Sets, Functions and Relations – CA Foundation Maths Study Material

This Sets, Functions and Relations – CA Foundation Maths Study Material is designed strictly as per the latest syllabus and exam pattern.

Sets, Functions and Relations – CA Foundation Maths Study Material

Previous Year Exam Questions

Question 1.
Out of 20 members in a family, 11 like to take tea and 14 like coffee. Assume that each one likes atleast one of the two drinks. F ind how many like both coffee and tea: [1 Mark, Nov. 2006]
(a) 2
(b) 3
(c) 4
(d) 5
Answer:
(d) Given That n(T)= 11 ;n(C) = 14
and n(T ∪ C) =20
n(T ∩ C) = n(T) + n(c) – n(T ∪ C)
= 11 + 14 – 20 = 5
(d) is correct

Question 2.
In a group of 70 people, 45 speak Hindi, 33 speak English and 10 speak neither Hindi nor English. Find how many can speak both English as well as Hindi: [1 Mark, Feb. 2007]
(a) 13
(b) 19
(c) 18
(d) 28.
Answer:
(c) n(H) = 45 ; n(E) = 33
n(H ∪ E)’= 10 ⇒ n(H ∪ E) = 70 – 10 = 60
∴ (H ∪ E) = n(H) + n(E) – n(H ∪ E)
= 45 + 33 – 60= 18
(c) is correct

Question 3.
Let R is the set of real numbers, such that the function f: R → R and g : R → R are defined by f(x) = x2 + 3x + 1 and g(x) = 2x – 3. Find (fog) : [1 Mark, Feb. 2007]
(a) 4x2 + 6x + 1
(b) x2 + 6x + 1
(c) 4x2 – 6x +1
(d) x2 – 6x + 1.
Answer:
(c) f(x) = x2 + 3x + 1
g (x) = 2x – 3
fog = f{g(x)}
= f (2x – 3)
= (2x – 3)2 + 3(2x – 3) + 1
= 4x2 – 2.2x3 + 9 + 6x – 9 + 1
= 4x2 – 6x +1
(c) is correct

Sets, Functions and Relations – CA Foundation Maths Study Material

Question 4.
In a survey of 300 companies, the number of companies using different media – Newspapers (N), Radio (R) and Television (T) are as follows:
n(N) = 200, n(R) = 100, n(T) = 40, n(N ∩ R) = 50, n(R ∩ T) = 20, n(N ∩ T) = 25 and n(N ∩ R ∩ T) = 5.
Find the numbers of companies using none of these media: [1 Mark, May 2007]
(a) 20 companies
(b) 250 companies
(c) 30 companies
(d) 50 companies.
Answer:
(d) n(N ∪ R ∪ T) = n(N) + n(R) + n(T) – n(N ∩ R) – n(N ∩ T) – n(R ∩ T) + n(N ∩ R ∩ T)
= 200 + 100 + 40 – 50 – 20 – 25 + 5
= 250
No. of Companies using no media = 300 – n(N ∪ R ∪ T)
= 300 – 250 = 50
(d) is correct

Question 5.
If R is the set of real numbers such that the function f : R → R is defined by f(x) = (x + 1)2, then find (fof). [1 Mark, May 2007]
(a) (x + 1)2 + 1
(b) x2 + 1
(c) {(x + 1)2 + 1}2
(d) None
Answer:
(c) f(x) = (x + 1)2
fof = f {f(x)} = f{(x + 1)2}
= {(x + 1)2 + 1}2
(c) is correct

Question 6.
If f: R → R, f(x) = 2x + 7, then the inverse of f is : [1 Mark, Aug. 2007]
{a) f-1(x) = (x – 7)/2
(b) f-1(x) = (x + 7)/2
(c) f-1(x) = (x – 3)/2
(d) None.
Answer:
(a)
Let y = f(x) = 2x + 7
or 2x = y – 7
or x = \(\frac{y-7}{2}\)
f-1(x) = \(\frac{x-7}{2}\)

Question 7.
In a town of20,000 families it was found that 40% families buy newspaper A, 20% families buy newspaper B and 10% families buy newspaper C, 5% families buy A and B, 3% buy B and C and 4% buy A and C. If 2% families buy all the three newspapers, then the number of families which buy A only is : [1 Mark, Nov. 2007]
(a) 6600
(b) 6300
(c) 5600
(d) 600.
Answer:
(a) Given
n(A) = 40% ; n(B) = 20% ; n(C) = 10% n(A ∩ B) = 5%; n(B ∩ C) = 4% n(∩) = 4 %; n(A ∩ B ∩ C) = 2%
No. of families which buy only A
= n(A) – n(A ∩ B) – n(A ∩ C) + n(A ∩ B ∩ C)
= 40 – 5 – 4 + 2 = 33%
= 20,000 × 33% = 6600
∴ (a) is correct

Question 8.
Let f: R → R be such that f(x) = 2x, then f(x + y) equals: [1 Mark, Nov. 2007]
(a) f(x) + f(y)
(b) f(x). f(y)
(c) f(x) ÷ f(y)
(d) None of these
Answer:
(b) f(x) = 2x
f(x+y) = 2x+y = 2x.2y = f(x).f(y)
∴ (b) is correct

Question 9.
Out of total 150 students, 45 passed in Accounts, 30 in Economics and 50 in Maths, 30 in both Accounts and Maths, 32 in both Maths and Economics, 35 in both Accounts and Economics, 25 students passed in all the three subjects. Find the numbers who passed atleast in anyone of the subjects: [1 Mark, Feb. 2008]
(a) 63
(b) 53
(c) 73
(d) None.
Answer:
(b) Total students = 150
n(A) = 45 ; n(E) = 30 ; n(M) = 50;
n(A ∩ M) = 30;n(M ∩ E) = 32 n(A ∩ M) = 35; n(M ∩ E ∩ M) = 25
∴ n(A ∪ E ∪ M) = 45 + 30 + 50 – 30 – 32 – 35 + 25 = 53
∴ (b) is correct

Sets, Functions and Relations – CA Foundation Maths Study Material

Question 10.
If f(x) = \(\frac{2+x}{2-x}\), then f-1(x):
(a) \(\frac{2(x-1)}{x+1}\)
(b) \(\frac{2(x+1)}{x-1}\)
(c) \(\frac{x+1}{x-1}\)
(d) \(\frac{x-1}{x+1}\)
Answer:
(a)
∵ f(x) = \(\frac{2+x}{2-x}\) = y(let)
2 + x = 2y – xy
or x + xy = 2y – 2
or x(1 + y) = 2 (y – 1)
or x = \(\frac{2(y-1)}{1+y}\)
f-1(x) = \(=\frac{2(x-1)}{x+1}\); (a) is correct

Question 11.
If A = {1, 2, 3, 4,} ; B = {2, 4, 6, 8,} f(1) = 2, f(2) = 4, f(3) = 6 and f(4) = 8, And f: A → B then f-1 is : [1 Mark, Dec. 2008]
(a) {(2, 1), (4, 2), (6, 3), (8, 4)}
(b) {(1, 2), (2, 4), (3; 6), (4, 8)}
(c) {(1, 4), (2, 2), (3, 6), (4, 8)}
(d) None of these
Answer:
(a)
Sets, Functions and Relations – CA Foundation Maths Study Material 1
f-1 = {(2, 1) (4, 2), (6, 3) (8, 4)}
(a) is correct.

Question 12.
If f (x) = x2 + x – 1 and 4f(x) = f(2x) then find ‘x’. [1 Mark, Dec. 2008]
(a) 4/3
(b) 3/2
(c) -3/4
(d) None of these
Answer:
(b) f(x) = x2 + x – 1
and 4f(x) = f (2x)
or 4(x2 + x – 1) = (2x)2 +2x – 1
or 4x2 + 4x – 4 = 4x2 + 2x – 1
or 4x – 2x = 4 – 1
or 2x = 3 x = 3/2
∴ (b) is correct

Question 13.
If A = {p, q, r, s}, B = {q, s, t}, C = {m, q, n} Find C – (A ∩ B). [1 Mark, Dec. 2008]
(a) {m, n}
(b) {p, q}
(c) {r, s}
(d) {p, r}
Answer:
(a)
C – (A ∩ B) = { m, q, n} – ({p, q, r, s} ∩ {q, s, t}
= {m, q, n} – {q, s}
= {m; n}
∴ (a) is correct.

Question 14.
X = {x, y, w, z} y = {l,2,3,4}; H = {(x, 1); (y, 2); (y, 3); (z, 4); (x, 4. [1 Mark, Dec. 2009]
(a) H is a function from x to y
(b) H is not a function from x to y
(c) H is a relation from y to x
(d) None of these
Answer:
(b) H is not a function from x to y because x has 2 images 1 & 4
(b) is correct

Question 15.
Given the function f(x) = (2x + 3), then the value of f(2x) – 2f(x) +3 will be : [1 Mark, Dec. 2009]
(a) 3
(b) 2
(c) 1
(d) 0
Answer:
(d) ∵ f(x) = 2x + 3
f(2x) – 2 f(x) + 3
= 2(2x) + 3 – 2(2x + 3) + 3
= 4x + 3 – 4x – 6 + 3 = 0
∴ (d) is correct

Question 16.
If f(x) = 2x + h then find f(x + h) – 2f(x). [1 Mark, Dec. 2009]
(a) h – 2x
(b) 2x – h
(c) 2x + h
(d) None of these
Answer:
(a) f(x) = 2x + h
f(x + h) – 2f(x)
= 2(x + h) + h – 2(2x + h)
= 2x + 2h + h – 4x – 2h
= h – 2x
∴ (a) is correct

Question 17.
If A = {X :X2 – 3X + 2 = o]
B = {X :X2 – 4X + 12 = o),
Then B – A is equal to: [1 Mark, June 2010]
(a) {-6}
(b) {1}
(c) {1, 2}
(d) {2, -6}
Answer:
(a) ∵ x2 – 3x + 2 = 0
or x2 – 2x – x + 2 = 0
or x(x – 2) -1 (x – 2) = 0
or (x- 2) (x – 1) = 0
∵ x = 1 ;2
A = {1, 2}
And x2 + 4x – 12 = 0
or x2 + 6x – 2x – 12 = 0
or x(x +6) -2 (x + 6) = 0
or (x + 6) (x – 2) = 0
x = -6 ; 2
∵ B = {-6 ; 2}
B – A = {-6 ; 2} – {1 ; 2}
= {-6}
(a) is correct

Question 18.
If F: A → R is a real-valued function defined by f(x) = \(\frac{1}{x}\) then: [1 Mark, June 2010]
(a) R
(b) R-{1}
(c) R-{o}
(d) R-N
Answer:
(c)
f(x) = \(\frac{1}{x}\) is defined at all x ∈ R except x = 0
A = R – { 0}
∴ (c) is correct

Sets, Functions and Relations – CA Foundation Maths Study Material

Question 19.
In the set N of all natural numbers the relation R defined by a R b “if and only if, a divide b”, then the relation R is: [1 Mark, June 2010]
(a) Partial order relation
(b) Equivalence relation
(c) Symmetric relation
(d) None of these.
Answer:
(a) It is transitive relation, i.e. partial order relation

Question 20.
For any two sets A and B, A ∩ (A’ ∪ B) = _______, where A’ represent the compliment of the set A. [1 Mark, Dec. 2010]
(a) A ∩ B
(A) A ∪ B
(c) A ∪ B
(d) None of these
Answer:
Tricks : Take an example and then decide the answer
Let U = {0, 1, 2, 3, 4, 5}
A = {0, 1, 2, 3}
B = {2, 3, 4, 5}
A’ = U – A
= (4; 5}
A‘ ∪ S = {4, 5} ∪ {2, 3, 4, 5}
= {2, 3, 4, 5}
∴ A ∩ (A’ ∪ B)
= {0, 1, 2, 3} ∩ {2, 3, 4, 5}
= {2, 3}
= A ∩ B
∴ (a) is correct
IInd method = (A ∩ A’) ∪ (A ∩ B) = { } ∪ (A ∩ B) = A ∩ B

Question 21.
If f : R → R, f(x) = x + 1, g :R → R g(x) = x2 + 1 then fog(-2) equals to : [1 Mark, Dec. 2010]
(a) 6
(b) 5
(c) -2
(d) None
Answer:
(a) f(x) = x + 1
g(x) = x2 + 1. ⇒ g(-2) = (-2)2 + 1 = 5
fog(-2) = f{g(-2)} = f(5)
= 5 + 1 = 6
∴ (a) is correct

Question 22.
If A ⊂ B, then following is true: [1 Mark, Dec. 2010]
(a) A ∩ B = B
(b) A ∪ B = B
(c) A ∩ B = A’
(d) A ∩ B
Answer:
(b)

Question 23.
If f(x – 1)= x2 – 4x + 8, then f(x + 1) = . [1 Mark, Dec. 2010]
(a) x2 + 8
(b) 2 + 7
(c) x2 + 4
(d) x2 – 4x
Answer:
(c); f(x – 1) = x2 – 4x + 8
= (x – 1 + 1)2 – 4 (x – 1+ 1) + 8
f(x + 1) = (1 + 1 + 1)2 – 4(x + 1 + 1) + 8
= (x + 2)2 – 4 (x + 2) + 8 = x2 + 4x + 4 – 4x – 8 + 8
= x2 +4
∴ (c) is correct.

Question 24.
There are 40 students, 30 of them passed in English, 25 of them passed in Maths and 15 of them passed in both. Assuming that every Student has passed atleast in one subject. How many student’s passed in English only but not in Maths. [1 Mark, June 2011]
(a) 15
(b) 20
(c) 10
(d) 25
Answer:
(a) Total students = 40
n(E) = 30 ;
n(M) = 25 (E ∩ M) = 15
No. of stds. passed in English only n(E) – n(E ∩ M)
= 30 – 15 = 15
∴ (a) is correct

Question 25.
If A = {±2,±3} B = -{1, 4, 9} AND F = {(2, 4) (-2, 4) (3, 9) (-3, 4)} then ‘F’ is defined as: [1 Mark, June 2011]
(a) One to one function from A into B
(b) One to one function from A onto B
(c) Many to one function from A onto B.
(d) Many to one function from A into B.
Answer:
(c)
Sets, Functions and Relations – CA Foundation Maths Study Material 2

Sets, Functions and Relations – CA Foundation Maths Study Material

Question 26.
If f(x) = \(\frac{x}{\sqrt{1+x^2}}\) and g(x) = \(\frac{x}{\sqrt{1-x^2}}\) Find fog ? [1 Mark, June 2011]
(a) x
(b) \(\frac{1}{x}\)
(c) \(\frac{x}{\sqrt{1-x^2}}\)
(d) x\(\sqrt{1-x^2}\)
Answer:
(a)
Sets, Functions and Relations – CA Foundation Maths Study Material 3
∴ (a) is correct

Question 27.
f(x) = 3 + x, for -3 < x < 0 and 3 – 2x for 0 < x < 3, then value of f(2) will be: [1 Mark, Dec. 2011]
(a) -1
(b) 1
(c) 3
(d) 5
Answer:
(a) is correct
f(x) = 3 + 2x ; when -3 < x < 0
= 3 – 2x ; when 0 < x < 3
f(x = 2) = 3 – 2 × 2 = -1
∴ 2 lies in 2nd condition

Question 28.
If A = (1, 2, 3, 4, 5), B = (2, 4) and C = (1, 3, 5) then (A – C) × B is: [1 Mark, Dec. 2011]
(a) {(2, 2)(2, 4)(4, 2)(4, 4)(5, 2) (5, 4)}
(b) {(1, 2) (1, 4) (3, 2) (3, 4) (5, 2) (5, 4)}
(c) {(2, 2) (4, 2) (4, 4) (4, 5)}
(d) {(2, 2) (2, 4) (4, 2) (4, 4)}
Answer:
(d) is correct
(A – C) × B = {2, 4} × {2, 4} = {(2, 2), (2, 4); (4 ;2) ; (4 ; 4)}

Question 29.
For any two sets A and B the set (A ∪ B’)’ is Equal to (where’ denotes compliment of the set): [1 Mark, Dec. 2011]
(a) B – A
(b) A – B
(c) A’ – B’
(d) B’ – A’
Answer:
(a) is correct
Tricks: Let U = {0, 1, 2, 3, 4, 5}
A = {0, 1, 2} ; B = {1, 2, 3}
B’ = U – B = {0, 4, 5)
A ∪ B’ = (0, 1, 2, 4, 5)
(A ∪ B’)’ = ∪ – (A ∪ B’) = {3}
Then Go by choices
For (a) B – A = {1, 2, 3} – {0, 1, 2} = {3}
(A ∪ B’)’ = ∪ – (A ∪ B’) = {3}
II nd method (A ∪ B’)’
= A’ ∩ (B’)’
= A’ ∩ B = B – A ∩ B = B – A

Question 30.
The number of proper sub-set of the set {3, 4, 5, 6, 7} is: [1 Mark, June 2012]
(a) 32
(b) 31
(c) 30
(d) 25
Answer:
(b) No. of proper subsets = 2n – 1 = 25 – 1 = 31

Question 31.
On the set of lines, being perpendicular is a : [1 Mark, June 2012]
(a) Reflexive
(b) Symmetric
(c) Transitive
(d) None of these
Answer:
(b) is correct
It is symmetric relation Because it x is perpendicular to y
Then y is also perpendicular to x

Question 32.
The range of the function f :N → N; f(x) = (-1)x-1, is: [1 Mark, June 2012]
(a) {0, -1}
(b) {1,-1}
(c) {1, 0}
(d) {1, 0, -1}
Answer:
(b) is correct
f(x) = (-1)x-1
If x = odd No. f(x) = 1
It x = 0; even No. f(x) = -1
∴ Range = {1 ; -1}
Domain = {any real No.}

Sets, Functions and Relations – CA Foundation Maths Study Material

Question 33.
For a group of 200 persons, 100 are interested in music, 70 in photography and 40 in swimming, Further more 40 are interested in both music and photography, 30 in both music and swimming, 20 in photography and swimming and 10 in all the three. How many are interested in photography but not in music and swimming ? [1 Mark, Dec. 2012]
(a) 30
(b) 15
(c) 25
(d) 20
Answer:
(d) is correct
Let A = No. of persons interested in Music
B = No. of persons interested in photography
C = No. of persons interested in Swimming
n(A) = 100; n(B) = 70 ; n(C) = 40; n( A ∩ B) = 40; n( A ∩ C) = 30; n(B ∩ C) = 20; n(A ∩ B ∩ C) = 10.
∴ n(B ∩ A’ ∩ C’) = n(B) – n(B ∩ A) – n(B ∩ C) + n(A ∩ B ∩ C)
= 70 – 40 – 20 + 10
= 20

Question 34.
If f : R → R is a function, defined by f(x) = 10x – 7, if g(x) = f-1(x), then the value of g(x) is equal to: [1 Mark, Dec. 2012]
(a) \(\frac{1}{10 x-7}\)
(b) \(\frac{1}{10 x+7}\)
(c) \(\frac{x+7}{10}\)
(d) \(\frac{x-7}{10}\)
Answer:
(c) is correct
Let y = f(x) = 10x – 7
or 10x = y + 7
∴ x = \(\frac{y+7}{10}\)
∴ f-1(x) = \(\frac{x+7}{10}\)
∴ g(x) = \(\frac{x+7}{10}\)

Question 35.
The No. of elements in range of constant function is: [1 Mark, Dec. 2012]
(a) One
(b) Zero
(c) Infinite
(d) None
Answer:
(a) is correct
Let f(x) = c (where c= constant)
Sets, Functions and Relations – CA Foundation Maths Study Material 4
Domain = {x/x ∈ R}
Range = {c}

Question 36.
If f(x) = x + 2, g(x) = 7x then go f(x) = _______. [1 Mark, June 2013]
(a) 7x.x + 2.7x
(b) 7x+2
(c) (7x) + 2
(d) none
Answer:
f(x) = x+2 ; g(x) =7x
gof (x) = g{f(x)} = g(x + 2) = 7x+2
∴ (b) is correct

Question 37.
If f(x) = log\(\left(\frac{1+x}{1-x}\right)\) then f\(\left(\frac{2 x}{1+x^2}\right)\). [1 Mark, June 2013]
(a) f(x)
(b) 2 f(x)
(c) 3 f(x)
(d) -f(x)
Answer:
Sets, Functions and Relations – CA Foundation Maths Study Material 5
(b) is correct

Sets, Functions and Relations – CA Foundation Maths Study Material

Question 38.
If A= {1,2,3} then the relation R={(1,1), (2,3), (2,2), (3,3), (1,2)} on A is: [1 Mark, June 2013]
(a) Reflexive
(b) Symmetric
(c) Transitive
(d) Equivalence
Answer:
(a) is correct.
Reflexive Relation.
xRx ; (x ; x) ∈ R
Here (1,1), (2, 2), (3, 3) ∈ R
So; It is Reflexive

Question 39.
Of the 200 candidates who were interviewed for a position at call center, 100 had a two wheeler, 70 had a credit card and 140 had a mobile phone 40 of them had both a two wheeler and a credit card, 30 had both a credit card and mobile phone, 60 had both a two wheeler and a mobile phone and 10 had all the three. How many candidates had none of them ? [1 Mark, Dec. 2013]
(a) 0
(b) 20
(c) 10
(d) 18
Answer:
(c) is correct
Let n(A) = No. of Candidates having two wheeler
n(B) = No. of candidates having credit cards
n(C) = No. of candidates having mobile phone.
Given
n(A) = 100 ; n(B) = 70 ; n(c) = 140
n(A ∩ B) = 40; n(B ∩ C) = 30; n(C ∩ A) = 60 n(A ∩ B ∩ C) = 10.
∴ n(A ∪ B ∪ C) = 100 + 70 + 140 – 40 – 30 – 60 + 10 = 190
No. of candidates having none = 200 – 190 = 10

Question 40.
If f(x) = \(\frac{x^2-25}{x-5}\) then f(5). [1 Mark, Dec. 2013]
(a) 0
(b) 1
(c) 10
(d) Undefined
Answer:
(d) is correct
f(5) = \(\frac{x^2-25}{x-5}=\frac{5^2-25}{5-5}=\frac{0}{0}\)
∴ Undefined

Question 41.
f(x) = (a – xn)1/n, a > 0 and n is positive integer then f'[f(x)]. [1 Mark, Dec. 2013]
(a) x
(b) a
(c) x1/n
(d) a1/n
Answer:
(a) is correct
f{f(x)} = f{a-nn)1/n]
Sets, Functions and Relations – CA Foundation Maths Study Material 6

Question 42.
In a class of 50 students 35 opted for Maths, 37 opted for commerce. The number of such student who opted for both maths and commerce is: [1 Mark, June 2014]
(a) 13
(b) 15
(c) 22
(d) 28
Answer:
(c) is correct
n(M) = No. of students opted for Maths = 35
n(C) = No. of Student opted for Commerce = 37
So;(M ∪ C) = 50
n(M ∩ C) = 35 + 37 – 50 = 22

Sets, Functions and Relations – CA Foundation Maths Study Material

Question 43.
The range of the relation {(1,0) (2, 0) (3, 0) (4, 0) (0, 0)} is: [1 Mark, June 2014]
(a) (1,2, 3, 4,0}
(b) {0}
(c) {1,2, 3, 4}
(d) None
Answer:
(b) is correct
Range = {0}

Question 44.
If A = {1, 2, 3} and B = {4, 6, 7} then the relation R = {(2, 4) (3, 6)} is: [1 Mark, June 2014]
(a) A function
(b) A function from A to B
(c) Both (a) and (b)
(d) Not a function
Answer:
(d) is correct.
Note:- 1 has no image

Question 45.
A = (2, 3), B = (4, 5), C = (5, 6) then A × (B ∩ C). [1 Mark, Dec. 2014]
(a) [(5, 2), (5, 3)]
(b) [(2, 5), (3, 5)]
(c) [(2, 4), (5, 3)
(d) [(3, 5), (2, 6)]
Answer:
(b) is correct
B ∩ C = {4, 5} ∩{5, 6} = {5}
∴ A × (B∩ C) – {2, 3} × {5}
= {(2,5); (3, 5)}

Question 46.
If a relation S = ((1,1), (2,2), (1,2), (2,1)) is symmetric and .[1 Mark, Dec. 2014]
(a) Reflexive but not transitive
(b) Reflexive as well as transitive
(c) Transitive but not reflexive
(d) Neither transitive nor reflexive
Answer:
If S = {1, 2, 3} then
Then relation {(1,1); (2,2); (1,2); (2,1)} is symmetric and transitive but not Reflexive.

Question 47.
If f(x) = \(\frac{x}{x-1}\), then \(\frac{f(x / y)}{f(y / x)}\) = _______. [1 Mark, Dec. 2014]
(a) x/y
(b) y/x
(c) -x/y
(d) -y/x
Answer:
f(x) = \(\frac{x}{x-1}\)
Sets, Functions and Relations – CA Foundation Maths Study Material 7

Question 48.
Let N be the set of all Natural number; E be the set of all even natural numbers then the function f: N → E defined as f(x) =2x ; ∀ x ∈ N is : [1 Mark, Dec. 2014]
(a) One-one into
(b) One-one onto
(c) Many-one into
(d) Many-one onto
Answer:
(b) is correct
N = {1, 2, 3, …………. ;n}
E = {2, 4, 6, …………. ;2n}
Sets, Functions and Relations – CA Foundation Maths Study Material 8
Clearly it is one-one onto mapping

Question 49.
Which of these is a function from A → B; A= {x, y, z}; B = {a, b, c, d}. [1 Mark, Dec. 2015]
(a) {(x, a) (x, b) (y, c)}
(b) {(x, a) (x, b) (y, c) (z, d)}
(c) {(x, a) (y, b) (z, d)}
(d) {(a, x) (b, z) (c, y)}
Answer:
(c) is correct.
Sets, Functions and Relations – CA Foundation Maths Study Material 9

Question 50.
f(x) = 2x + 2, g(x) = x2, fog(4) = ? [1 Mark, Dec. 2015]
(a) 100
(b) 10
(c) 34
(d) None of these
Answer:
(c) is correct
fog(x) = f{g(x)}
= f(x2) = 2.x2 + 2
fog (4) = 2 × 42 + 2 = 34

Question 51.
In a class of 80 students, 35% play only cricket, 45% only Tennis, How many play Cricket? [1 Mark, Dec. 2015]
(a) 86
(b) 54
(c) 36
(d) 44
Answer:
(d) is correct
Given n(C-T) = n(C) – n(C ∩ T) = 35%
n(T – C) = n(T) – n(C ∩ T) = 45%
n(CUT) = n(C) + n(T) – n(C ∩ T) = 100
or; 35 + n(C ∩ T) + 45 + n(C ∩ T) – n(C ∩ T) = 100
or 80 + n(C ∩ T) = 100 n(C ∩ T) = 20%
n(C) = 35 + n(C ∩ T) = 35 + 20 = 55%
= 80 × 55% = 44

Sets, Functions and Relations – CA Foundation Maths Study Material

Question 52.
If Set A = {x: \(\frac{x}{2}\) ∈ Z, 0 ≤ x ≤ 10}
B = {x : x is one digit prime number} and
C = {x : \(\frac{x}{3}\) ∈ N, x ≤ 12} then A ∩ (B ∩ C) = .[1 Mark, June 2016]
(a) Φ
(b) Set A
(c) Set B
(d) Set C
Answer:
(a)
A = {2, 4, 6, 8, 10}
B = {2, 3, 5, 7}
C = {3, 6, 9, 12}
A ∩(B ∩ C)
= A ∩ (B ∩ C) = Φ
No Common element in all 3 sets.

Question 53.
The domain (D) and range (R) of the function: [1 Mark, June 2016]
f (x) = 2 – |x + 1| is
(a) D = Real numbers, R = (2, ∞)
(b) D = Integers, R = (0, 2)
(c) D = Integers, R = (- ∞, ∞)
(d) D = Real numbers, R = (- ∞, 2)
Answer:
(d) is correct, let y = f(x) = 2 – |x + 1l
For any real values of x; f (x) is defined.
∴ Domain = D Real numbers
Minimum value of |x + 1| is Zero
Maximum value of Range = 2 – 0 = 2
∴ Range = – ∞ < y < 2
= (- ∞; 2]

Question 54.
Let A be the set of the squares of natural numbers and x ∈ A, y ∈ A . Then [1 Mark, June 2016]
(a) x + y ∈ A
(b) x – y ∈ A
(c) \(\frac{x}{y}\) ∈ A
(d) xy ∈ A
Answer:
(d) is correct.
A = {x / x is the squares of natural Nos.}= {1, 4, 9, 16, 25, …………..}
Tricks: then Go by Choices let x = 1; y = 4 ∈ A.
x + y = 1 + 4 = 5 ∉ A.
x – y = 1 – 4 = -3 ∉ A.
\(\frac{x}{y}=\frac{1}{4}\) = ∉ A.
But xy = 1 × 4 = 4 ∈ A.
∴ (d) is correct.

Question 55.
The number of sub-sets formed from the letters of the word “ALLAHABAD”. [1 Mark, June 2016]
(a) 128
(b) 16
(c) 32
(d) None
Answer:
(C) is correct.
Let X = {Letters of word ALLAHABAD}
= {A, L, H, B, D }
No. of sub-sets = 25 = 32

Question 56.
If f(x)=100 A then f-1(x) = .[1 Mark, June 2016]
(a) \(\frac{x}{100}\)
(b) \(\frac{1}{100 x}\)
(c) \(\frac{1}{100}\)
(d) None of these
Answer:
(a) is correct
Let y = f(x) = 100x
x = \(\frac{y}{100}\); So, f-1(x) = \(\frac{x}{100}\)

Question 57.
f : R → R is defined by f (x) = 2x then f is [1 Mark, June 2016]
(a) One – one and onto
(b) One – one and into
(c) Many to one
(d) One to many
Answer:
(B) is correct.

Question 58.
In a class, 80 students speak Hindi, 60 students speak English and 40 students speak both Hindi and English then the number of students in the class is _______. [1 Mark, June 2017]
(a) 100
(b) 120
(c) 140
(d) 180
Answer:
Let H = Students speak Hindi
E = Students speak English
Given
n(H) = 80 ; n (E) = 60
and n (H ∩ E) = 40
n(H ∪ E) = n(H) + n(E) – n(H ∩ E)
= 80 + 60 – 40 = 100.
Option (a) is correct

Question 59.
If f(x) = \(\frac{x-1}{x}\) and g(x) = \(\frac{1}{1-x}\) then fog(x) = ∪
(a) x – 1
(b) x
(c) 1 – x
(d) – x
Answer:
Sets, Functions and Relations – CA Foundation Maths Study Material 10
Option (b) is correct

Sets, Functions and Relations – CA Foundation Maths Study Material

Question 60.
The Range of the function f is defined by f(x) = \(\frac{x}{x^2+2}\) is _______.[1 Mark, June 2017]
Sets, Functions and Relations – CA Foundation Maths Study Material 11
Answer:
Let y = \(\frac{x}{x^2+2}\) = f(x)
or ; yx2 + 2y = x
or yx2 – x + 2y = 0
It is a quadratic equation in terms of x.
Discriminant = D = b2 – 4ac
= (-1)2 – 4. y ,2y = 1 – 8y2
To be Real solutions ;
D ≥ 0 ⇒ 1 – 8y2 ≥ 0
or 1 ≥ 8y2 ⇒ 8y2 ≤ 1
Sets, Functions and Relations – CA Foundation Maths Study Material 12
(c) is correct.

Question 61.
In a class of 35 students, 16 students play football and 24 students play cricket. Assume that each one play atleast one game, then number of students who play both the games is :
(a) 5
(b) 11
(c) 12
(d) 17
Answer:
n(F ∩ C) = n(F) + n(C) – n(F ∪ C)
= 16 + 24 – 35 = 5
option (a) is correct.

Question 62.
If f(x) = \(\frac{x+1}{x+2}\) = then f[f\(\left(\frac{1}{x}\right)\)] = _______. [1 Mark, Dec. 2017]
(a) \(\frac{2 x+3}{3 x+5}\)
(b) \(\frac{2 x+5}{3 x+2}\)
(c) \(\frac{3 x+2}{5 x+3}\)
(d) \(\frac{5 x+2}{2 x+3}\)
Answer:
(c)
Sets, Functions and Relations – CA Foundation Maths Study Material 13

Question 63.
If A = {Φ, {Φ}} then the Power Set of A is: [1 Mark, June 2018]
(a) {Φ},{0}
(b) {Φ, {Φ}, {{Φ}}, A}
(c) A
(d) {A}, {Φ}
Answer:
(b)
A= {Φ;{Φ}}
p(A)={{}}; {Φ};{{Φ}}; {Φ; {Φ}}
= {Φ; {Φ};{{Φ}}; A}}

Question 64.
If A = {x / x =3n – 2n – 1, where n ∈ N}, B = {x/x = 4(n – 1), where n ∈ N}. Then
(A) A ⊂ B
(b) B ⊂ A
(c) A = B
(d) None
Answer:
(a)
Putting n = 1, 2, 3, ……….. ; we get
A = {x / x = 3n – 2n -1}
= {0 ; 4 ; 20 ; ……………..}
B = {A / x = 4 (n – 1)}
= {0 ; 4 ; 8 ; 12 ; 16 ; 20 ; …………}
Clearly; A ⊂ B

Question 65.
The range of the function \(\frac{x^6}{x^{12}+1}\) is: [1 Mark, June 2018]
(a) (0, ∞)
(b) [0,
(c) (-∞, 0) ∪ [2, ∞)
(d) (0, \(\frac{1}{2}\))
Answer:
(b)
Let y = \(\frac{x^6}{x^{12}+1}\)
or yx12 + y = x6
let z = x6
yz2 + y = z ⇒ yz2 – z + y = 0
It is a Quadratic Eqn. in terms of Z. for real solns.
D = b2 – 4ac = (-1)2 – 4.y.y
= 1 – 4 y2
D > 0
or; 1 – 4y2 ≥ 0 ⇒ 1 ≥ 4y2
or ; 4y2 ≤ 1
or y2 ≤ \(\frac{1}{4}\) If y2 = \(\frac{1}{4}\) ⇒ y ± \(\frac{1}{2}\)
–\(\frac{1}{2}\) ≤ y ≤ \(\frac{1}{2}\)
FromQts. 0, [∵ y is always positive.]

Question 66.
Let N be the set of all natural numbers; E be the set of all even natural numbers then the function; [1 Mark, May 2018]
(a) One – one – into
(b) Many – one – into
(c) One – one – one
(d) Many – one – onto
Answer:
(c)
∵ N = set of Natural Numbers
= {1, 2, 3, ………….}
f{x} = 2x ; ∀ X ∈ N
So; f(1) = 2 × 1 = 2
f(2) = 2 × 2 = 4
f(3) = 2 × 3 = 6
So;
Sets, Functions and Relations – CA Foundation Maths Study Material 14
Clearly ; It is one-one and onto mapping.

Sets, Functions and Relations – CA Foundation Maths Study Material

Question 67.
In a town of 20,000 families it was found that 40% families buy newspaper. A, 20% families buy newspaper B and 10% families buy newspaper C. 5% families buy A and B, 3% buy B and C and 4% buy A and C if 2% families buy all the three newspapers, then the number of families which buy A only is : [1 Mark, May 2018 ]
(a) 6600
(b) 6300
(c) 5600
(d) 600
Answer:
(a)
Sets, Functions and Relations – CA Foundation Maths Study Material 15
n(A) = 40% ; n(B) = 20% n(c) = 10% ;
n(A ∩ B) = 5% n(B ∩ C) = 3% ;
n(C ∩ A)=4% n(A ∩ B ∩ C) = 2%
n(A ∩ B̄ ∩ C̄) = Only A
= n(A) – n(A ∩ B) – n(A ∩ C)+ n(A ∩ B ∩ C)
= 20000 × 33%
= 6600.

Question 68.
The numbers of proper sub-sets of the set {3, 4, 5, 6, 7} is : [1 Mark, May 2018]
(a) 32
(b) 31
(c) 30
(d) 25
Answer:
(b)
Formula
No. of proper sub-sets = 2n – 1 = 25 – 1 = 31.

Question 69.
A is {1, 2, 3, 4} and B is {1, 4, 9, 16, 25} if a function f is defined from set A to B where f(x) = x2 then the range of f is: [1 Mark, Nov. 2018]
(a) {1, 2, 3, 4}
(b) {1, 4, 9, 16}
(c) {1, 4, 9, 16, 25}
(d) None of these
Answer:
(b)
∵ f(x) = x2
Sets, Functions and Relations – CA Foundation Maths Study Material 16
Range = {1, 4, 9, 16}

Question 70.
If A = {1, 2} and B = {3, 4}. Determine the number of relations from A and B: [1 Mark, Nov. 2018]
(a) 3
(b) 16
(c) 5
(d) 6
Answer:
(b)
No. of Relations = 2n(A×B)
= 2(2×2)
= 16.

Question 71.
If A = {1, 2, 3, 4, 5, 6, 7} and B = {2, 4, 6, 8}. Cardinal number of A – B is: [1 Mark, Nov. 2018]
(a) 4
(b) 3
(c) 9
(d) 7
Answer:
A ∩ B = {1, 2, 3, 4, 5, 6, 7} ∩ {2,4, 6, 8}
= {2, 4, 6} ⇒n(A ∩ B) = 3
n(A – B) = n(A) – n(A ∩ B)
= 7 – 3 = 4

Question 72.
Identify the function from the following: [1 Mark, Nov. 2018]
(a) {(1,1), (1,2), (1,3)}
(b) {(1,1), (2,1), (2,3)}
(c) {(1,2), (2,2), (3,2), (4,2)}
(d) None of these
Answer:
(c)
Go by choices
Sets, Functions and Relations – CA Foundation Maths Study Material 17

Sets, Functions and Relations – CA Foundation Maths Study Material

Question 73.
If A= {1, 2, 3, 4, 5, 6, 7, 8, 9}; B = {1, 3, 4, 5, 7, 8}; C = {2, 6, 8}. [1 Mark, June 2019]
Then find (A – B) ∪ C
(a) {2, 6}
(b) {2, 6, 8}
(c) {2, 6, 8, 9}
(d) None
Answer:
(c)
A – B = A – (A ∩ B)
= {1, 2, 3, 4, 5, 6, 7, 8, 9}
= {1, 3, 4, 5, 7, 8}
= {2, 6, 9}
(A – B) ∪ C = {2; 6; 9} ∪{2; 6; 8}
= {2 ; 6 ; 8 ; 9}
(c) is correct.

Question 74.
A= {1, 2, 3, 4 ………….. 10} a relation on A, R= {(x,y)/x + y= 10, x ∈ A, Y ∈ A, x ≥ Y} then domain of R-1 is: [1 Mark, June 2019]
(a) {1,2, 3, 4, 5}
(b) {0,3, 5, 7, 9}
(c) {1, 2, 4, 5, 6, 7}
(d) None
Answer:
(a)
Given; A = {1, 2, 3, ………………10}
R={x; y)/x + y = 10; X ∈ A; Y ∈ A; x > y}
⇒ R = (5 ; 5) ; (6 ; 4) ; (7 ; 3); (8 ; 2); (9 ; 1)
⇒ R-1 = (5; 5);(4; 6) ; (3 ; 7); (2 ; 8); (1 ; 9)
Domain of R-1 = (5 ; 4 ; 3 ; 2 ; 1)

Question 75.
The no. of sub-sets of the set {3, 4, 5} is :
(a) 4
(b) 8
(c) 16
(d) 32
Answer:
(b)
No. of sub-sets = 2n = 23 = 8.

Question 76.
If f(x) = x2 and g(x) = √x then
(a) go f(3) = 3
(b) go f(-3) = 9
(c) go f(9) = 3
(d) go f(-9) = 3
Answer:
(a)
∵ f(x) = x2; g(x)= √x
go f(x) = g {f(x)}
\(\sqrt{f(x)}=\sqrt{x^2}\)
= x
gof{ 3) = 3
(a) is correct.

Sets, Functions and Relations – CA Foundation Maths Study Material

Question 77.
If A = {a, b, c, d}; B = {p, q, r, s} which of the following relation is a function from A to B: [1 Mark, June 2019]
(a) R1 = {(a, p), (b, q), (c, s)}
(b) R2 = {(p, a), (b, r), (d, s)}
(c) R3 = {(b, p), (c, s), (b, r)}
(d) R4 = {(a, p), (b, r), (c, q), (d, s)}
Answer:
(d)
GBC
(a) All have one and only one solution of A in B except d i.e. d has no solution in B
⇒ Clearly R, is not Function.
Similarly (b) & (c) are not functions or mappings.
Sets, Functions and Relations – CA Foundation Maths Study Material 18
∴ R4 is a function or mapping from A to B.

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