IISER Mathematics - Set Theory, Relations and Mappings
Exam Duration: 45 Mins Total Questions : 30
The smallest set Y such that \(Y\cup \{ 1,2\} =\{ 1,2,3,5,9\} \) is given by
- (a)
{1,2,3,5,9}
- (b)
{3,5,9}
- (c)
{1,2}
- (d)
none of these
{3, 5, 9} \(\cup\) {1, 2} = {1, 2, 3, 4, 5, 9}
A relation from set A to B is
- (a)
a universal set of \(A\times B\)
- (b)
a set equal to \(A\times B\)
- (c)
a set equivalent to \(A\times B\)
- (d)
a subset of \(A\times B\)
See definition
If f and g are mapping from R into R defined as \(f:x\rightarrow x+1\) and \(g:x\rightarrow { e }^{ x }\) ,then fog equals
- (a)
ex+1
- (b)
ex+1
- (c)
ex
- (d)
none of these
fog = f[g(x)] = f[ex] = ex + 1
Let E={1,2,3,4} and F={1,2}. Then number of onto functions from E to F is
- (a)
14
- (b)
16
- (c)
12
- (d)
8
The number of onto mappings
2n(E) - 2 = 24 - 2 = 14
Hence the correct alternative is (a).
Let A={2,4,6,9}, B={4,6,18,27,54}. Then, the set of all ordered pairs such that a is a factor of b and a < b, where \(a\epsilon A\) , and \(b\epsilon B\) , is
- (a)
{(2,6),(2,18),(4,27),(6,27),(9,54)}
- (b)
{(2,4),(2,6),(2,18),(2,54),(6,18),(6,54),(9,18),(9,27),(9,54)}
- (c)
{(2,4),(2,6),(2,18),(2,54),(6,18),(6,54),(9,6),(9,18)}
- (d)
{(2,4),(2,6),(2,18),(2,54),(4,6),(4,18),(4,27)}
Here, A = {2, 4, 6, 9} and B = {4, 6, 18, 27, 54}
Now, we have to find a set of ordered pairs (a, b) such that a < b and a is a factor of b, where a \(\in\) A and b \(\in\) B.
Now, 2 is a factor of 4 and 2 < 4.
So, (2, 4) is one such ordered pair.
Also, 2 is a factor of 6 and 2 < 6.
So, (2, 6) is another such ordered pair.
Similarly, (2, 18), (2, 54), (6, 18), (6, 54), (9, 18), (9, 27) and (9, 54) are other such ordered pairs.
Hence, the required set is {(2, 4), (2, 6), (2, 18), (2, 54), (6, 18), (6, 54), (9, 18), (9, 27), (9, 54)}.
Let A and B be the sets such that \(A\cap X=B\cap X=\phi \) and \(A\cup X=B\cup X\) for some set X. Then,
- (a)
A=B
- (b)
A=X
- (c)
B=X
- (d)
\(A\cup B=X\)
Given that, A\(\cup\)X = B\(\cup\)X
⇒ A\(\cap\)(A\(\cup\)X) = A\(\cap\)(B\(\cup\)X)
⇒ (A\(\cap\)A)\(\cup\)(A\(\cap\)X) = (A\(\cap\)B)\(\cup\)(A\(\cap\)X) [distributive law]
⇒ A\(\cup\)\(\phi\) = (A\(\cap\)B)\(\cup\)\(\phi\)
⇒ A = A\(\cap\)B ....(i)
Again, consider A\(\cup\)X = B\(\cup\)X
⇒ B\(\cap\)(A\(\cup\)X) = B\(\cap\)(B\(\cup\)X)
⇒ (B\(\cap\)A)\(\cup\)(B\(\cap\)X) = (B\(\cap\)B)\(\cup\)(B\(\cap\)X)
⇒ (B\(\cap\)A\(\cup\)\(\phi\)) = B\(\cup\)\(\phi\) [∵ BX =
⇒ A\(\cap\)B = B ....(ii)
Thus, from Eqs. (i) and (ii), we get A = B
Let R be the set of real numbers. Statements I \(A=\{ (x,y)\epsilon R\times R:y-x\) is an integer} is an equivalence relation on R. Statement II \(B=\{ (x,y)\epsilon R\times R:x=\alpha y\) for some rational number \(\alpha \) } is an equivalence relation on R.
- (a)
Statement I is true, Statement II is true; Statement II is the correct explanation for Statement I
- (b)
Statement I is true, Statement II is true; Statement II is not the correct explanation for Statement I
- (c)
Statement I is true, Statement II is false
- (d)
Statement I is false, Statement II is true
Statement I A = {(x, y) R x R : y - x in an interger}
(a) reflexive (x, x) A ⇒ (x - x) is an integer,
i.e., true
ஃ reflexive
(b) Symmetric (x, y)\(\in\) A
⇒ (y - x) is an integer
⇒ -(y - x) is an integer
⇒ (x - y) is an integer
⇒ (y, x)\(\in\) A
ஃ Symmetric
(c) Transitive (x, y) A and (y, z) A
⇒ (y - x) is an integer and (z-y) is an integer.
⇒ (y-x) + (z-y) is an integer
⇒ (z-x) is an integer
⇒ (x, z)\(\in\) A
ஃ Transitive
Hence it is an equivalance relation
Statement II B = {x, y) \(\in\) R x R : x = \(\alpha\)y,
(a) Reflexive Clearly, for any x \(\in\) R, x = 1.x, where 1 is a rational number.
⇒ B is reflexive
(b) Symmetric Note that (x, y)\(\in\) B ⇏ (y, x)\(\in\) B
e.g.,(0, 1)\(\in\) B as 0 = 0 - 1 but (1, 0) \(\in\) B as 1 ⇏ - 0 for any rational numbern \(\alpha\).
⇒ B is symmetric
Hence, B is not an equivalance relation
Which of the following is a finite set?
- (a)
Set of all points in a plane
- (b)
Set of all lines in plane
- (c)
{x:x ∈ R and 0 < x < 1}
- (d)
Set of all persons on the earth
Which of the following sets are equal?
- (a)
A={a, b, c, d}, B={d, c, b, a}
- (b)
A={4, 8, 12, 16}, B={8, 4, 16, 18}
- (c)
A={x:x is a multiple of 10}, B={10, 15, 20, 25, 30, ...}
- (d)
None of these
(a) Here, all the elements of the sets A and B are same which are a, b, c, and d. Thus, A=B
(b) Here, 12 is an element of A but not the element of B and 18 is an element of B but not the element of A thus all the elements of the sets A and B are not sam. Hence A≠B
(c) Here, A={x:x is a multiple of 10}={10, 20,30, ...} and B={10, 15, 20, 25, 30, ...} Clearly A≠B
Let A={1,2,3}, B={3,4} and C={4,5,6}, then Ax(B⋂C)=
- (a)
{(1,4), (2,4), (3,4)}
- (b)
{(4,4), (3,4), (1,4)}
- (c)
{(3,4), (2,4)}
- (d)
{(1,2), (1,4), (1,6), (3,4)}
B∩C={4} ⇒Ax(B⋂C)={(1,4), (2,4), (3,4)}
Let A be the set of first ten natural numbers and let R be a relation in A defined by R={(x,y):x+2y=10, x,y∈N}. Which of the following is false?
- (a)
R={(2,4), (4,3), (6,2), (8,1)}
- (b)
Domain of R={2,4,6,8}
- (c)
Range of R={1,2,3,4}
- (d)
None of these
We have R={(x,y):x,y∈N, x+2y=10, 1≤x, y≤10}
x+2y=10 ⇒ y=\(\frac{10-x}{2}\)
x=1⇒y=\(\frac{9}{2}\)∉N; x=2⇒y=4
x=3⇒y=\(\frac{7}{2}\)∉N; x=4⇒y=3
x=5⇒y=\(\frac{5}{2}\)∉N; x=6⇒y=2
x=7⇒y=\(\frac{3}{2}\)∉N; x=8⇒y=1
x=9⇒y=\(\frac{1}{2}\)∉N; x=10⇒y=0∉N.
ஃ R={(2,4), (4,3), (6,2), (8,1)}
Domain of R={x:(x,y)∈R}={2,4,6,8}
Range of R={y:(x,y)∈R}={4,3,2,1}
If A is the void set Φ, then P(A) has just one element Φ i.e P(Φ). So number of elements of P[P(P(Φ))] is
- (a)
4
- (b)
3
- (c)
5
- (d)
24
We have P(Φ) = {Φ}
∴ P(A) = {Φ, {1}}P(B)={Φ, {2}, {3}, {2, 3}}
P(AUB)={Φ, {1}, {2}, {3}, {2, 3}, {1, 2}, {3, 1}, {1, 2, 3}} and P(A∩B)={Φ}
If a relation R is defined from a set A={2,3,4,5} to a set B={3,6,7,10} as follows (x,y) ∈ R ⇔ x divides y. Expression of R-1 is represented by
- (a)
{(6,2), (10,2), (3,3), (6,3)}
- (b)
{(6,2), (3,3), (10,5), (10,2)}
- (c)
{(6,2),(10,2),(3,3), (6,3), (10,5)}
- (d)
None of these
Recall that a/b stands for a divides b. For the elements of the given sets A and B, we find 2/6, 2/10, 3/3, 3/6, 5/10
⇒ R={(2,6), (2,10), (3,3), (3,6), (5,10)}
ஃ R-1={(6,2), (10,2), (3,3), (6,3), (10,5)}
Let A={a,b,c} and B={4,5}. Consider a relation R defined from set A to set B, then R is equal to
- (a)
A
- (b)
B
- (c)
AxB
- (d)
BxA
If A and B are two non-empty sets, the relation from set A to set B is denoted by AxB.
Let R be a relation from N to N defined by R={(a,b):a,b ∈ N and a=b2}. Which of the following is true?
- (a)
(a,b) ∈ R, for all a∈N
- (b)
(a,b)∈R, implies (b,a)∈R.
- (c)
(a,b)∈R, (b,c)∈R implies (a,c)∈R.
- (d)
None of these
We are given R={(a,b):a,b ∈ N and a=b2}
={(b2,b):b∈N}
(a) False
True, only when a=1, (a,a)=(1,1)=(12,1)∈R.
(b) False
If (a,b)∈R ⇒ a=b2 ⇒ b=a2
ஃ (a,b) ∈ R ⇒ (b,a)∉R.
(c) False
If(a,b) ∈ R ⇒ a=b2 ...(1) and (b,c) ∈ R ⇒b=c2...(2)
From (1) and (2), a=(c2)2=c4 ⇏ a=c2
⇒ (a,b) ∈ R and (b,c) ∈ R but (a,c)∉R.
The relation R defined on set A={x:|x|<3, x∈I} by R={(x,y):y=|x|} is
- (a)
{(-2,2), (-1,1), (0,0), (1,1), (2,2)}
- (b)
{(-2,-2), (-2,2), (-1,1), (0,0), (1,-2), (1,2), (2,-1), (2,-2)}
- (c)
{(0,0), (1,1), (2,2)}
- (d)
None of these
Given, A={x:|x|<3, x∈I}
A={x:-3<x<3, x∈I}={-2,-1,0,1,2}
Also, R={(x,y):y=|x|}
ஃ R={(-2,2), (-1,1), (0,0), (1,1), (2,2)}
The domain of the function \(f(x)=log_2(-log_{1/2}(1+\frac{1}{\sqrt[4]{x}})-1)\)is
- (a)
(0,1)
- (b)
(0,1]
- (c)
[1,∞)
- (d)
(1,∞)
We have f(x)=\(log_2(-log_{1/2}(1+\frac{1}{\sqrt[4]{x}})-1)\)
f(x) is defined if \(-log_{1/2}(1+\frac{1}{\sqrt[4]{x}})-1>0\)
or if \(log_{1/2}(1+\frac{1}{\sqrt[4]{x}})<-1\)or if \((1+\frac{1}{\sqrt[4]{x}})>(1/2)^{-1}\)
or if \(1+\frac{1}{\sqrt[4]{x}}>2\) or if \(\frac{1}{\sqrt[4]{x}}>1\)
or if x1/4<1 or if 0<x<1
ஃ D(f)=(0,1).
The range of the function f(x)=\(\frac{x^2-x+1}{x^2+x+1}\) where x∈R is
- (a)
(-∞,3]
- (b)
(-∞,∞)
- (c)
[3,∞)
- (d)
\([\frac{1}{3},3]\)
We have, \(y=\frac{x^2-x+1}{x^2+x+1}\)
⇒ yx2+yx+y-x2+x-1=0
⇒ x2(y-1)+x(y+1)+(y-1)=0
ஃ x=\(\frac{-(y+1)\pm\sqrt{(y+1)^2-4(y-1)^2}}{2(y-1)}\)
=\(\frac{-(y+1)\pm\sqrt{-3y^2+10y-3}}{2(y-1)}\)
If y=1 then original equation gives x=0. Also, -3y2+10y-3≥0
⇒ 3y2-10y+3≤0
⇒ 3y2-9y-y+3≤0 ⇒ (3y-1)(y-3)≤0
⇒ y∈\([\frac{1}{3},3]\)
ஃ Range is \([\frac{1}{3},3]\)
The domain of the function f(x)=log4(log5(18x-x2-77))) is
- (a)
x∈(4,5)
- (b)
x∈(0,10)
- (c)
x∈(8,10)
- (d)
x∈(8,10]
Since log x is defined ∀ x>0
f(x)=log4(log5(18x-x2-77)))
⇒ log5(log3(18x-x2-77))>0
⇒ log3(18x-x2-77)>50 ⇒ log3(18x-x2-77)>1
⇒ 18x-x2-77>31 ⇒ (x-8)(x-10)<0
⇒ 8<x<10
ஃ x∈(8,10)
If A ⊆ B, then B'-A' is equal to
- (a)
A'
- (b)
B'
- (c)
A-B
- (d)
Φ
x∈B' ⇒x∉B ⇒ x∉A ⇒ x∈A'
⇒ B'⊆A' ⇒ B' - A'=Φ
Let \(f=\{(x,\frac{x^2}{1+x^2}), x∈R\}\) be a function from R into R. The range of f is
- (a)
[0,1]
- (b)
[0,1)
- (c)
R
- (d)
(-∞,0]
We have, \(f=\{(x,\frac{x^2}{1+x^2}), x∈R\}\)
Clearly, domain of f is R.
Let y=\(\frac{x^2}{1+x^2}\) . It is clear that \(\frac{x^2}{1+x^2}\ge0\)
and x2<1+x2 ⇒ \(\frac{x^2}{1+x^2}<1\)
So this implies 0≤y<1. Hence, range of f is [0,1)
The domain of the function f given by f(x)=\(\frac{1}{\sqrt{[x]^2-[x]-6}}\) is
- (a)
(-∞,-2)
- (b)
(-∞,-2)U[4,∞)
- (c)
[4,∞)
- (d)
(-∞,2)U[4,∞)
Given that f(x)=\(\frac{1}{\sqrt{[x]^2-[x]-6}}\), f is defined if [x]2-[x]-6>0. or ([x]-3)([x]+2)>0,
⇒ [x]<-2 or [x]>3 ⇒ x<-2 or x≥4
Hence, domain=(-∞,-2)U[4,∞)
If f(x)=x2-x-2, x∈R-{0}, then \(f(\frac{1}{x})\) is equal to
- (a)
f(x)
- (b)
-f(x)
- (c)
\(\frac{1}{f(x)}\)
- (d)
(f(x))2
f(x)=x2-\(\frac{1}{x^2}\)⇒ \(f(\frac{1}{x})=(\frac{1}{x})^2-\frac{1}{(1/x)^2}\)
=\(\frac{1}{x^2}-x^2=-(x^2-\frac{1}{x^2})=-f(x)\)
If n(U)=700, n(A)=200, n(B)=240, n(A∩B)=100 then n(A'UB') equals
- (a)
260
- (b)
560
- (c)
360
- (d)
600
n(A'UB')=n(A∩B)'
=n(U)-n(A∩B)=700-100=600
If n(A)=1000, n(B)=500, n(A∩B) > 1 and n(AUB=p, then
- (a)
500 < p < 1000
- (b)
1001< p< 1498
- (c)
100< p< 64
- (d)
1000< p< 1499
We know that, n(AUB)=n(A)+n(B)-n(A∩B)
⇒ p=1000+500-n(A∩B) ⇒1 < n(A∩B) < 500
Hence, p=1000+500-1=1499
an p=1000+500-500=1000
∴ 1000 < p < 1499
Let f(x)=√x-x and g(x)=x2 be two functions defined over the set of non-negative real numbers, then (fg)(x)=
- (a)
\(x^{\frac{2}{5}}-x^3\)
- (b)
x3-√x
- (c)
\(x^3-x^{\frac{2}{5}}\)
- (d)
\(x^{-\frac{5}{2}}-x^3\)
(fg)(x)=f(x)g(x)=(√x-x)x2=x5/2-x3
Let f(x)=\(\sqrt{1+x^2}\), then
- (a)
f(xy)=f(x).f(y)
- (b)
f(xy)≥f(x).f(y)
- (c)
f( xy )≤f(x).f(y)
- (d)
None of these
\(f(x)=\sqrt{1+x^2}\)
Now, \(f(xy)=\sqrt{1+x^2y^2}\) ...(i)
and \(f(x)f(y)=\sqrt{1+x^2}\sqrt{1+y^2}\)
\(\Rightarrow f(x)f(y)=\sqrt{1+x^2y^2+x^2+y^2}\)...(ii)
ஃ From (i) and (ii), we get
f(x)f(y)≥f(xy)
In a school, there are 20 teachers who teach mathematics or physics. Of these, 12 teach mathematics and 4 teachers teach both mathematics and physics. Then the number of teachers teaching only physics are
- (a)
12
- (b)
16
- (c)
8
- (d)
4
Let M= set of mathematics teachers and P = set of physics teachers.
n(only mathematics teacher) = n(M) - n(M⋂P)=12-4=8
Also, n(MUP)=n(only mathematics teachers)+n(M⋂P)+n (only physics teachers)
ஃ 20 = 8 + 4 + n (only physics teachers)
ஃ n(only physics teachers) = 20 - 12 = 8
Statement-I: Let n(U) = 200, n(A) = 120 and n(A⋂B)=30 then n(A⋂B')=90
Statement II: n(X-Y)=n(X)-n(X⋂Y)
- (a)
If both Statement-I and Statement-II are true and Statement-II is the correct explanation of Statement-I.
- (b)
If both Statement-I and Statement-II are true but Statement-II is not the correct explanation of Statement-I.
- (c)
If Statement-I is true but Statement-II is false.
- (d)
If Statement -I is false and Statement -II is true.
n(A∩B')=n(A-B)=n(A)-n(A∩B)=120-30=90
Statement-I: n(U) = 600, n(A) = 450, n(B) = 250 and n(A⋂B)=50, then given data's are correct.
Statement II: n(AUB)≤n(U)
- (a)
If both Statement-I and Statement-II are true and Statement-II is the correct explanation of Statement-I.
- (b)
If both Statement-I and Statement-II are true but Statement-II is not the correct explanation of Statement-I.
- (c)
If Statement-I is true but Statement-II is false.
- (d)
If Statement -I is false and Statement -II is true.
Given, n(U)=600, n(A)=450, n(B)=250 and n(A∩B)=50
Now, n(AUB)=n(A)+n(B)-n(A⋂B)=450+250-50
ஃ n(AUB)=700-50=650≰ n(U)
Data is incorrect
Statement I is false and the fact n(AUB)≤n(U) is true.