Question 1.

Show that the relation R in the set is given by A = {x ∈ Z, 0 = x = 12}

R = {(a,b): a – b is a multiple of 4}

Answer:

Given; R = {(a,b): a – b is a multiple of 4}

Clearly a – a = 0 is a multiple of 4. So

(a, a) ∈ R, hence Reflexive

(a,b) ∈ R ⇒ a -b is a multiple of 4, imply that b – a is a multiple of 4 ⇒ (b, a) ∈ R So Symmetric.

(a, b), (b, c) ∈ R ⇒ a – b and b – c are multiple of 4.

a – b + b – c is a multiple of 4

a – c is a multiple of 4 ⇒ (a, c) ∈ R So Transitive.

Hence R is an equivalence Relation.

Question 2.

Let R be a Relation in the set A = {1, 2, 3, 4, 5, 6} defines as R = {(x,y): y = 2x – 1}

- Write R in roster form and find it’s domain and range
- Is R an equivalence relation ? Justify.

Answer:

1. R ={(1, 1), (2, 3), (3, 5)}

Domain = {1, 2, 3} ; Range ={1, 3, 5}

2. Since (2, 2) ∉ R, R is not reflexive

(2, 3) ∈ R but (3, 2) ∉ R

R is not symmetric.

(2, 3) ∈ R, (3, 5) ∉ R but, (2, 5) ∉ R

R is not transitive

∴ R is not an equivalence relation.

Question 3.

Answer:

Given;

Question 4.

Let ‘ * ’ is a binary operation on the set Q of rational numbers as follows;

- a * b = a – b
- a * b = a + ab
- a * b = a2 + b2
- a * b = (a – b)2

Find which of the binary operation are commutative and associative. Also find the identity element if exists.

Answer:

1. Given; a * b = a – b

b * a = b – a ≠ a – b = a * b

Hence not commutative.

Now; a * (b * c) = a * (b – c)

= a – (b – c) – a – b + c

(a * b) * c = (a – b) * c = a – b – c

a * (b * c) ≠ (a * b) * c

Hence not associative.

Now; a * e = a ⇒ a – e = a ⇒ e = 0

e * a = a ⇒ e – a = a ⇒ e = 2a

a * e ≠ e * a.

So identity element does not exist.

2. Given; a * b = a + ab

b * a = b + ab ≠ a + ab = a * b

Hence not commutative.

Now; a * (b * c) = a * (b + bc)

= a + a(b + be) = a + ab + abc

(a * b) * c = (a + ab) * c

= a + ab + c(a + ab)

= a + ab + ac + abc

a * (b * c) * (a * b) * c

Hence not associative.

Now; a * e = a ⇒ a + ae = a ⇒ e = 0

e * a = a ⇒ e + ea = a ⇒ e = a/1+a

a * e ≠ e * a.

So identity element does not exist.

3. Given; a * b = a2 + b2

b * a = b2 + a2 = a2 + b2 = a * b

Hence commutative.

Now; a * (b * c) = a* (b2 + c2) = a2 + (b2 + c2)2

(a * b) * c = (a2 + b2) * c

= (a2 + b2)2 + c2

a * (b * c) ≠ (a * b) * c

Hence not associative.

Now; a * e – a ⇒ a2 + e2 = a

So identity element does not exist.

4. Given; a * b = (a – b)2

b * a = (b – a)2 = (a – b)2 = a * b

Hence commutative.

Now; a * (b * c) = a * (b – c)2 (a – (b-c)2)2

(a * b) * c – (a – b)2 * c = ((a – b)2 – c)2

a * (b * c) ≠ (a * b) * c

Hence not associative.

Now; a * e = √a ⇒ e = a – √a ∉ Q

So identity element does not exist.

Question 5.

Show that the relation R on the set of natural numbers defined as R: { (x, y): y – x is a multiple of 2} is an equivalance relation

Answer:

Since x – x = 0 is multiple of 2, (x, x)∈ R

Therefore reflexive.

If y – x is a multiple of 2 then x – y is also a multiple of 2. Therefore (x, y) ∈ R ⇒ (y, x) ∈ R. Hence symmetric.

If y – x is a multiple of 2 and z-y is a multiple of 2, then their sum y – x + z – y = z – x is a multiple of 2. Therefore (x, y), (y, z) ∈ R ⇒ (x, z) ∈ R

Hence transitive.

Therefore R is an equivalance relation.

Question 6.

Consider f : [3, ∞) → [1, ∞) given by f(x) = x2 – 6x + 10. Find f-1

Answer:

f(x) = (x – 3)2 + 1

y = (x – 3)2 + 1

⇒ (x – 3)2 = y – 1

Question 7.

‘*’ be a binary operation on N × N defined as (a, b) * (c, d) – (ac, bd)

- Show that * is commutative.
- Find the identity element of * if any
- Write an element of N × N which has an inverse.

Answer:

1. (a, b) * (c, d) = (ac, bd) = (ca, db)

= (c, d) * (a, b)

Hence * is commutative.

2. (a, b) * (e, e) = (a, b)

⇒ (ae, be) = (a, b)

⇒ ae = a ⇒ e = 1

⇒ be = b ⇒ e = 1

Therefore (1,1) is the identity element.

3. Invertible element is (1,1).

Question 8.

Show that the relation R in the set R of Real numbers defined asR = {(a, b): a ≤ b2} is neither reflexive, nor symmetric nor transitive.

Answer:

a ≤ a2 is not true for all a. So (a, a) ∉ R hence not Reflexive.

(a, b) ∈ R ⇒ a ≤ b2 does not imply that b ≤ a2. So (b, a) ∉ R, hence not Symmetric. (a, b),(b, c) ∈ R ⇒ a ≤ b2 and b ≤ c2 does not imply that a ≤ c2. So (a, c) ∉ R, hence not Transitive.

### Plus Two Maths Relations and Functions Four Mark Questions and Answers

Question 1.

The relation R defined in the set A = {-1, 0, 1} as R = {(a,b):a = b2}

- Check whether R is reflexive, symmetric and transitive.
- Is R an equivalence relation

Answer:

1. (-1, -1) ∉ R, R is not reflexive

(-1,1) ∈ R but (1, -1) ∉ R, R is not symmetric

(-1,1) ∈ R, (1, 1) ∈ R and (-1, 1) ∈ R,

R is transitive.

2. R is not reflexive, not symmetric and not transitive. So R is not an equivalence relation.

Question 2.

Let A = {1, 2, 3}. Give an example of a relation on A which is

- Symmetric but neither reflexive nor transitive.
- Transitive but neither reflexive nor symmetric.

Answer:

1. R = {(1, 2),(2, 1)}

(1, 1) ∉ R ⇒ R is not reflexive

(1, 2) ∈ R ⇒ (2, 1) ∈ R, R is symmetric

(1, 2) ∈ R, (2, 1) ∈ R but (1,1) ∉ R,

R is not transitive

2. R = {(1, 2),(1, 3),(2, 3)}

(1, 1) ∉ R ⇒ R is not reflexive

(1, 2) ∈ R ⇒ but (2,1) ∉ R, R is not symmetric

(1, 2) ∈ R, (2, 3) ∈R ⇒ (1, 3) ∈ R,

R is transitive.

Question 3.

Find fog and gof if

- f(x) – |x| and g(x) – |3x + 4|
- f(x) = 16x4 and g(x) = x ¼

Answer:

1. Given; f(x) = |x| and g(x) = |3x + 4|

fog(x) = f(g(x))

= f(|3x + 4|) = ||3x + 4|| = |3x + 4|

gof(x) = g(f(x)) = g(|x|) = |3|x| + 4|

2. fog(x) = f(g(x)) = g(x¼) = 16(x¼)4 = x

gof (x) = g(f(x)) = g(16x4) = (16x4)¼= 4x.

Question 4.

Consider the binary operation* : Q → Q where Q is the set of rational numbers is defined as a * b = a + b – ab

- Find 2 * 3 (1)
- Is identity for * exist? If yes, find the identity element. (2)
- Are elements of Q invertible? Is yes, find the inverse of an element in Q (1)

Answer:

1. 2 * 3 = 2 + 3 – 6 = -1.

2. a * e = a ⇒ a + e – ae = a

⇒ e(1 – a) – 0 ⇒ e = 0

e * a = a ⇒ e + a – ea = a

⇒ e(1 – a) = 0 ⇒ e = 0 is the identity element.

3. a * b = 0 ⇒ a + b – ab = 0 ⇒ b = a/a−1

Question 5.

‘*’ is a binary operation on R defined as a * b = 2ab

- Determine whether * is commutative and associative
- Find the identity element if exists.
- Find the inverse element, if exists

Answer:

1. a * b = 2ab = 2ba = b * a

Therefore commutative

a * (b * c) = a * (2bc) = 4abc

(a * b) * c = (2ab) * c = 4 abc

Therefore is associative.

2. a * e = a ⇒ 2ae = a ⇒ e = 1/2

e * a = a ⇒ 2ea = a ⇒ e = 1/2

Therefore identity element is 1/2.

3. a * b = 1/2 ⇒ 2ab = 1/2 ⇒ b = 1/4a, a ≠ 0.

Question 7.

If f :R → R is a function defined by f(x) = 3x – 2

- Show that f is one-one.
- Find fof(x).
- Find the inverse of f if exists.

Answer:

1. f(x1) = f(x2)

⇒ 2x1 – 3 = 2x2 – 3 ⇒ x1 = x2

Therefore one-one.

2. fof(x) = f(f(x)) = f(3x – 2)

f (x) = 3(3x – 2) – 2 = 9x – 6 – 2 = 9x – 8.

Question 8.

Let A = N × N and be the binary operation. On A defined by (a, b) * (c, d) = (a+c, b+d). Show that ‘ * ’ is commutative and associative. Find the identity for ‘ * ’ on A if any.

Answer:

Given; A = N × N and

(a, b) * (c, d) = (a + c, b + d)

(c, d) * (a, b) = (c + a, d + b)

= (a + c, b + d) = (a, b) * (c, d))

Hence commutative.

Now; (a, b) * [(c, d) * (e, f)] = (a, b) * [c + e, d+f]

= (a + c + e, b + d + f)

[(a, b) * (c, d)] * (e, f) = [a + c, b + d] * (e, f)

= (a + c + e, b + d + f)

Hence associative.

(a, b) * (e, e) = (a, b) ⇒ (a + e, b + e) = (a, b)

⇒ a + e = a, b + e = b ⇒ e=0, e = 0 ⇒ (0,0) ∉ A

So identity element does not exist.

Question 10.

1. Let f be a function defined by f(x) = √x is a function if it defined from (1)

(a) f : N → N

(b) f : R → R

(c) f : R → R+

(d) f : R+ → R+

2. Check the injectivity and surjectivity of the following functions (3)

(a) f : N → N defined by f(x) = x3

(b) f : R → R given by f(x) = [x]

Answer:

1.

(d) f : R+ → R+

2.

(a) For x, y ∈ N,

f(x) = f(y) ⇒ x3 = y3 ⇒ x = y

Therefore, f is injective

For 2 ∈ N, there does not exist x in the domain N such that f(x) = x3 = 2.

∴ f is not surjective.

(b) f : R → R given by f(x) = [x]

It seen that f(1.1) = 1 and f(1.8) = 1;

But 1.1 ≠ 1.8;

∴ f is not injective

There does not exist any element x ∈ R

such that f(x) = 0.7

∴ f is not surjective.

### Plus Two Maths Relations and Functions Six Mark Questions and Answers

Question 2.

1. Show that the function f: R → R defined by f(x) = 2x – 3 is one-one and onto. Find f-1

2. Which of the following figure represents the graph of a function on R which is onto but not one-one.

3. Write a function on R which is onto but not one-one

Answer:

1. f(x) = 2x – 3

f(x1) = f(x2)

⇒ 2x1 – 3 = 2x2 – 3 ⇒ x1 = x2

Therefore f is one-one.

Let f(x) = yeR, then

y = 2x – 3 ⇒ 2x = y + 3 ⇒ x = y+3 /2 ∈ R

Therefore f is onto. Then f-1(x) = x+3/2.

2. Option (c)

[(b),(c),(d) are not one-one since for different values of x, we have same value of y. ie; horizontal line meets at more than one point, (b) and (d) are not onto since range and codomain are different].

3. f(x) = x2, f : R → [0, ∞) (Any other function).

Question 3.

A = {1, 2, 3, 4, 6}, * is a binary operation on A is defined as a * b = HCF of a and b.

- Represent * with the help of an operation table.
- Find the identity element.
- Write a commutative binary operation on A with 3 as the identity element. (Hint: Operation table may be used.

Answer:

1.

2. From the table it is clear that the identity element is 6.

3. We can write any operation table which is commutative with 3 as the identity element.

Question 4.

Let * be a binary operation on the set of all real numbers R defined by a * b = a + b + a2b for a, b R.

- Find 2 * 6 and 6 * 2. (2)
- Prove that * is neither commutative nor associative. (2)
- Find the identity elements in R if exists. (2)

Answer:

1. a * b = a + b + a2b

2 * 6 = 2 + 6 + 22 × 6 = 8 + 24 = 32

6 * 2 = 6 + 2 + 62 × 2 = 8 + 72 = 80.

2. a * b = a + b + a2b a * b ≠ b * a

b * a = b + a + b2a

∴ is not commutative.

(ab)c = (a + b + a2b) c

= a + b + a2b + c + (a + b + a2b)2c

a * (b * c) = a * (b + c + b2c)

= a + b + c + b2c + a2(b + c + b2c)

∴ is not associative.

3. Let e is the identity element.

∴ a * e = a ⇒ a + e + a2e = a ⇒ e(1 + a2) = 0

∴ e = 0

∴ e * a = a ⇒ e + a + e2a = a ⇒ e + e2a = 0

⇒ e(1 + ae) = 0

⇒ e = 0; (1 + ae) = 0

ea = -1 ⇒ e = – 1a

Identity element does not exists.

Question 5.

- Consider f: {3, 4, 5, 6} → {8, 10, 12, 13, 14} And f = { (3, 8), (4, 10), (5, 12), (6, 14)}. State whether f has inverse ? Give reason. (2)
- Consider f : R → R given by f(x) = 3x + 2 Show that f is invertible. Find the inverse of f

Answer:

1. Distinct elements in set {3, 4, 5, 6} has distinct images, under f.

∴ f is one- one

But 14 in the codomain has no pre image.

∴ f is not onto.

∴ f has no inverse.

2. f(x) = 3x + 2; then

f(x1) = f(x2) = 3x2 + 2 ⇒ x1= x2

Hence F is one – one

For y ∈ R, let y = 3x + 2 ⇒ x = y−2/3 ∈ R