# Free Objective Test 01 Practice Test - 11th and 12th

Let X = {1,2,3,4,5} and Y = {1,3,5,7,9}. Which of the following is/are relations from X to Y

A.

R1={(x,y)y=2+x,xX,yY}

B.

R2={(1,1),(2,1),(3,3),(4,3),(5,5)}

C.

R3={(1,1),(1,3),(3,5),(3,7),(5,7)}

D.

R4={(1,3),(2,5),(2,4),(7,9)}

#### SOLUTION

Solution : A, B, and C

If  'R' is a relation from 'A' to 'B' , then 'R' is defined as {(x,y)| x A and y B}

In all the ordered pairs in the realtions of R1, R2, R3 first component X and second component Y.  So, R1, R2, R3are relations from X to Y.

R4 is not a relation from X to Y , because in ordered pain (7,9) the first component 7 X.

With reference to a universal set, the inclusion of a subset in another, is relation, which is

A.

Symmetric only

B.

Equivalence relation

C.

Reflexive only

D.

None of these

#### SOLUTION

Solution : D

Since A A . Relation ' ' is relfexive

Since A B , B C

Relation '  ' is transitive.

But A ' ' B , B   A . relation is not symmetric.

If R be a relation < from A={1,2,3,4} to B={1,3,5} i.e., (a,b)  R  a<b, then RoR1 is

A.

{(1, 3), (1, 5), (2, 3), (2, 5), (3, 5), (4, 5)}

B.

{(3, 1) (5, 1), (3, 2), (5, 2), (5, 3), (5, 4)}

C.

{(3, 3), (3, 5), (5, 3), (5, 5)}

D.

{(3, 3) (3, 4), (4, 5)}

#### SOLUTION

Solution : C

We have, R={(1,3);(1,5);(2,3);(2,5);(3,5);(4,5)}
R1 = {(3,1);(5,1);(3,2);(5,2);(5,3);(5,4)}
Hence RoR1 = {(3,3);(3,5);(5,3);(5,5)}

The relation R is defined on the set of natural numbers as {(a,b) : a = 2b}. Then R1 is given by

A.

{(2, 1), (4, 2), (6, 3).....}

B.

{(1, 2), (2, 4), (3, 6)....}

C.

R1 is not defined

D.

None of these

#### SOLUTION

Solution : B

R = {(2,1),(4,2),(6,3),.....}.

So, R1 = {(1,2),(2,4),(3,6),......}.

The relation R defined on the set of natural numbers as {(a, b) : a differs from b by 3}, is given by

A.

{(1, 4, (2, 5), (3, 6),.....}

B.

{(4, 1), (5, 2), (6, 3),.....}

C.

{(1, 3), (2, 6), (3, 9),..}

D.

None of these

#### SOLUTION

Solution : B

R={(a,b):a,bN,ab=3}={((n+3),n):nN}

={(4,1),(5,2),(6,3), .......} .

Given two finite sets A  and B  such that n(A) = 3, n(B) = 3. Then total number of relations from A to B is

A.

4

B.

8

C.

512

D.

6

#### SOLUTION

Solution : C

Here n(A × B) = 3 × 3 = 9

Since every subset of A × B defines a relation from A to B, the number of relations from A to B is equal to the number of subsets of A × B = 2n(A×B)
= 29
= 512

Let A = {1, 2, 3}. The total number of distinct relations that can be defined over A  is

A.

29

B.

6

C.

8

D.

None of these

#### SOLUTION

Solution : A

n ( A × A) = n(A)n(A) = 32 = 9

So, the total number of subsets of A × A is 29

and a subset of A × A is a relation over the set A .

Let A = {1,2,3}, B = {1,3,5}. A relation R:A B is

defined by R = {(1,3),(1,5),(2,1)}. Then R1 is defined by

A.

{(1,2), (3,1), (1,3), (1,5)}

B.

{(1, 2), (3, 1), (2, 1)}

C.

{(1, 2), (5, 1), (3, 1)}

D.

None of these

#### SOLUTION

Solution : C

( x,y) R ⇔ (y,x) R1,       R1={(3,1),(5,1),(1,2)}.

If A = {x:x25x+6=0}, B = {2,4} , C = {4,5}, then A×(B C) is ___

A.

{(2, 4), (3, 4)}

B.

{(4, 2), (4, 3)}

C.

{(2, 4), (3, 4), (4, 4)}

D.

{(2,2), (3,3), (4,4), (5,5)}

#### SOLUTION

Solution : A

Given, A = {x:x25x+6=0}
The elements of A are the roots of x25x+6=0
x25x+6=0(x3)(x2)=0x=3 and 2
A = {2,3} , B = {2,4}, C = {4,5}

BC = {4}

A × (B C) = {2, 3} × {4}
={(2,4), (3,4)}

A relation from P to Q is

A.

A universal set of P × Q

B.

P × Q

C.

An equivalent set of P × Q

D.

A subset of P × Q

#### SOLUTION

Solution : D

A relation from P to Q is a subset of P × Q.

Range of the function f(x)=x2+x+2x2+x+1;xϵR is

A.

(1,)

B.

(1,117]

C.

(1,73]

D.

(1,75]

#### SOLUTION

Solution : C

We have,f(x)=x2+x+2x2+x+1=(x2+x+1)x2+x+1=1+1(x+12)2+34We can see here that as x,f(x)1 which is the min value of f(x).Also f(x) is max when (x+12)2+34is min which is so when x=12 and then 34.  fmax=1+134=73    Ri=(1,73]

If f(x) is a function whose domain is symmetric about the origin, then f(x) + f(–x) is

A.

One-one

B.

Even

C.

Odd

D.

Both even and odd

#### SOLUTION

Solution : B

(a, b)
g(x) = f(x) + f(–x)
g(–x) = f(–x) + f(x) = g(x)
therefore g(x) is even

Product of two odd functions is

A.

Even function

B.

Odd function

C.

Neither even nor odd

D.

Cannot be determined

#### SOLUTION

Solution : A

Let f(x), g(x) be odd
Let F(x) = f(x)g(x)
F(–x) = f(–x)g(–x) = F(x)
therefore F(x) is even

If f(x)+2f(1x)=x2+1  xϵR then f(x) is

A.

13(x2+4x3)

B.

23(x2+4x3)

C.

13(x24x+3)

D.

23(x24x+3)

#### SOLUTION

Solution : C

f(x)+2f(1x)=x2+1   .........(i)
Replacing x by 1 – x
f(1x)+2f(x)=(1x2)+1  .......(ii)
multiplying (ii) by 2 and subtracting it from (1), we get
3f(x)=x22(1x)21
3f(x)=2(1x)2+1x2=(1x)(22x+1+x)=(1x)(3x)=x24x+3
f(x)=13(x24x+3)

Let f be a function satisfying 2f(x)3f(1x)=x2 for any x0, then the value of f(2) is

A. -2
B. 74
C. 78
D. 4

#### SOLUTION

Solution : B

2f(x)3f(1x)=x2(i)Replacing x by 1x2f(1x)3f(x)=1x2(ii)
Solving (i) and (ii) we get
5f(x)=2x2+3x2f(x)=15(2x2+3x2)f(2)=15(8+34)=74