Chapter 1: Relations and Functions

We have,

f: R → R, f(x) = cos x

Now,

f (x₁) = f (x₂)

cos x₁ = cos x₂

x₁ = 2nπ ± x₂, n ∈ Z

It’s seen that the above equation has infinite solutions for x₁ and x₂

Hence, f(x) is many one function.

Also the range of cos x is [-1, 1], which is subset of given co-domain R.

Therefore, the given function is not onto.

Given, X = {1, 2, 3} and Y = {4, 5}

So, X × Y = {(1, 4), (1, 5), (2, 4), (2, 5), (3, 4), (3, 5)}

(i) f = {(1, 4), (1, 5), (2, 4), (3, 5)}

f is not a function as f(1) = 4 and f(1) = 5

Hence, pre-image ‘1’ has not unique image.

(ii) g = {(1, 4), (2, 4), (3, 4)}

It’s seen clearly that g is a function in which each element of the domain has unique image.

(iii) h = {(1,4), (2, 5), (3, 5)}

It’s seen clearly that h is a function as each pre-image with a unique image.

And, function h is many-one as h(2) = h(3) = 5

(iv) k = {(1, 4),(2, 5)}

Function k is not a function as ‘3’ has not any image under the mapping.

Given,

f: A → B and g: B → A satisfy g o f = I_A

It’s clearly seen that function ‘g’ is inverse of ‘f’.

So, ‘f’ has to be one-one and onto.

Hence, ‘g’ is also one-one and onto.

Given,

f(x) = 1/(2 – cos x) ∀ x ∈ R

Let y = 1/(2 – cos x)

2y – ycos x = 1

cos x = (2y – 1)/ y

cos x = 2 – 1/y

Now, we know that -1 ≤ cos x ≤ 1

So,

-1 ≤ 2 – 1/y ≤ 1

-3 ≤ – 1/y ≤ -1

1 ≤ – 1/y ≤ 3

1/3 ≤ y ≤ 1

Thus, the range of the given function is [1/3, 1].

Given ∀ a, b ∈ Z, aRb if and only if a – b is divisible by n.

Now, for

aRa ⇒ (a – a) is divisible by n, which is true for any integer a as ‘0’ is divisible by n.

Thus, R is reflective.

Now, aRb

So, (a – b) is divisible by n.

⇒ – (b – a) is divisible by n.

⇒ (b – a) is divisible by n

⇒ bRa

Thus, R is symmetric.

Let aRb and bRc

Then, (a – b) is divisible by n and (b – c) is divisible by n.

So, (a – b) + (b – c) is divisible by n.

⇒ (a – c) is divisible by n.

⇒ aRc

Thus, R is transitive.

So, R is an equivalence relation.

Long Answer (L.A.)

Given that, A = {1, 2, 3}.

(i) Let R₁ = {(1, 1), (1, 2), (1, 3), (2, 3), (2, 2), (1, 3), (3, 3)}

R₁ is reflexive as (1, 1), (2, 2) and (3, 3) lie is R₁.

R₁ is transitive as (1, 2) ∈ R₁, (2, 3) ∈ R₁ ⇒ (1, 3) ∈ R₁

Now, (1, 2) ∈ R₁ ⇒ (2, 1) ∉ R₁.

(ii) Let R₂ = {(1, 2), (2, 1)}

Now, (1, 2) ∈ R₂, (2, 1) ∈ R₂

So, it is symmetric,

And, clearly R₂ is not reflexive as (1, 1) ∉ R₂

Also, R₂ is not transitive as (1, 2) ∈ R₂, (2, 1) ∈ R₂ but (1, 1) ∉ R₂

(iii) Let R₃ = {(1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3)}

R₃ is reflexive as (1, 1) (2, 2) and (3, 3) ∈ R₁

R₃ is symmetric as (1, 2), (1, 3), (2, 3) ∈ R₁ ⇒ (2, 1), (3, 1), (3, 2) ∈ R₁

Therefore, R₃ is reflexive, symmetric and transitive.

Given function: R = {(x, y): x ∈ N, y ∈ N, 2x + y = 41}.

So, the domain = {1, 2, 3, ….., 20} [Since, y ∈ N ]

Finding the range, we have

R = {(1, 39), (2, 37), (3, 35), …., (19, 3), (20, 1)}

Thus, Range of the function = {1, 3, 5, ….., 39}

R is not reflexive as (2, 2) ∉ R as 2 x 2 + 2 ≠ 41

Also, R is not symmetric as (1, 39) ∈ R but (39, 1) ∉ R

Further R is not transitive as (11, 19) ∉ R, (19, 3) ∉ R; but (11, 3) ∉ R.

Thus, R is neither reflexive nor symmetric and nor transitive.

Given, A = {2, 3, 4}, B = {2, 5, 6, 7}

(i) Let f: A → B denote a mapping

f = {(x, y): y = x + 3} or

f = {(2, 5), (3, 6), (4, 7)}, which is an injective mapping.

(ii) Let g: A → B denote a mapping such that g = {(2, 2), (3, 2), (4, 5)}, which is not an injective mapping.

(iii) Let h: B → A denote a mapping such that h = {(2, 2), (5, 3), (6, 4), (7, 4)}, which is one of the mapping from B to A.

(i) Let f: N → N, be a mapping defined by f (x) = x²

For f (x₁) = f (x₂)

Then, x₁² = x₂²

x₁ = x₂ (Since x₁ + x₂ = 0 is not possible)

Further ‘f’ is not onto, as for 1 ∈ N, there does not exist any x in N such that f (x) = 2x + 1.

(ii) Let f: R → [0, ∞), be a mapping defined by f(x) = |x|

Then, it’s clearly seen that f (x) is not one-one as f (2) = f (-2).

But |x| ≥ 0, so range is [0, ∞].

Therefore, f (x) is onto.

(iii) Let f: R → R, be a mapping defined by f (x) = x²

Then clearly f (x) is not one-one as f (1) = f (-1). Also range of f (x) is [0, ∞).

Therefore, f (x) is neither one-one nor onto.

Given,

A = R – {3}, B = R – {1}

And,

f : A → B be defined by f (x) = x – 2/ x – 3 ∀ x ∈ A

Hence, f (x) = (x – 3 + 1)/ (x – 3) = 1 + 1/ (x – 3)

Let f(x₁) = f (x₂)

So, f (x) is an injective function.

Now let y = (x – 2)/ (x -3)

x – 2 = xy – 3y

x(1 – y) = 2 – 3y

x = (3y – 2)/ (y – 1)

y ∈ R – {1} = B

Thus, f (x) is onto or subjective.

Therefore, f(x) is a bijective function.