# NCERT Solutions for Class 12 Maths Chapter 1 Miscellaneous

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## 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.

It is given that fR → R is defined as f(x) = 10x + 7.

One-one:

Let f(x) = f(y), where xy ∈R.

⇒ 10x + 7 = 10y + 7
⇒ x = y

∴ is a one-one function.

Onto:

For ∈ R, let y = 10x + 7.

Therefore, for any ∈ R, there exists such that

∴ is onto.

Therefore, is one-one and onto.

Thus, f is an invertible function.

Let us define gR → R as

Now, we have:

Hence, the required function gR → 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.

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 2+ 1 in co-domain is the image of 2in domain and any even number 2in co-domain is the image of 2+ 1 in domain N.

f is onto.

Hence, is an invertible function.

Let us define g: W → W as:

Now, when n is odd:

Thus, f is invertible and the inverse of 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)).

It is given that fR → 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 fR → {x ∈ R: −1 < x < 1} is defined as f(x) = R.

Suppose f(x) = f(y), where x∈ R.

It can be observed that if x is positive and y is negative, then we have:

Since 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 < < 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.

fR → R is given as f(x) = x3.

Suppose f(x) = f(y), where xy ∈ 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| )

Define fN → Z as f(x) = x and gZ → 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, gofN → Z is defined as

Let xy ∈ 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

Define fN → N by,

f(x) = x + 1

And, gN → 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, gofN → N is defined by,

Then, it is clear that for y ∈ N, there exists ∈ 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 AB in P(X), ARB if and only if A ⊂ B. Is R an equivalence relation on P(X)? Justify you 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 = {1, 2} and B = {1, 2, 3}, then it cannot be implied that B is related to A.

∴ R is not symmetric.

Further, if ARand BRC, then A ⊂ B and ⊂ 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.

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)}

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.

#### 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).

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 fA → B and g: A → B such that f(a) = g(a) &mnForE;a ∈A,, are called equal functions).

#### 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)?

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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