - b| and a o b = a, "a,b ∈ R. Show that * is commutative but not associative, o is associative but not commutative. Further, show that "a, b, c ∈ R, a*(b o c) = (a * b) o (a * c). [If it is so, we say that the operation * distributes over the operation o]. Does o distribute over *? Justify your answer.
Ans. It is given that *: × → and o: R × Ris defined as
a * b = |a - b| and a o b = a, "a, b ∈ R.
For a, b ∈ R, we have:
a * b = |a - b|
b * a = |b - a| = |-(a - b)| = |a - b|
∴ a * b = b * a
∴ The operation * is commutative.
It can be observed that,
(1 * 2) * 3 = (|1 - 2|) * 3 = 1 * 3 = |1 - 3| = 2
and 1 * (2 * 3) = 1 * (|2 - 3|) = 1 * 1 = |1 - 1| = 0
∴ (1 * 2) * 3 * 1 * (2 * 3) where 1, 2, 3 ∈ R.
Hence, the operation * is not associative.
Now. consider the operation o
It can be observed that 1 o 2 = 1 and 2 o 1 = 2.
∴ 1 o 2 ≠ 2 o 1 where 1, 2 ∈ R.
Hence, the operation o is not commutative.
Let a.b.c ∈ R. Then, we have
(a o b)o c = a o c = a
and a o (b o c) = a o b = a
∴ a o b) o c = a o (b o c). where a, b, c ∈ R
Hence, the operation o is associative.
Now. let a, b, c ∈ R. then we have
a * (b o c) = a * b =|a — b|
(a * b) o (a * c) = (|a — b|) o (|a — c|) = |a — b|
Hence, a * (b o c) = (a * b)o (a * c).
Now, 1 o (2 * 3) = 1 o (|2 - 3|) = 1 o 1 = 1
(1 o 2) * (1 o 3) = 1 * 1 = |1 - 1| = 0
∴ 1 o (2 * 3) # (1 o 2) * (1 o 3) where 1, 2, 3 ∈ R
Hence, the operation o does not distribute over *.

Q.13. Given a non -empty set X, let *: P(X) × P(X) → P(X) be defined as A * B = (A − B) ∪ (B − A), ∀ A, B ∈ P(X). Show that the empty set Φ is the identity for the operation * and all the elements A of P(X) are invertible with A−1 = A.
(Hint: (A − Φ) ∪ (Φ − A) = A and (A − A) ∪ (A − A) = A * A = Φ).

Ans. It is given that *: P(X) × P(X) → P(X) is defined as A *B = (A -B) U (B - A) ∀ A, B ∈ P (X).
Let A ∈ P(X). Then, we have
A * Φ = (A - Φ) U (Φ - A)= A U Φ = A
Φ * A = (Φ - A) U (A - Φ) = Φ U A = A
∴ A * Φ = A = Φ * A for all A ∈ P(X)
Thus, Φ is the identity element for the given operation *.
Now, an elements A ∈ P(X) will be invertible if there exists B ∈ P(X) such that
A * B = Φ = B * A. [As Φ is the identity element]
Now. we observed that
A * A = (A - A) U (A - A) = Φ U Φ = 0 for all A ∈ P(X).
Hence, all the elements A of P(X) are invertible with A-1 = A.

Q.14. Define a binary operation *on the set {0, 1, 2, 3, 4, 5} as 

NCERT Solutions (Miscellaneous Exercise) - Relations and Functions JEE Notes | EduRev
Show that zero is the identity for this operation and each element a ≠ 0 of the set is invertible with 6 − a being the inverse of a. 
Ans. Let X = {0, 1, 2, 3, 4, 5}.
The operation * on X is defined as
NCERT Solutions (Miscellaneous Exercise) - Relations and Functions JEE Notes | EduRev
An element e ∈ X is the identity element for the operation *. if
a * e = a = e * a for all a ∈ X
For a ∈ X, we have
a * 0 = a + 0 = a [a ∈ X ⇒ a + 0 < 6 ]
0 * a = 0 + a = a [a ∈ X ⇒ 0 + a < 6 ]
∴ a * 0 = a = 0 * a for all a ∈ X
Thus, 0 is the identity element for the given operation *.
An element a ∈ X is invertible if there exists b ∈ X such that a * b = 0 = b * a.

NCERT Solutions (Miscellaneous Exercise) - Relations and Functions JEE Notes | EduRev
⇒ a = - b or b = 6 - a
But, X = {0, 1, 2, 3, 4, 5} and a, b ∈ X. Then, a ≠ -b.
∴ b = 6 - a is the inverse o f a for all a E X .
Hence, the inverse o f an element a ∈ X. a ≠ 0 is 6 - a i.e., a-1 = 6 - a.

Q.15. Let A = {−1, 0, 1, 2}, B = {−4, −2, 0, 2} and f, ??: A → B be functions defined by f(x) = x2 − x, x ∈ A 
NCERT Solutions (Miscellaneous Exercise) - Relations and Functions JEE Notes | EduRev
Are f and g equal? Justify your answer. (Hint: One may note that two function f: A → B and ??: A → B such that f(a) = ??(a) ∀ a ∈ A, are called equal functions). 
Ans. It is given that A = {-1, 0, 1, 2}, B = {-4, -2, 0, 2}
Also, it is given that f, g : A → B are defined by
NCERT Solutions (Miscellaneous Exercise) - Relations and Functions JEE Notes | EduRev
It is observed that
NCERT Solutions (Miscellaneous Exercise) - Relations and Functions JEE Notes | EduRev
NCERT Solutions (Miscellaneous Exercise) - Relations and Functions JEE Notes | EduRev
NCERT Solutions (Miscellaneous Exercise) - Relations and Functions JEE Notes | EduRev
NCERT Solutions (Miscellaneous Exercise) - Relations and Functions JEE Notes | EduRev
Hence, the functions f and g are equal.

Q.16. Let A = {1, 2, 3}. Then number of relations containing (1, 2) and (1, 3) which are reflexive and symmetric but not transitive is
(a) 1

(b) 2
(c) 3
(d) 4
Ans. (a)
Solution. The given set is A = {1, 2, 3}.
The smallest relation containing (1, 2) and (1, 3) which is reflexive and symmetric, but not transitive is given by:
R = {(1, 1), (2, 2), (3, 3), (1, 2), (1, 3), (2, 1), (3, 1)}
This is because relation R is reflexive as (1, 1), (2, 2), (3, 3) ∈ R.
Relation R is symmetric since (1, 2), (2, 1) ∈ R and (1, 3), (3, 1) ∈ R.
But relation R is not transitive as (3, 1), (1, 2) ∈ R, but (3, 2) ∉ R.
Now, if we add any two pairs (3, 2) and (2, 3) (or both) to relation R, then relation R will become transitive.
Hence, the total number of desired relations is one.

Q.17. Let A = {1, 2, 3}. Then number of equivalence relations containing (1, 2) is
(a) 1

(b) 2
(c) 3
(d) 4
Ans. (b)
Solution. It is given that A = {1, 2, 3}.
The smallest equivalence relation containing (1, 2) is given by,
R1 = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1)}
Now, we are left with only four pairs i.e., (2, 3), (3, 2), (1, 3), and (3, 1).
If we odd any one pair [say (2, 3)] to R1, then for symmetry we must add (3, 2).
Also, for transitivity we are required to add (1, 3) and (3, 1).
Hence, the only equivalence relation (bigger than R1) is the universal relation.
This shows that the total number of equivalence relations containing (1, 2) is two.

Q.18. Let f: R → R be the Signum Function defined as  
NCERT Solutions (Miscellaneous Exercise) - Relations and Functions JEE Notes | EduRev
 and g: R → R be the Greatest Integer Function given by g(x) = [x], where [x] is greatest integer less than or equal to x. Then does fog and gof coincide in (0, 1]?
Ans.
It is given that,
f: RR is defined as
NCERT Solutions (Miscellaneous Exercise) - Relations and Functions JEE Notes | EduRev
Also, g: RR is defined as g(x) = [x], where [x] is the greatest integer less than or equal to x.
Now, let x ∈ (0, 1].
Then, we have:
NCERT Solutions (Miscellaneous Exercise) - Relations and Functions JEE Notes | EduRev
Thus, when x ∈ (0, 1), we have fog(x) = 0and gof (x) = 1.
Hence, fog and gof do not coincide in (0, 1].

Q.19. Number of binary operations on the set {a, b} are
(a) 10
(b) 16
(c) 20
(d) 8
Ans. (b)
Solution. A binary operation * on {a, b} is a function from {a, b} × {a, b} → {a, b} i.e., * is a function from {(a, a), (a, b), (b, a), (b, b)} → {a, b}.
Hence, the total number of binary operations on the set {a, b} is 24 i.e., 16.

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