NCERT Solutions for Class 12 Maths Chapter 1 Miscellaneous Exercise (Relations and Functions)
NCERT Solutions for Class 12 Maths Chapter 1 Miscellaneous Ex: Get accurate error free NCERT Solutions for Class 12 Maths chapter 1-Relations and Functions Miscellaneous Ex.
Krati Saraswat6 Jan, 2024
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NCERT Solutions for Class 12 Maths Chapter 1 Miscellaneous Exercise (Relations and Functions)
NCERT Solutions for Class 12 Maths Chapter 1 Miscellaneous Exercise of Relations and Functions is prepared by academic team of Physics Wallah. We have prepared NCERT Solutions for all exercise of chapter 1. Given below is step by step solutions of all questions given in NCERT textbook for Chapter 1 Relations and Functions.NCERT Solutions for class 12 Physics
NCERT Solutions for Class 12 Maths Chapter 1 Miscellaneous Exercise
Solve The Following Questions NCERT Solutions for Class 12 Maths Chapter 1 Miscellaneous Exercise (Relations and Functions):
Question 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. Solution :
NCERT Solutions for Class 12 Maths Chapter 1 Exercise 1.1
Question 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. Solution :
NCERT Solutions for Class 12 Maths Chapter 1 Exercise 1.2
Question 3. If f: R → R is defined by f(x) = x 2 − 3x + 2, find f(f(x)). Solution :
Question
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.
Solution :
NCERT Solutions for Class 12 Maths Chapter 1 Exercise 1.2
Question 5. Show that the function f: R → R given by f(x) = x 3 is injective. Ans f: R → R is given as f(x) = x 3 . Suppose f(x) = f(y), where x, y ∈ R. ⇒ x 3 = y 3 … (1) Now, we need to show that x = y. Suppose x ≠ y, their cubes will also not be equal. ⇒ x 3 ≠ y 3 However, this will be a contradiction to (1). ∴ x = y Hence, f is injective.NCERT Solutions for Class 12 Maths Chapter 1 Exercise 1.3
Question 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| ) Solution :
NCERT Solutions for Class 12 Maths Chapter 1 Exercise 1.4
Question 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
)
Solution :
Question
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:
Solution :
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.
Question
9. Given a non-empty set X, consider the binary operation *: P(X) × P(X) → P(X) given by A * B = A ∩ B & mn For E; 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*.
Solution :
Let S be a non-empty set and P(S) be its power set. Let any two subsets A and B of S.
Question
10. Find the number of all onto functions from the set {1, 2, 3, … , n) to itself.
Solution :
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!.
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)}
Solution :
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.
Question
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.
Solution :
Question
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).
Solution :
: 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.
Question
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
Solution :
| * | 0 | 1 | 2 | 3 | 4 | 5 |
| 0 | 0 | 1 | 2 | 3 | 4 | 5 |
| 1 | 1 | 2 | 3 | 4 | 5 | 0 |
| 2 | 2 | 3 | 4 | 5 | 0 | 1 |
| 3 | 3 | 4 | 5 | 0 | 1 | 2 |
| 4 | 4 | 5 | 0 | 1 | 2 | 3 |
| 5 | 5 | 0 | 1 | 2 | 3 | 4 |
Question
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
Solution :
It is clear that 1 is reflexive and symmetric but not transitive.
Therefore, option (A) is correct.
Question
17. Let A = {1, 2, 3}. Then number of equivalence relations containing (1, 2) is:
(A) 1
(B) 2
(C) 3
(D) 4
Solution :
2, Therefore, option (B) is correct.
Question
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)?
Solution :
Therefore, option (B) is correct.
Question
19. Number of binary operation on the set {a, b} are:
(A) 10
(b) 16
(C) 20
(D) 8
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 2
4
i.e., 16.
The correct answer is B.
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