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| Chapter | 1. Relations and Functions |
| Exercise | Miscellaneous |
| Class | Twelve |
| Subject | Maths |
| Category | NCERT Solutions for Class 12 |
NCERT Solutions for Class 12 Maths are available online for students to use and refer to while studying for their exams. The solutions provide step-by-step explanations for each problem in the textbook, making it easy for students to understand the concepts. With Class 12 Maths NCERT Solutions, students can be sure that they are getting the most accurate and up-to-date information available. The solutions are regularly updated by experts in the field, so students can be confident that they are using the most current information.
NCERT Solutions for Class 12 Maths Chapter 1 Miscellaneous
Relations and Functions Miscellaneous Solutions
Q 1) Let f: R → R be defined as f(x) = 10x + 7. Find the function g: R → R such that g o f = f o g = 1R.
Answer)
It is given that f: R → R 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.
Therefore, for any y ∈ R, there exists such that 
∴ f is onto.
Therefore, f is one-one and onto.
Thus, f is an invertible function.
Let us define g: R → R as 
Now, we have:
Hence, the required function g: R → R is defined as 
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.
Answer)
It is given that:
f: W → W is defined as 
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 N is the image of 2r in domain N and any even number 2r in co-domain N 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:
Now, when n is odd:
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)).
Answer)
It is given that f: R → R is defined as f(x) = x2 − 3x + 2.
Q 4) Show that function f: R → {x ∈ R: −1 < x < 1} defined by f(x) = 
, x ∈R is one-one and onto function.
Answer) It is given that f: R → {x ∈ R: −1 < x < 1} is defined as f(x) = , x ∈R.
Suppose f(x) = f(y), where x, y ∈ R.
It can be observed that if x is positive and y is negative, then we have:
Since x is positive and y is negative:
x > y ⇒ x − y > 0
But, 2xy is negative.
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:
When x and y are both negative, we have:
∴ f is one-one.
Now, let y ∈ R such that −1 < y < 1.
If x is negative, then there exists 
such that
∴ 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.
Answer)
f: R → R is given as f(x) = x3.
Suppose f(x) = f(y), where x, y ∈ R.
⇒ x3 = 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| )
Answer)
Define f: N → Z as f(x) = x and g: Z → Z 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: N → Z is defined as
Let x, y ∈ N such that gof(x) = gof(y).
= |x| = |y|
Since x and y ∈ N, both are positive.
Hence, gof is injective
Q 7) Given examples of two functions f: N → N and g: N → N such that gof is onto but f is not onto.
(Hint: Consider f(x) = x + 1 and 
Answer)
Define f: N → N by,
f(x) = x + 1
And, g: N → N by,
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: N → N is defined by,
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:
Answer)
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 since 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 &mnForE; 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*.
Answer) It is given that 
Thus, X is the identity element for the given binary operation *.
Now, an element 
is invertible if there exists 
such that
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.
Answer)
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!.
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)}
Answer)
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 ao b = a, & mn For E; a, b ∈ R. Show that * is commutative but not associative, o is associative but not commutative. Further, show that &mn For E;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.
Answer)
Q 13) Given a non-empty set X, let *: P(X) × P(X) → P(X) be defined as A * B = (A − B) ∪ (B − A), & mn For E; 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).
Answer)
It is given that ∗: P(X) × P(X) → P(X) be defined as
A * B = (A – B) ∪ (B – A), A, B ∈ P(X).
Now, let A ϵ P(X). Then, we get,
A * ф = (A – ф) ∪ (ф –A) = A∪ф = A
ф * A = (ф – A) ∪ (A – ф ) = ф∪A = A
A * ф = A = ф * A, A ϵ P(X)
Therefore, ф is the identity element for the given operation *.
Now, an element A ϵ P(X) will be invertible if there exists B ϵ P(X) such that
A * B = ф = B * A. (as ф is an identity element.)
Now, we can see that A * A = (A –A) ∪ (A – A) = ф∪ф = ф AϵP(X).
Therefore, all the element A of P(X) are invertible with A-1 = A.
Q 14) Define binary operation * on the set {0, 1, 2, 3, 4, 5} as
Show that zero is the identity for this operation and each element a ≠ 0 of the set is invertible with 6 – abeing the inverse of a
Q 15) Let A = {–1, 0, 1, 2}, B = {–4, –2, 0, 2} and f; g: A→B be the functions defined by f(x) = x2 − x, x ∈ A and g(x) = 2x – 1/2 – 1. x ∈ A. Are f and g equal? Justify your answer.
(Hint: One may note that two functions f: A → B and g: A → B such that f(a) = g(a) &mnForE;a ∈A,, are called equal functions).
Answer)
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
Answer) It is clear that 1 is reflexive and symmetric but not transitive.
Therefore, option (A) is correct.
Q 17) Let A = {1, 2, 3}. Then number of equivalence relations containing (1, 2) is:
(A) 1
(B) 2
(C) 3
(D) 4
Answer) 2, Therefore, option (B) is correct.
Q 18) Let R → R be the Signum Function defined as
and R → R be the Greatest 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)?
Answer)
Therefore, option (B) is correct.
Q 19) Number of binary operation on the set {a, b} are:
(A) 10
(b) 16
(C) 20
(D) 8
Answer) 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.
The correct answer is B.
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