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CBSE Questions for Class 11 Commerce Applied Mathematics Relations Quiz 16 - MCQExams.com
CBSE
Class 11 Commerce Applied Mathematics
Relations
Quiz 16
Let
R
:
N
→
N
be defined by
R
=
{
(
a
,
b
)
:
a
,
b
∈
N
and
a
=
b
2
}
then, which of the following is true?
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0%
(
a
,
a
)
∈
R
,
∀
a
∈
N
0%
(
a
,
b
)
∈
R
,
⇒
(
b
,
a
)
∈
R
0%
(
a
,
b
)
∈
R
,
(
b
,
c
)
∈
R
⇒
(
a
,
c
)
∈
R
0%
None of the above
Let
A
be set of first ten natural numbers and
R
be a relation on
A
defined by (x , y)
∈
R
⇒
x + 2y = 10 , then domain of
R
is
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0%
{
1
,
2
,
3
,
.
.
.
.
.
.
.
.
.
.
.
.
,
10
}
0%
{
2
,
4
,
6
,
8
}
0%
{
1
,
2
,
3
,
4
}
0%
{
2
,
4
,
6
,
8
,
10
}
Explanation
A
=
{
1
,
2
,
3
,
⋯
⋯
,
10
}
R
=
{
(
x
,
y
)
:
x
+
2
y
=
10
,
x
,
y
∈
A
}
∴
Correct option is (B).
If
h=\left\{ ((x,y),(x-y, x+y))/ x,y \epsilon N \right\}
is a relation on NxN, then domain of h
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0%
$$\left\{ (x,y):\quad x
0%
\left\{ (x,y):\quad x\le y,x,y\epsilon N \right\}
0%
\left\{ (x,y):\quad x>y,x,y\epsilon N \right\}
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\left\{ (x,y):\quad x \ge y,x,y\epsilon N \right\}
If
A=\left\{ x:{ x }^{ 2 }-3x+2=0 \right\}
, and R is a universal relation on A, then R is
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0%
{(1, 1), (2, 2)}
0%
{(1, 1)}
0%
\{ \phi \}
0%
None of these.
Let
R_1, R_2
are relation defined on
Z
such that
aR_1b \Longleftrightarrow (a - b)
is divisible by
3
and
a \,R_2 b \Longleftrightarrow (a - b)
is divisible by
4
. Then which of the two relation
(R_1 \cup R_2), (R_1 \cap R_2)
is an equivalence relation?
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0%
(R_1 \cup R_2)
Only
0%
(R_1 \cap R_2)
Only
0%
Both
(R_1 \cup R_2), (R_1 \cap R_2)
0%
Neither
(R_1 \cup R_2)
nor
(R_1 \cap R_2)
Explanation
Givn relations defined on z are,
R_1=aR_1b\leftrightarrow (a-b)
is divisible by 3.
and
R_2=aR_2b\leftrightarrow (a-b)
is divisible by 4.
Consider,
aR_1b\leftrightarrow a-b
is divisible by 3.
Reflexive
:
aR_1a\leftrightarrow a-a
is divisible by 3
True
\therefore
the given relation
R_1
is a reflexive relation.
Symmetric:
aR_1b\leftrightarrow a-b
is divisible by 3.
\Rightarrow a-b=3k
; where
k
is an integer.
then
bR_1a\leftrightarrow b-a
is divisible by 3.
because
b-a=-(a-b)=-3k
which is divisible by 3.
\therefore
the given relation
R
is a symmetric relation.
Transitive:
aR_1b\leftrightarrow a-b
is divisible by 3.
\Rightarrow a-b=3k; k
is an integer.
b R_1c\leftrightarrow b-c
is divisible by 3.
\Rightarrow b-c=3p;p
is integer.
Then
aR_1c\leftrightarrow a-c
is divisible by 3.
Because
(a-b)+(b-c)=a-c
\Rightarrow 3k+3p=3(k+p)
which is divisible by 3.
\therefore
The given relation
R_1
is transitive relation.
\therefore R_1
satisfies the reflexive, symmetric and transitive relation properties.
\therefore R_1
is an equivalence relation.
Similarly
R_2
is also an equivalence relation.
And as we know that,
"If
R_1
and
R_2
are equivalence relations then
R_1\cap R_2
is also an equivalence relation."
\therefore R_1\cap R_2
is an equivalence relation.
Let
R_1
be a relation defined by
R_1 =
{
(a, b)| a >b , a, b \epsilon R
}. Then
R_1
is
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0%
an equivalence relation on R
0%
reflexive, transitive but not symmetric
0%
symmetric, transitive but nor reflective
0%
neither transitive nor reflective but symmetric
If a relation { R } on the set { 1,2,3 } be defined
by R = { ( 1,2 ) } then { R } is
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0%
reflexive
0%
transitive
0%
Symmetric
0%
None of these
The relation R defined in N as
aRb \Leftrightarrow b
is divisible by a is
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0%
Reflexive but not symmetric
0%
Symmetric but not transitive
0%
Symmetric and transitive
0%
Equivalence relation
If
(x^{2}-2)+(y+3)i=7+4i
then x and y are
Report Question
0%
\pm 3 and 4
0%
\pm \sqrt{5} and 4
0%
\pm 3 and 1
0%
None of these
Let T be the set of all triangles in the Euclidean plane, and let a relation R' on T be defined as aRb if a is congruent to b for all a, b
\in
T. Then, R is?
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0%
Reflexive but not symmetric
0%
Transitive but not symmetric
0%
Equivalence
0%
None of these
Let
R = \left \{(2, 3), (3, 3), (2, 2), (5, 5), (2, 4), (4, 4), (4, 3)\right \}
be a relation on the set
\left \{2, 3, 4, 5\right \}
, then
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0%
R
is reflexive and symmetric but not transitive
0%
R
is reflexive and transitive but not symmetric
0%
R
is symmetric and transitive but not reflexive
0%
R
is an equivalence relation
If
A = \left \{1, 2, 3, 4, 5\right \}
then
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0%
The number of relation on
A
is
2^{25}
0%
The number of relation on
A
is
2^{20}
0%
The number of reflexive relation on
A
is
20^{20}
0%
The number of reflexive relation on
A
is
2^{15}
If
A=\left\{1,2,4 \right\}
,
B=\left\{1,4,6,9 \right\}
and
R
is a relation from
A
to
B
defined by
x
is greater than
y
. The rangle of
R
is :
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0%
\left\{1,4,6,9 \right\}
0%
\left\{4,6,9 \right\}
0%
\left\{ 1 \right\}
0%
none of these
If
g(f(x))= |sinx|
and
g(f(x))= sin^2 \sqrt{x},
then
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0%
f(x)= \sqrt{x}, \ g(x)= sin^2x
0%
f(x) = sin \sqrt{x}, \ g(x)= x^2
0%
f(x)= \sqrt{sinx}, \ g(x)= x^2
0%
f(x)= |x|, \ g(x)= sin^2 x
If
A=\{ 1,2,3\}
then the number of equivalance relation containing the element {1,2} is :
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0%
1
0%
2
0%
3
0%
14
Let S be the set of all straight lines in a plane. Let R be a relation on S defined by
a
R
b\Leftrightarrow a||b
. Then, R is?
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0%
Reflexive and symmetric but not transitive
0%
Reflexive and transitive but not symmetric
0%
Symmetric and transitive but not reflexive
0%
An equivalence relation
If
f(x) = -{ \frac { x\left| x \right| }{ 1+x^{ 2 } } }
then
f^{-1} (x)
equals
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0%
\sqrt { \frac { x\left| x \right| }{ 1-\left| x \right| } }
0%
(Sgn x) \sqrt { \frac { x\left| x \right| }{ 1-\left| x \right| } }
0%
- \sqrt { \frac {x} {1-x}}
0%
None of these
Explanation
\textbf{Step-1: Apply the concept of function & simplify}
\text{Let y = f(x)}
= \dfrac{-x^2}{1+x^2},x\geq0
= \dfrac{x^2}{1+x^2}, x<0
f^{-1}(y)=
\sqrt{\dfrac{-y}{1+y}},y<0
= -\sqrt{\dfrac{y}{1-y}},y>0
f'^{-1}(x) = \sqrt{\dfrac{-x}{1+x}}, x<0
=\sqrt{\dfrac{-x}{1-x}},x>0
\text{= (sgn x)} \sqrt{\dfrac{|x|}{1-|x|}}
\textbf{Hence, option B}
From the following relations defined on set Z of integers, which of the relation is not equivalence relation
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0%
aR_{1} b \Leftrightarrow ( a + b)
is an even integer
0%
aR_{1} b \Leftrightarrow ( a - b)
is an even integer
0%
aR_{3} b \Leftrightarrow ( a + b) a < b
0%
aR_{4} b \Leftrightarrow ( a + b) a = b
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Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers
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