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,, g∶R → R 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
Justify your answer
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}