**Question 1**

**Question 2**

**Question 3**

**Question 4**

**5. Find the domain and the range of the real function f defined by f (x) = |x –1| **

We know that absolute or modulus of a function is defined for all of Real values of x

Hence Domain of f(x) = |x – 1| is Real Numbers *R*

As f(x) = |x – 1| can only exist for non-negative values.

Hence

Range of f(x) = |x – 1| = *R ^{+} ∪ {0}*

**Question 6**

**Question 7**

**8. Let f = {(1,1), (2,3), (0,–1), (–1, –3)} be a function from Z to Z defined by f(x) = ax + b, for some integers a, b. Determine a, b**

**9. Let R be a relation from N to N defined by R = {(a, b) : a, b ∈ N and a = b ^{2}}. Are the following true?(i) (a,a) ∈ R, for all a ∈ N**

We are given that

R : N → N defined as R = {(a, b) : a, b ∈ N and a = b

^{2}}

If a = 1 then a

^{2}= 1

If a = 2 then a

^{2}= 4

As R = {(a, b) : a, b ∈ N and a = b

^{2}} is not True for all possibles of a, b ∈ N.

Thus R is not a relation.

**(ii) (a,b) ∈ R, implies (b,a) ∈ R**

We are given that

R : N → N defined as R = {(a, b) : a, b ∈ N and a = b

^{2}}

(a,b) ∈ R, implies (b,a) ∈ R

It is alos false as if a = b

^{2}

but b = a

^{2}is not True

For example: 4 = 2

^{2}i which means (2, 4) ∈ R

But 2 ≠ 4

^{2}which means (4, 2)

**∉**R

Hence R is not a relation.

**(iii) (a,b) ∈ R, (b,c) ∈ R implies (a,c) ∈ R**

We are given that

R : N → N defined as R = {(a, b) : a, b ∈ N and a = b

^{2}}

It is false as a = b

^{2}and b = c

^{2}

Which means a = (c

^{2})

^{2}= c

^{4}

a = c

^{4}which means a ≠ c

^{2}

Hence R is not a relation.

**10. Let A ={1,2,3,4}, B = {1,5,9,11,15,16} and f = {(1,5), (2,9), (3,1), (4,5), (2,11)} Are the following true?**

Let’s first find out what’s A ✕ B = ?

A ✕ B = {1, 2, 3, 4} ✕ {1, 5, 9, 11, 15, 16}*A ✕ B = {(1, 1), (1, 5), (1, 9), (1, 11), (1, 15), (1, 16), (2, 1), (2, 5), (2, 9), (2, 11), (2, 15), (2, 16), (3, 1), (3, 5), (3, 9), (3, 11), (3, 15), (3, 16), (4, 1), (4, 5), (4, 9), (4, 11), (4, 15), (4, 16)}***(i) f is a relation from A to B**

f = {(1,5), (2,9), (3,1), (4,5), (2,11)}

As f contains corresponding ordered pair where values from set A are mapped to values from set B

Hence it’s a Relation from A to B.**(ii) f is a function from A to B**

f = {(1,5), (2,9), (3,1), (4,5), (2,11)}

As order pairs (2, 9) and (2, 11) have same value 2 from set A

Hence f is not a Function

**11. Let f be the subset of Z × Z defined by f = {(ab, a + b) : a, b ∈ Z}. Is f a function from Z to Z? Justify your answer.**

**12. Let A = {9,10,11,12,13} and let f : A→N be defined by f (n) = the highest prime factor of n. Find the range of f.**

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