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CBSE Questions for Class 12 Commerce Maths Relations And Functions Quiz 3 - MCQExams.com
CBSE
Class 12 Commerce Maths
Relations And Functions
Quiz 3
Let
S
be set of all rational numbers. The functions
f
:
R
→
R
,
g
:
R
→
R
are
defined as
f
(
x
)
=
{
0
,
x
∈
S
1
,
x
∉
S
g
(
x
)
=
{
−
1
x
∈
S
0
x
∉
S
then,
(
f
o
g
)
(
π
)
+
(
g
o
f
)
(
e
)
=
Report Question
0%
−
1
0%
0
0%
1
0%
2
Explanation
f
(
x
)
=
{
0
,
x
is rational
1
,
x
is irrational
g
(
x
)
=
{
−
1
x
is rational
0
x
is irrational
f
o
g
(
π
)
=
f
(
g
(
π
)
)
=
f
(
0
)
=
0
so
g
o
f
(
e
)
=
g
(
f
(
e
)
)
=
g
(
1
)
=
−
1
∴
f
o
g
(
π
)
+
g
o
f
(
e
)
=
−
1
Set
A
has
n
elements. The number of functions that can be defined from
A
into
A
is:
Report Question
0%
n
2
0%
n
!
0%
n
n
0%
n
Explanation
The number of functions that can be defined from a set into another set can be found by
(Number of elements in co-domain)
Number of elements in domain
Since, set
A
has
n
elements, we have
Number of elements in co-domain
=
Number of elements in domain
=
n
(
A
)
=
n
.
Therefore, number of functions that can be defined from
A
into
A
=
n
(
A
)
n
(
A
)
=
n
n
.
If
n
≥
1
is any integer,
d
(
n
)
denotes the number of positive factors of
n
, then for any prime number
p
,
d
(
d
(
d
(
p
7
)
)
)
=
Report Question
0%
1
0%
2
0%
3
0%
4
Explanation
For a number being
p
n
the total number of positive factors excluding
1
and the number itself will be
n
−
1
.
Therefore total number of positive factors will be
n
−
1
+
2
=
n
+
1
Hence for
p
7
we will have
8
factors.
Now for
8
=
2
3
.
Hence
8
will have
4
factors.
Now
4
=
2
2
, hence the number of factors will be
3
.
Let
f
(
x
)
=
x
2
−
x
+
1
,
x
≥
(
1
2
)
then the solution of the equation
f
(
x
)
=
f
−
1
(
x
)
is
Report Question
0%
x
=
1
0%
x
=
2
0%
x
=
1
2
0%
None of these
Explanation
y
=
x
2
−
x
+
1
⇒
x
2
−
x
+
(
1
−
y
)
=
0
Here,
a
=
1
,
b
=
−
1
and
c
=
1
−
y
x
=
1
±
√
1
−
4
(
1
−
y
)
2
[
∵
x
=
−
b
±
√
b
2
−
4
a
c
2
a
]
∵
x
>
1
2
,
∴
x
=
1
2
+
√
y
−
3
4
⇒
f
−
1
(
x
)
=
1
2
+
√
x
−
3
4
Now,
x
2
−
x
+
1
=
1
2
+
√
x
−
3
4
Since the graphs of the original and inverse functions can intersect only on the straight line
y
=
x
,
therefore
x
=
f
(
x
)
⇒
x
=
x
2
−
x
+
1
⇒
x
2
−
2
x
+
1
=
0
⇒
(
x
−
1
)
2
=
0
⇒
x
=
1
lf
f
:
[
−
6
,
6
]
→
R
is defined by
f
(
x
)
=
x
2
−
3
for
x
∈
R
then
(
f
o
f
o
f
)
(
−
1
)
+
(
f
o
f
o
f
)
(
0
)
+
(
f
o
f
o
f
)
(
1
)
=
Report Question
0%
f
(
4
√
2
)
0%
f
(
3
√
2
)
0%
f
(
2
√
2
)
0%
f
(
√
2
)
Explanation
f
o
f
o
f
(
−
1
)
=
f
o
f
(
−
2
)
=
f
(
1
)
=
−
2
f
o
f
o
f
(
1
)
=
f
o
f
(
−
2
)
=
f
(
1
)
=
−
2
f
o
f
o
f
(
0
)
=
f
o
f
(
−
3
)
=
f
(
0
)
=
33
∴
f
o
f
o
f
(
−
1
)
+
f
o
f
o
f
(
1
)
+
f
o
f
o
f
(
0
)
=
29
=
32
−
3
=
(
4
√
2
)
2
−
3
⇒
f
(
4
√
2
)
=
(
4
√
2
)
2
−
3
lf
f
:
R
→
R
is defined by
f
(
x
)
=
{
x
+
4
x
<
−
4
3
x
+
2
−
4
≤
x
<
4
x
−
4
x
≥
4
then the correct matching of list I to List II is.
List - I
List - II
A
)
f
(
−
5
)
+
f
(
−
4
)
=
i
)
14
B
)
f
(
|
f
(
−
8
)
|
)
=
ii
)
4
C
)
f
(
f
(
−
7
)
+
f
(
3
)
)
=
i
i
i
)
−
11
D
)
f
(
f
(
f
(
f
(
0
)
)
)
+
1
=
i
v
)
−
1
v)
1
vi)
0
Report Question
0%
A-iii , B-vi , C-ii , D- v
0%
A-iii , B-iv , C-ii , D- vi
0%
A-iv , B-iii , C-ii , D- i
0%
A-ii , B-vi , C-v , D- ii
Explanation
f
(
−
5
)
=
−
5
+
4
=
−
1
;
f
(
−
4
)
=
3
(
−
4
)
+
2
=
−
10
∴
f
(
−
5
)
+
f
(
4
)
=
−
11
f
(
|
f
(
−
8
)
|
)
=
f
(
(
−
8
+
4
)
)
=
f
(
4
)
=
4
−
4
=
0
f
(
f
(
−
7
)
)
t
f
(
3
)
=
f
(
−
7
+
4
)
+
3
(
3
)
+
2
=
3
(
−
3
)
+
2
+
3
(
3
)
+
2
=
4
f
(
f
(
f
(
f
(
0
)
)
)
)
+
1
=
f
(
f
(
f
(
2
)
)
)
+
1
=
f
(
f
(
8
)
)
+
1
=
f
(
4
)
+
1
=
0
+
1
=
1
lf
g
(
f
(
x
)
)
=
|
sin
x
|
,
f
(
g
(
x
)
)
=
(
sin
√
x
)
2
, then
Report Question
0%
f
(
x
)
=
sin
2
x
,
g
(
x
)
=
√
x
0%
f
(
x
)
=
sin
x
,
g
(
x
)
=
|
x
|
0%
f
(
x
)
=
x
2
,
g
(
x
)
=
sin
√
x
0%
f
,
g
cannot be determined
Explanation
g
(
f
(
x
)
)
=
|
sin
x
|
=
√
sin
2
x
.......(i)
f
(
g
(
x
)
)
=
sin
2
√
x
.......(ii)
Comparing
i
and
i
i
⇒
g
(
f
(
x
)
)
=
|
sin
x
|
=
√
sin
2
x
⇒
g
(
sin
2
x
)
=
√
sin
2
x
=
g
(
f
(
x
)
)
And
⇒
f
(
g
(
x
)
)
=
sin
2
√
x
⇒
f
(
√
x
)
=
sin
2
√
x
=
f
(
g
(
x
)
)
Therefore
⇒
g
(
x
)
=
√
x
⇒
f
(
x
)
=
sin
2
x
lf
f
(
x
)
=
x
−
x
2
+
x
3
−
x
4
+
…
.
.
∞
when
|
x
|
<
1
, then the ascending order of the following is
a)
f
(
1
/
2
)
b)
f
−
1
(
1
/
2
)
c)
f
(
−
1
/
2
)
d)
f
−
1
(
−
1
/
2
)
Report Question
0%
a, b, c, d
0%
c, d, a, b
0%
b, a, d, c
0%
d, c, a, b
Explanation
f
(
x
)
=
x
−
x
2
+
x
3
−
x
4
f
(
x
)
=
x
1
+
x
(
x
)
<
1
f
(
1
2
)
=
1
3
f
(
−
1
2
)
=
−
1
f
(
x
)
=
x
1
+
x
y
=
x
1
+
x
y
+
x
y
=
x
y
(
u
−
1
)
=
x
f
−
1
x
=
−
x
x
−
1
f
−
1
(
1
2
)
=
1
f
−
1
(
−
1
2
)
=
−
1
3
Let
f
be an injective function with domain
{
x
,
y
,
z
}
and range
{
1
,
2
,
3
}
such that exactly one of the follwowing statements is correct and the remaining are false :
f
(
x
)
=
1
,
f
(
y
)
≠
1
,
f
(
z
)
≠
2
,
then the value of
f
−
1
(
1
)
is
Report Question
0%
x
0%
y
0%
z
0%
none
Explanation
It gives three cases
Case (1) When
f
(
x
)
=
1
is true.
In this case remaining two are false
f
(
y
)
=
1
and
f
(
z
)
=
2
This means
x
and
y
have the same image, so
f
(
x
)
is not an injective, which is a contradiction.
Case (2) when
f
(
y
)
≠
1
is true
If
f
(
y
)
≠
1
is true then the remaining statements are false.
∴
f
(
x
)
≠
1
and
f
(
z
)
=
2
i.e., both
x
and
y
are not mapped to
1.
So, either both associate to
2
or
3
,
Thus, it is not injective.
Case (3) When
f
(
z
)
≠
2
is true
If
f
(
z
)
≠
2
is true then remaining statements are fales
∴
If
f
(
x
)
≠
1
and
f
(
y
)
=
1
But
f
is injective
Thus, we have
f
(
x
)
=
2
,
f
(
y
)
=
1
and
f
(
z
)
=
3
Hence,
f
−
1
(
1
)
=
y
If
f
:
R
→
R
is defined by
f
(
x
)
=
2
x
−
2
,
then
(
f
∘
f
)
(
x
)
+
2
=
Report Question
0%
f
(
x
)
0%
2
f
(
x
)
0%
3
f
(
x
)
0%
−
f
(
x
)
Explanation
Let
f
:
R
→
R
is defined by
f
(
x
)
=
2
x
−
2
.
Then
(
f
∘
f
)
(
x
)
+
2
=
f
[
f
(
x
)
]
+
2
=
f
(
2
x
−
2
)
+
2
=
2
(
2
x
−
2
)
−
2
+
2
=
2
f
(
x
)
If
f
(
x
)
=
x
√
1
−
x
2
,
g
(
x
)
=
x
√
1
+
x
2
then
(
f
∘
g
)
(
x
)
=
Report Question
0%
x
√
1
−
x
2
0%
x
√
1
+
x
2
0%
1
−
x
2
√
1
−
x
2
0%
x
Explanation
Let
x
=
tan
θ
(
f
∘
g
)
(
x
)
=
f
[
g
(
x
)
]
=
f
[
g
(
tan
θ
)
]
.
Since
g
(
x
)
=
x
√
1
+
x
2
, we have
f
[
g
(
tan
θ
)
]
=
f
[
tan
θ
√
1
+
tan
2
θ
]
=
f
[
tan
θ
sec
θ
]
=
f
(
sin
θ
)
Also since
f
(
x
)
=
x
√
1
−
x
2
, we have
f
(
sin
θ
)
=
sin
θ
√
1
−
sin
2
θ
=
tan
θ
=
x
If
f
(
x
)
=
log
x
,
g
(
x
)
=
x
3
then
f
[
g
(
a
)
]
+
f
[
g
(
b
)
]
=
Report Question
0%
f
[
g
(
a
)
+
g
(
b
)
]
0%
f
[
g
(
a
b
)
]
0%
g
[
f
(
a
b
)
]
0%
g
[
f
(
a
)
+
f
(
b
)
]
Explanation
Given that
g
(
x
)
=
x
3
. Therefore,
f
[
g
(
a
)
]
+
f
[
g
(
b
)
]
=
f
(
a
3
)
+
f
(
b
3
)
.
Since
f
(
x
)
=
log
x
, we have
f
(
a
3
)
+
f
b
3
=
log
a
3
+
log
b
3
=
log
(
a
3
b
3
)
=
log
(
(
a
b
)
3
)
=
f
[
g
(
a
b
)
]
If
X
=
{
1
,
2
,
3
,
4
,
5
}
and
Y
=
{
1
,
3
,
5
,
7
,
9
}
, determine which of the following sets represent a relation and also a mapping.
Report Question
0%
R
1
=
{
(
x
,
y
)
:
y
=
x
+
2
,
x
∈
X
,
y
∈
Y
}
0%
R
2
=
{
(
1
,
1
)
,
(
2
,
1
)
,
(
3
,
3
)
,
(
4
,
3
)
,
(
5
,
5
)
}
0%
R
3
=
{
(
1
,
1
)
,
(
1
,
3
)
,
(
3
,
5
)
,
(
3
,
7
)
,
(
5
,
7
)
}
0%
R
4
=
{
(
1
,
3
)
,
(
2
,
5
)
,
(
4
,
7
)
,
(
5
,
9
)
,
(
3
,
1
)
}
Explanation
R
1
=
(
1
,
3
)
,
(
2
,
4
)
,
(
3
,
5
)
,
(
4
,
6
)
,
(
5
,
7
)
Since
4
and
6
do not belong to
Y
(
2
,
4
)
,
(
4
,
6
)
∉
R
1
R
1
=
(
1
,
3
)
,
(
3
,
5
)
,
(
5
,
7
)
⊂
A
×
B
Hence
R
1
is a relation but not a mapping as the elements
2
and
4
do not have any image.
R
2
: It is certainly a mapping and since every mapping is a relation, it is a relation as well.
R
3
: It is a relation being a subset of
A
×
B
but the elements
1
and
3
do not have a unique image and hence it is not mapping.
R
4
: It is both a mapping and a relation. Each element in A has a unique image. It is also one-one and onto mapping and hence a bijection
If
f
:
R
→
R
and
g
:
R
→
R
are defined by
f
(
x
)
=
2
x
+
3
and
g
(
x
)
=
x
2
+
7
, then the values of
x
such that
g
(
f
(
x
)
)
=
8
are:
Report Question
0%
1
,
2
0%
−
1
,
2
0%
−
1
,
−
2
0%
1
,
−
2
Explanation
Since
f
(
x
)
=
2
x
+
3
,
g
[
f
(
x
)
]
=
8
⇒
g
(
2
x
+
3
)
=
8
.
Also since
g
(
x
)
=
x
2
+
7
,
g
(
2
+
3
x
)
=
8
⇒
(
2
x
+
3
)
2
+
7
=
8
⇒
2
x
+
3
=
±
1
⇒
x
=
−
1
or
−
2
.
If
f
(
x
)
=
2
x
+
5
x
2
+
x
+
5
, then
f
[
f
(
−
1
)
]
is equal to
Report Question
0%
149
155
0%
155
147
0%
155
149
0%
147
155
Explanation
Given expression is
f
(
x
)
=
2
x
+
5
x
2
+
x
+
5
∴
f
(
−
1
)
=
2
×
(
−
1
)
+
5
(
−
1
)
2
+
(
−
1
)
+
5
=
3
5
∴
f
(
f
(
−
1
)
)
=
2
×
3
5
+
5
(
3
5
)
2
+
3
5
+
5
=
155
149
Set A has 3 elements and set B has 4 elements. The number of injections that can be defined from A into B is :
Report Question
0%
144
0%
12
0%
24
0%
64
Explanation
If
A
B
C
∼
P
Q
R
, then AB:PQ =
Report Question
0%
A
C
:
P
B
0%
A
C
:
P
R
0%
A
B
:
P
R
0%
A
C
:
R
Q
If
f
:
R
→
R
and
g
:
R
→
R
are defined by
f
(
x
)
=
x
−
[
x
]
and
g
(
x
)
=
[
x
]
for
x
∈
R
, where
[
x
]
is the greatest integer not exceeding
x
, then for every
x
∈
R
,
f
(
g
(
x
)
)
=
Report Question
0%
x
0%
0
0%
f
(
x
)
0%
g
(
x
)
Explanation
Since
f
(
x
)
=
x
−
[
x
]
, we have
f
[
g
(
x
)
]
=
g
(
x
)
−
[
g
(
x
)
]
Also since
g
(
x
)
=
[
x
]
,we have
g
(
x
)
−
[
g
(
x
)
]
=
[
x
]
−
[
[
x
]
]
.
Here,
[
x
]
is the greatest integer not exceeding
x
. Therefore,
[
[
x
]
]
=
[
x
]
and hence,
[
x
]
−
[
[
x
]
]
=
[
x
]
−
[
x
]
=
0
If
y
=
f
(
x
)
=
2
x
−
1
x
−
2
, then
f
(
y
)
=
Report Question
0%
x
0%
y
0%
2
y
−
1
0%
y
−
2
Explanation
f
(
y
)
=
2
y
−
1
y
−
2
=
2
(
2
x
−
1
x
−
2
)
−
1
(
2
x
−
1
x
−
2
)
−
2
=
4
x
−
2
−
x
+
2
2
x
−
1
−
2
x
+
4
=
3
x
3
=
x
If
f
(
g
(
x
)
)
is one-one function, then
Report Question
0%
g(x) must be one-one
0%
f(x) must be one-one
0%
f(x) may not be one-one
0%
g(x) may not be one-one
Explanation
It is fundamental concept that, If
f
(
g
(
x
)
)
is one-one function then,
f
(
x
)
must be one-one function and
g
(
x
)
may be one-one or many one
Which of the following functions are one-one?
Report Question
0%
f
:
R
→
R
given by
f
(
x
)
=
2
x
2
+
1
for all
x
∈
R
0%
g
:
Z
→
Z
given by
g
(
x
)
=
x
4
for all
x
∈
R
0%
h
:
R
→
R
given by
h
(
x
)
=
x
3
+
4
for all
x
∈
R
0%
ϕ
:
C
→
C
given
ϕ
(
z
)
=
2
z
6
+
4
for all
x
∈
R
Explanation
For all even powered function,
f
(
x
1
)
=
f
(
x
2
)
where
x
1
=
±
(
x
2
)
Hence, the only possible one-one function is
f
(
x
)
=
x
3
+
4
f
(
x
)
is a strictly increasing function. So, it is one-one.
A mapping function
f
:
X
→
Y
is one-one, if
Report Question
0%
f
(
x
1
)
≠
f
(
x
2
)
for all
x
1
,
x
2
∈
X
0%
f
(
x
1
)
=
f
(
x
2
)
⇒
x
1
=
x
2
for all
x
1
,
x
2
∈
X
0%
x
1
=
x
2
⇒
f
(
x
1
)
=
f
(
x
2
)
for all
x
1
,
x
2
∈
X
0%
none of these
Explanation
For a function to be one one
f
(
x
1
)
=
f
(
x
2
)
Implies that
x
1
=
x
2
Hence the answer is
option B
Let
f
:
R
→
A
=
{
y
:
0
≤
y
<
π
2
}
be a function such that
f
(
x
)
=
tan
−
1
(
x
2
+
x
+
k
)
,
where
k
is a constant. The value of
k
for which
f
is an onto function is
Report Question
0%
1
0%
0
0%
1
4
0%
none of these
Explanation
For
f
to be an onto function, Range and co-domain of
f
should be equal
⇒
f
(
x
)
≥
0
∀
x
∈
R
⇒
tan
−
1
(
x
2
+
x
+
k
)
≥
0
For the above equation to be valid for all
x
We must have, the discriminant of
x
2
+
x
+
k
=
0
, is zero
b
2
−
4
a
c
=
0
⇒
1
−
4
k
=
0
⇒
k
=
1
4
If
f
:
R
→
R
given by
f
(
x
)
=
x
3
+
(
a
+
2
)
x
2
+
3
a
x
+
5
is one-one, then
a
belongs to the interval
Report Question
0%
(
−
∞
,
1
)
0%
(
1
,
∞
)
0%
(
1
,
4
)
0%
(
4
,
∞
)
Explanation
If
f
(
x
)
is one one then
f
′
(
x
)
should be either positive or negative for all
x
.
f
′
(
x
)
=
3
x
2
+
2
(
a
+
2
)
x
+
3
a
⇒
D
=
4
(
a
+
2
)
2
−
4
×
3
×
3
a
<
0
⇒
a
2
−
5
a
+
4
<
0
⇒
(
a
−
1
)
(
a
−
4
)
<
0
Hence,
a
∈
(
1
,
4
)
Let R be the relation in the set N given by
=
{
(
a
,
b
)
:
a
=
b
−
2
,
b
>
6
}
. Choose the correct answer.
Report Question
0%
(
2
,
4
)
∈
R
0%
(
3
,
8
)
∈
R
0%
(
6
,
8
)
∈
R
0%
(
8
,
7
)
∈
R
Explanation
(
a
,
b
)
∈
R
only if
a
=
b
−
2
and
b
>
6
(
6
,
8
)
∈
R
as
6
=
8
−
2
and
8
>
6
Hence
(
c
)
is the correct alternative
Let
f
:
{
x
,
y
,
z
}
→
{
a
,
b
,
c
}
be a one-one function and only one of the conditions
(
i
)
f
(
x
)
≠
b
,
(
i
i
)
f
(
y
)
=
b
,
(
i
i
i
)
f
(
z
)
≠
a
is true then the function
f
is given by the set
Report Question
0%
{
(
x
,
a
)
,
(
y
,
b
)
,
(
z
,
c
)
}
0%
{
(
x
,
a
)
,
(
y
,
c
)
,
(
z
,
b
)
}
0%
{
(
x
,
b
)
,
(
y
,
a
)
,
(
z
,
c
)
}
0%
{
(
x
,
c
)
,
(
y
,
b
)
,
(
z
,
a
)
}
Explanation
f
:
{
x
,
y
,
z
}
→
{
a
,
b
,
c
}
is a one-one function
⇒
each element in
{
x
,
y
,
z
}
will have exactly one image in
{
a
,
b
,
c
}
and no two elements of
{
x
,
y
,
z
}
will have same image in
{
a
,
b
,
c
}
Coming to the given 3 conditions, only one is true.
1) if
f
(
x
)
≠
b
is true then
f
(
y
)
=
b
is false which makes
f
(
z
)
≠
a
true
⟹
f
(
x
)
≠
b
is false.
2) if
f
(
y
)
=
b
is true then
f
(
x
)
≠
b
will also be true
⟹
f
(
y
)
=
b
is false
∴
f
(
z
)
≠
a
is the true condition and remainig two are false conditions.
∴
f
(
x
)
=
b
,
f
(
y
)
=
a
,
f
(
z
)
=
c
hence
f
=
{
(
x
,
b
)
,
(
y
,
a
)
,
(
z
,
c
)
}
Which of the following function is one-one?
Report Question
0%
f
:
R
→
R
given by
f
(
x
)
=
|
x
−
1
|
for all
x
∈
R
0%
g
:
[
−
π
2
,
π
2
]
→
R
given by
g
(
x
)
=
|
s
i
n
x
|
for all
x
∈
[
−
π
2
,
π
2
]
0%
h
:
[
−
π
2
,
π
2
]
∈
R
given by
h
(
x
)
=
s
i
n
x
for all
x
∈
[
−
π
2
,
π
2
]
0%
ϕ
:
R
→
R
given by
f
(
x
)
=
x
2
−
4
for all
x
∈
R
Explanation
Option A - Consider
x
=
2
and
x
=
0
, the value of
f
(
x
)
is same. Hence it is not one-one
Option B - If we replace
x
by
−
x
, then the value of
g
(
x
)
remains the same. Hence it is not one-one
Option C -
h
(
x
)
is an increasing function for the given values of
x
. Hence it is one-one function
Option D -
f
(
x
)
is an even function. So it is not one-one for the given values of
x
If
f
and
g
are one-one functions from
R
→
R
, then
Report Question
0%
f
+
g
is one-one
0%
f
g
is one-one
0%
f
o
g
is one-one
0%
none of these
Explanation
Let
x
1
,
x
2
∈
R
be two distinct elements, then
g
(
x
1
)
≠
g
(
x
2
)
, as
g
is one-one function. Similarly
f
(
g
(
x
1
)
)
≠
f
(
g
(
x
2
)
)
as
f
is also one-one function.
Hence,
f
∘
g
is one-one function.
Note that
f
+
g
and
f
⋅
g
may not one-one functions even if
f
and
g
are one one. For example consider
f
(
x
)
=
x
and
g
(
x
)
=
−
x
, then
f
+
g
and
f
⋅
g
are not one-one.
If
f
:
R
+
→
A
, Where
A
=
{
x
:
−
5
<
x
<
∞
}
be defined by
f
(
x
)
=
x
2
−
5
. Then
f
−
1
(
7
)
=
Report Question
0%
only
−
2
√
3
0%
only
2
√
3
0%
both options A and B
0%
none
Explanation
The value
f
−
1
(
7
)
will be the roots of the equation
x
2
−
5
=
7
⇒
x
2
=
12
⇒
x
=
±
2
√
3
But only positive.
∴
x
=
2
√
3
If the function
f
:
[
1
,
∞
)
→
[
1
,
∞
)
is defined by
f
(
x
)
=
3
x
(
x
−
1
)
then
f
−
1
(
x
)
is
Report Question
0%
(
1
2
)
x
(
x
−
1
)
0%
1
2
(
1
+
√
1
+
4
log
3
x
)
0%
1
2
(
1
−
√
1
+
4
log
3
x
)
0%
not defined
Explanation
For the given domain, the function is one one.
Thus
,
f
(
x
)
=
2
x
(
x
−
1
)
y
=
2
x
2
−
x
l
n
y
=
(
x
2
−
x
)
l
n
2
l
o
g
2
y
=
x
2
−
x
l
o
g
2
y
=
(
x
−
1
2
)
2
−
1
4
l
o
g
2
y
+
1
4
=
(
x
−
1
2
)
2
±
1
2
(
√
1
+
4
l
o
g
2
y
)
=
x
−
1
2
Since
f
:
[
1
,
∞
)
→
[
1
,
∞
)
Thus domain is
[
1
,
∞
)
Hence domain is positive so negative sign will be ignored.
S
o
x
−
1
2
=
1
2
(
√
1
+
4
l
o
g
2
y
)
x
=
1
2
[
1
+
√
1
+
4
l
o
g
2
y
]
Replacing x and y, gives us
f
−
1
(
x
)
=
1
2
[
1
+
√
1
+
4
l
o
g
2
x
]
If the function
f
:
R
→
R
be such that
f
(
x
)
=
x
−
[
x
]
,
where
[
y
]
denotes the greatest integer less than or equal to
y
, then
f
−
1
(
x
)
is
Report Question
0%
1
x
−
[
x
]
0%
[
x
]
−
x
0%
not defined
0%
none of these
Explanation
f
(
x
)
=
x
−
[
x
]
=
{
x
}
is not one-one function from R to R. Hence
f
−
1
(
x
)
can not be defined.
If
f
:
(
3
,
4
)
→
(
0
,
1
)
is defined by
f
(
x
)
=
x
−
[
x
]
where
[
x
]
denotes the greatest integer function then
f
−
1
(
x
)
is
Report Question
0%
1
x
−
[
x
]
0%
[
x
]
−
x
0%
x
−
3
0%
x
+
3
Explanation
Note that
f
(
x
)
=
x
−
[
x
]
=
{
x
}
. Hence every number
x
∈
(
3
,
4
)
is mapped to its fractional part.
Hence, the inverse of the function can be written by adding the integer part of the number i.e.,
3
to each of the fraction part.
Hence the function
f
−
1
(
x
)
=
x
+
3
.
Let
f
:
(
−
∞
,
1
]
→
(
−
∞
,
1
]
such that
f
(
x
)
=
x
(
2
−
x
)
.
Then
f
−
1
(
x
)
is
Report Question
0%
1
+
√
1
−
x
0%
1
−
√
1
−
x
0%
√
1
−
x
0%
none of these
Explanation
We have,
f
(
x
)
=
x
(
2
−
x
)
y
=
x
(
2
−
x
)
−
y
=
x
(
x
−
2
)
−
y
=
x
2
−
2
x
1
−
y
=
x
2
−
2
x
+
1
1
−
y
=
(
x
−
1
)
2
±
√
1
−
y
=
x
−
1
Now
f
:
(
−
∞
,
1
]
→
(
−
∞
,
1
]
x
−
1
=
−
√
1
−
y
x
=
1
−
√
1
−
y
Hence
f
−
1
(
x
)
=
1
−
√
1
−
x
If
f
(
x
)
=
a
x
+
b
and
g
(
x
)
=
c
x
+
d
, then
f
(
g
(
x
)
)
=
g
(
f
(
x
)
)
implies
Report Question
0%
f
(
a
)
=
g
(
c
)
0%
f
(
b
)
=
g
(
b
)
0%
f
(
d
)
=
g
(
b
)
0%
f
(
c
)
=
g
(
a
)
Explanation
f
(
g
(
x
)
)
=
a
(
c
x
+
d
)
+
b
=
a
c
x
+
a
d
+
b
...(i)
g
(
f
(
x
)
)
=
c
(
a
x
+
b
)
+
d
=
a
c
(
x
)
+
b
c
+
d
...(ii)
f
(
g
(
x
)
)
=
g
(
f
(
x
)
)
From (i) and (ii)
a
c
(
x
)
+
b
c
+
d
=
a
c
x
+
a
d
+
b
c
b
+
d
=
a
d
+
b
g
(
b
)
=
f
(
d
)
If
f
(
x
)
=
1
1
−
x
,
x
≠
0
,
1
then the graph of the function
y
=
f
{
f
(
f
(
x
)
)
}
,
x
>
1
,
is
Report Question
0%
a circle
0%
an ellipse
0%
a straight line
0%
a pair of straight lines
Explanation
f
(
f
(
x
)
)
=
1
1
−
1
1
−
x
=
−
1
−
x
x
=
g
(
x
)
f
(
g
(
x
)
)
=
1
1
+
1
−
x
x
=
x
x
+
1
−
x
=
x
y
=
x
is an equation of straight line.
Hence, 'C' is correct.
If
f
(
x
)
=
3
x
−
5
then
f
−
1
(
x
)
=
Report Question
0%
1
3
x
−
5
0%
x
+
5
3
0%
does not exist because
f
is not one-one
0%
does not exist because
f
is not onto
Explanation
f
(
x
)
=
3
x
−
5
Since
f
(
x
)
is a purely linear function (straight line), it is therefore a one-one function.
Hence
f
(
x
)
is revertible for all real
x
.
y
=
3
x
−
5
y
+
5
=
3
x
x
=
y
+
5
3
Replacing
x
by
y
, gives us
f
−
1
(
x
)
=
x
+
5
3
If
f
and
g
are two functions such that
(
f
g
)
(
x
)
=
(
g
f
)
(
x
)
for all
x
. Then
f
and
g
may be defined as
Report Question
0%
f
(
x
)
=
√
x
,
g
(
x
)
=
cos
x
0%
f
(
x
)
=
x
3
,
g
(
x
)
=
x
+
1
0%
f
(
x
)
=
x
−
1
,
g
(
x
)
=
x
2
+
1
0%
f
(
x
)
=
x
m
,
g
(
x
)
=
x
n
where
m
,
n
are unequal integers
Explanation
A
)
f
(
x
)
=
√
x
,
g
(
x
)
=
c
o
s
x
t
h
e
n
(
f
g
)
(
x
)
=
√
cos
x
a
n
d
(
g
f
)
(
x
)
=
cos
√
x
.They are not equal.
B)
f
(
x
)
=
x
3
,
g
(
x
)
=
x
+
1
t
h
e
n
(
f
g
)
(
x
)
=
(
x
+
1
)
3
a
n
d
(
g
f
)
(
x
)
=
x
3
+
1
.They are not equal.
C)
f
(
x
)
=
x
1
,
g
(
x
)
=
x
2
+
1
t
h
e
n
(
f
g
)
(
x
)
=
x
2
a
n
d
(
g
f
)
(
x
)
=
x
2
−
2
x
. They are not equal.
D)
f
(
x
)
=
x
m
,
g
(
x
)
=
x
n
t
h
e
n
(
f
g
)
(
x
)
=
x
n
+
m
a
n
d
(
g
f
)
(
x
)
=
x
n
+
m
.Hence
(
f
g
)
(
x
)
=
(
g
f
)
(
x
)
If
f
(
x
)
=
x
n
,
n
∈
N
and
(
g
o
f
)
(
x
)
=
n
g
(
x
)
then
g
(
x
)
can be
Report Question
0%
n
|
x
|
0%
3.
3
√
x
0%
e
x
0%
log
|
x
|
Explanation
f
(
x
)
=
x
n
g
(
f
(
x
)
)
=
n
g
(
x
)
...
(
i
)
log
(
f
(
x
)
)
=
n
log
(
x
)
...
(
i
i
)
Taking
log
(
|
x
|
)
as
g
(
x
)
the above expression is reduced to eq
i
.
Hence
g
(
x
)
=
log
(
|
x
|
)
.
The composite mapping
f
o
g
of the map
f
:
R
→
R
,
f
(
x
)
=
sin
x
and
g
:
R
→
R
,
g
(
x
)
=
x
2
is
Report Question
0%
x
2
sin
x
0%
(
sin
x
)
2
0%
sin
x
2
0%
sin
x
x
2
Explanation
Composite mapping will be
f
(
g
(
x
)
)
=
f
(
x
2
)
=
sin
(
x
2
)
Hence option
′
C
′
is the answer.
Let
S
be a set containing
n
elements. Then the total number of binary operations on
S
is
Report Question
0%
n
n
0%
2
n
2
0%
n
n
2
0%
n
2
Explanation
No of binary operations imply
S
×
S
→
S
=
n
(
S
)
[
n
(
S
×
S
)
]
=
n
n
×
n
=
n
n
2
Let
f
:
R
→
R
be defined by
f
(
x
)
=
3
x
−
4
then
f
−
1
(
x
)
is
Report Question
0%
1
3
(
x
+
4
)
0%
1
3
(
x
−
4
)
0%
3
x
+
4
0%
not defined
Explanation
f
(
x
)
is invertible for all real values of
x
Hence
y
=
3
x
−
4
y
+
4
=
3
x
y
+
4
3
=
x
f
−
1
(
x
)
=
x
+
4
3
Let
f
(
x
)
=
a
x
x
+
1
, where
x
≠
−
1
. Then for what value of
a
is
f
(
f
(
x
)
)
=
x
always true
Report Question
0%
√
2
0%
−
√
2
0%
1
0%
−
1
Explanation
f
(
f
(
x
)
)
=
a
a
x
x
+
1
a
x
x
+
1
+
1
=
a
2
x
x
+
1
a
x
+
x
+
1
x
+
1
=
a
2
x
a
x
+
x
+
1
Since,
f
(
f
(
x
)
)
=
x
, we have,
a
2
x
a
x
+
x
+
1
=
x
.
Simplifying the equation we get,
a
2
x
=
(
a
+
1
)
x
2
+
x
∴
(
a
+
1
)
x
2
+
(
1
−
a
2
)
x
=
0
or
(
a
+
1
)
x
(
x
+
1
−
a
)
=
0
Hence the only possible value is
a
=
−
1
A function
y
=
f
(
x
)
is invertible only when
Report Question
0%
y
=
f
(
x
)
is monotonic increasing
0%
y
=
f
(
x
)
is bijective
0%
y
=
f
(
x
)
is monotonic decreasing
0%
invertible
Explanation
Invertible function is defined as, the function
f
applied to an input
x
gives a result of
y
, then applying its inverse function
g
to
y
gives the result
x
, and vice versa. i.e.,
f
(
x
)
=
y
⇒
g
(
y
)
=
x
.
And bijective function has the same definition as that of an Invertible function
Let
f
:
R
→
R
and
g
:
R
→
R
be defined by
f
(
x
)
=
x
2
+
2
x
−
3
,
g
(
x
)
=
3
x
−
4
then
(
g
o
f
)
(
x
)
=
Report Question
0%
3
x
2
+
6
x
−
13
0%
3
x
2
−
6
x
−
13
0%
3
x
2
+
6
x
+
13
0%
−
3
x
2
+
6
x
−
13
Explanation
Given that,
f
(
x
)
=
x
2
+
2
x
−
3
and
g
(
x
)
=
3
x
−
4
(
g
o
f
)
(
x
)
=
g
(
f
(
x
)
)
=
3
(
x
2
+
2
x
−
3
)
−
4
(
g
o
f
)
(
x
)
=
3
x
2
+
6
x
−
13
Hence, option A.
If
f
:
[
1
,
+
∞
]
→
[
2
,
+
∞
)
is given by
f
(
x
)
=
x
+
1
x
then
f
−
1
(
x
)
equals
Report Question
0%
x
+
√
x
2
+
4
2
0%
x
1
+
x
2
0%
x
+
√
x
2
−
4
2
0%
1
+
√
x
2
−
4
Explanation
Let
y
=
x
+
1
x
⇒
x
2
−
y
x
+
1
=
0
⇒
x
=
y
+
√
y
2
−
4
2
=
f
−
1
(
y
)
Hence
f
−
1
(
x
)
=
x
+
√
x
2
−
4
2
Note: -ve term in step two is discarded using given domain and co-domain.
f
:
R
→
R
is a function defined by
f
(
x
)
=
10
x
−
7
. If
g
=
f
−
1
, then
g
(
x
)
is equals
Report Question
0%
1
10
x
−
7
0%
1
10
x
+
7
0%
x
+
7
10
0%
x
−
7
10
Explanation
y
=
10
x
−
7
It is a one one function over the given domain.
y
+
7
=
10
(
x
)
x
=
y
+
7
10
Hence
f
−
1
will be
g
(
x
)
g
(
x
)
=
x
+
7
10
Hence, option 'C' is correct.
Set
A
has
3
elements and set
B
has
4
elements. The number of injections that can be defined from
A
to
B
is
Report Question
0%
144
0%
12
0%
24
0%
64
Explanation
If
f
:
R
→
R
such that
f
(
x
)
=
l
o
g
3
x
then
f
−
1
x
is equal to
Report Question
0%
l
o
g
x
3
0%
3
x
0%
3
−
x
0%
3
1
/
x
Explanation
f
(
x
)
is invertible for all positive
x
.
y
=
l
o
g
3
(
x
)
Taking antilogarithm on both sides, we get
3
y
=
x
.
Hence,
f
−
1
(
x
)
=
3
x
Hence, option 'B' is correct.
The inverse of
f
(
x
)
=
e
3
x
−
e
−
3
x
e
3
x
+
e
−
3
x
is
Report Question
0%
1
6
log
10
(
1
+
x
1
−
x
)
0%
1
6
log
10
(
x
1
−
x
)
0%
1
6
log
e
(
1
+
x
1
−
x
)
0%
1
6
log
e
(
1
−
x
1
+
x
)
Explanation
y
=
e
3
x
−
e
−
3
x
e
3
x
+
e
−
3
x
⇒
y
=
e
6
x
−
1
e
6
x
+
1
⇒
y
+
1
y
−
1
=
e
6
x
−
1
+
e
6
x
+
1
e
6
x
−
1
−
e
6
x
−
1
⇒
y
+
1
y
−
1
=
2
e
6
x
−
2
⇒
1
+
y
1
−
y
=
e
6
x
Taking
log
e
we get
log
e
(
1
+
y
1
−
y
)
=
6
x
x
=
1
6
(
log
e
(
1
+
y
1
−
y
)
)
Hence
f
−
1
=
1
6
(
log
e
(
1
+
x
1
−
x
)
)
If
X
=
{
1
,
2
,
3
,
4
,
5
}
and
Y
=
{
1
,
3
,
5
,
7
,
9
}
then which of the following sets are relation from
X
to
Y
Report Question
0%
R
1
=
{
(
x
,
a
)
:
a
=
x
+
2
,
x
∈
X
,
a
∈
Y
}
0%
R
2
=
{
(
1
,
1
)
,
(
2
,
1
)
,
(
3
,
3
)
,
(
4
,
3
)
,
(
5
,
5
)
}
0%
R
3
=
{
(
1
,
1
)
,
(
1
,
3
)
,
(
3
,
5
)
,
(
3
,
7
)
,
(
5
,
7
)
}
0%
R
4
=
{
(
1
,
3
)
,
(
2
,
5
)
,
(
4
,
7
)
,
(
5
,
9
)
,
(
3
,
1
)
}
Explanation
R
1
=
{
(
1
,
3
)
,
(
2
,
4
)
,
(
3
,
5
)
,
(
4
,
6
)
,
(
5
,
7
)
}
since
4
and
6
do not belong to
y
.
∴
(
2
,
4
)
,
(
4
,
6
)
∉
R
1
∴
R
1
=
{
(
1
,
3
)
,
(
3
,
5
)
,
(
5
,
7
)
}
⊂
X
×
Y
Hence,
R
1
is a relation but not a mapping as the elements
2
and
4
do not have any image.
R
2
:
It is certainly a mapping and since every mapping is a relation, it is a relation as well.
R
3
:
It is a relation being a subset of
X
×
Y
.
R
4
:
It is both mapping and a relation. Each element in
X
has a unique image. It is also one-one and on-to mapping and hence a bijection.
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Practice Class 12 Commerce Maths Quiz Questions and Answers
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