# Relations and Functions

## Misscellanous Questions

Question 1: The function of f is defined by: The relation g is defined by Show that f is a function and g is not a function.

Solution: Given,

The function of f is defined by: Therefore,
f (0) = 0
f (1) = 1
f (2) = 4
f (3) = 9
f (4) = 12
f (5) = 15
f (6) = 18
f (7) = 21
f (8) = 24
f (9) = 27
f (10) = 30

Here, each element of domain which correspond exactly one element of co-domain. Therefore, it is clear that f is a function.

Now,

The relation g is defined by Therefore,
g (0) = 0,
g (1) = 1
g (2) = 4
g (3) = 6

Here ‘g’ is not a function as (2,4) and (2,6) have the same first component.

Clearly, g is Clearly, g is not a function.

Question 2: If f(x)=x^2, find (f(1.1)-f(1))/(1.1-1)

Answer: Given, f(x)=x^2

Hence, (f(1.1)-f(1))/(1.1-1)

=((1.1)^2-1^2)/(1.1-1)

=(1.21-1)/(0.1)=(0.21)/(0.1)

=(21)/(10)=2.1

Question 3: Find the domain of the function f(x)=(x^2+2x+1)/(x^2-8x+12)

Answer: f(x)=(x^2+2x+1)/(x^2-8x+12)

=(x^2+2x+1)/(x^2-6x-2x+12)

=(x^2+2x+1)/(x(x-6)-2(x-6))

=(x^2+2x+1)/((x-6)(x-2))

Here, the function f is defined as all real numbers except x=6 and x=2

Therefore, domain of f is R-{2, 6}

Question 4: Find the domain and the range of the real function f defined by f(x)=sqrt((x-1))

Answer: Given, f(x)=sqrt((x-1))

It is clear that sqrt((x-1)) is defined if (x-1)≥0

This means x≥1

Hence, the domain of g is the set of all real numbers greater than or equal to 1.

That is the domain of f=(1, ∞)

As x≥1

Hence, (x-1)≥0

Or, sqrt(x-1)≥0

Hence, the range of f is the set of all real numbers which are greater than or equal to 0.

Hence, range of f=(0, ∞)

Question 5: Find the domain and the range of the real function f defined by; f (x)= | x-1 |

Solution:

Given, the real function is  f (x)= | x-1 |

From the above function it is clear that | x - 1| is defined for all real numbers.

Therefore, Domain of f = R

Now, For x ∈ R, | x-1 | assumes all real numbers.

Therefore, the range of f is the set of all non-negative real numbers.

Hence, Domain = R

Range = All non-negative real numbers

Question 6: Let f={(x, (x^2)/(1+x^2)):x∈R} be a function from R into R. Determine the range of f.

Solution: Here,

f={(x, (x^2)/(1+x^2)):x∈R}

=(0,0), (±0.5,1/5), (±1,1/2),(±1.5,(9)/(13)), (±2,4/5), (3, (9)/(10)), (4, (16)/(17)…..

It is clear that the range of f is the set of all second elements. SIt is clear that the range of  f  is the set of all second elements. So, it can be observed that all these elements are greater than or equal to 0 but less than 1(Because all denominators are greater than numerator).

Therefore, the range of  f = [ 0, 1)

Or, Range = Any positive real number x is such that 0 ≤ x < 1

Question 7: Let f,, gRR be defined, respectively by f (x)= x +1, g (x) = 2x-3.

Find f+g, f-g and f/g

Solution: Let f (x) = x + 1

g (x) = 2x – 3

Therefore,

f + g = f (x) + g (x)– 3

= 3x – 2

f – g = f (x) – g (x)

= (x + 1) – (2x – 3)

= x + 1 – 2x + 3

= – x + 4

f/g=(f(x))/(g(x))

=(x+1)/(2x-3), x≠3/2

Hence, f+g=3x-2

f-g=-x+4

f/g=(x+1)/(2x-3), x≠3/2

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

Solution: Given, f = {( 1, 1 ), ( 2, 3 ), ( 0, - 1 ), ( -1, -3)}

f (x) = ax + b ⇒ f (1) = 1

Therefore, a ×1 + b = 1

⇒ a + b = 1

As (1,1)∈ f

⇒ f (0) = -1

⇒ b= -1

Now, after substituting b = - 1 in a + b = 1, we get

a + ( - 1) = 1

⇒ a-1 =1

Therefore, the values of

a = 2 and b = - 1

Question 9: Let R be a relation from N to N defined by

R = { ( a,b) ∶ a,b ∈ N and a = b2
Are the following true?

(i) (a,a) ∈ R, f or all a∈ N

(ii) (a,a) ∈ R,implies (b,a)∈ R

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

Solution:

(i) (a,a) ∈ R, for all a ∈ N

It is not true, because

When a = 2, >2

⇒ 2 = 4

Which is not true.

(ii) (a,a)∈R,implies (b,a)∈R
Which is not true.

(ii) (a,a)∈R,implies (b,a)∈R

It is not true

As,(a,b)∈R implies b= aich is not true,

Therefore,

(b,a) ∉ R

(iii) (a,b)∈R,(b,c)∈R implies (a,c)∈R

No, it is (b,a) ∉ R

(iii) (a,b)∈R,(b,c)∈R implies (a,c)∈R

No, it is also not true

Because,(a,b) ∈ R, (b,c)∈R implies a = b

Qeustion 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?

(i) f is a relation from A to B

(ii) f is a function from A to B

Solution:

(i) f is a relation from A to B

It is true.in f have the same first components.

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

Solution: Here,1 x 6 = 6 and 2 x 3 = 6nd let f : A → N be defined by f(n) = highest prime factor of n. Find the range of f.

HereSo, f is not a function from Z to Z.

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

Solution: Here, A = {9, 10, 11, 12, 13}

f : A → N is defined as f (n) = highest prime factor of n

Prime factor of 9 = 3

f (11) = The highest prime factor of 11 = 11

f (12) = The highest prime factor of 12 = 3

f (13) = The highest prime factor of 13 = 13

The range of f is the set of all f (n), where n ∈ A

Therefore,

Range of f = {3, 5, 11, 13}