Processing math: 63%
MCQExams
0:0:1
CBSE
JEE
NTSE
NEET
Practice
Homework
×
CBSE Questions for Class 12 Commerce Maths Relations And Functions Quiz 7 - MCQExams.com
CBSE
Class 12 Commerce Maths
Relations And Functions
Quiz 7
If
f
(
x
)
=
a
x
+
b
and
g
(
x
)
=
c
x
+
d
, then
f
(
g
(
x
)
)
=
g
(
f
(
x
)
)
⇔
Report Question
0%
f
(
a
)
=
g
(
c
)
0%
f
(
b
)
=
g
(
b
)
0%
f
(
d
)
=
g
(
b
)
0%
f
(
x
)
=
g
(
a
)
Explanation
We have
f
(
x
)
=
a
x
+
b
,
g
(
x
)
=
c
x
+
d
Therefore,
f
{
g
(
x
)
}
=
g
{
f
(
x
)
}
⇔
f
(
c
x
+
d
)
=
g
(
a
x
+
b
)
⇔
a
(
c
x
+
d
)
+
b
=
c
(
a
x
+
b
)
+
d
⇔
a
d
+
b
=
c
b
+
d
⇔
f
(
d
)
=
g
(
b
)
The inverse of the function
f
(
x
)
=
log
(
x
2
+
3
x
+
1
)
,
x
ϵ
[
1
,
3
]
, assuming it to be an onto function, is
Report Question
0%
−
3
+
√
5
+
4
e
x
2
0%
−
3
±
√
5
+
4
e
x
2
0%
−
3
−
√
5
+
4
e
x
2
0%
None of the above
Explanation
Given,
f
(
x
)
=
log
(
x
2
+
3
x
+
1
)
Therefore,
f
′
(
x
)
=
2
x
+
3
(
x
2
+
3
x
+
1
)
>
0
∀
x
ϵ
[
1
,
3
]
which is a strictly increasing function. Thus,
f
(
x
)
is injective, given that
f
(
x
)
is onto. Hence, the given function
f
(
x
)
is invertible.
Now,
f
{
f
−
1
(
x
)
}
=
x
⇒
log
{
f
−
1
(
x
)
}
2
+
3
{
f
−
1
(
x
)
+
1
}
=
x
⇒
{
f
−
1
(
x
)
}
2
+
3
{
f
−
1
(
x
)
}
+
1
−
e
x
=
0
Thus
f
−
1
(
x
)
=
−
3
±
√
9
−
4.1
(
1
−
e
x
)
2
=
−
3
±
√
5
+
4
e
x
2
=
f
−
1
(
x
)
=
−
3
±
√
5
+
4
e
x
2
[
∵
f
−
1
(
x
)
ϵ
[
1
,
3
]
]
Hence,
f
−
1
(
x
)
=
−
3
+
√
5
+
4
e
x
2
Let
f
(
x
)
=
x
3
−
3
x
+
1
. The number of different real solutions of
f
(
f
(
x
)
)
=
0
Report Question
0%
2
0%
4
0%
5
0%
7
If
f
(
x
)
and
g
(
x
)
are two functions with
g
(
x
)
=
x
−
1
x
and
f
∘
g
(
x
)
=
x
3
−
1
x
3
, then
f
′
(
x
)
is equal to
Report Question
0%
3
x
2
+
3
0%
x
2
−
1
x
2
0%
1
+
1
x
2
0%
3
x
2
+
3
x
4
Explanation
f
∘
g
(
x
)
=
x
3
−
1
x
3
Writing
x
3
−
1
x
3
using
(
a
−
b
)
3
=
a
3
−
b
3
−
3
a
b
(
a
−
b
)
, we have
(
x
−
1
x
)
3
=
x
3
−
1
x
3
−
3
x
⋅
1
x
(
x
−
1
x
)
⇒
x
3
−
1
x
3
=
(
x
−
1
x
)
3
+
3
(
x
−
1
x
)
We have,
f
(
g
(
x
)
)
=
x
3
−
1
x
3
=
(
x
−
1
x
)
3
+
3
(
x
−
1
x
)
As
g
(
x
)
=
x
−
1
x
, this yields
f
(
x
−
1
x
)
=
(
x
−
1
x
)
3
+
3
(
x
−
1
x
)
On putting
x
−
1
x
=
t
, we get
f
(
t
)
=
t
3
+
3
t
Thus,
f
(
x
)
=
x
3
+
3
x
and
f
′
(
x
)
=
3
x
2
+
3
If
f
:
R
→
R
,
g
:
R
→
R
are defined by
f
(
x
)
=
5
x
−
3
,
g
(
x
)
=
x
2
+
3
, then
(
g
o
f
−
1
)
(
3
)
=
Report Question
0%
25
3
0%
111
25
0%
9
25
0%
25
111
Explanation
f
:
R
→
R
g
:
R
⟶
R
f
(
x
)
=
5
x
−
3
g
(
x
)
=
x
2
+
3
(
g
o
f
−
1
)
(
3
)
f
(
x
)
=
5
x
−
3
Let
f
(
x
)
=
y
⇒
x
=
f
−
1
(
y
)
f
(
x
)
=
y
=
5
x
−
3
⇒
x
=
y
+
3
5
⇒
f
−
1
(
y
)
=
y
+
3
5
f
−
1
(
x
)
=
x
+
3
5
and
g
(
x
)
=
x
2
+
3
(
g
o
f
−
1
)
(
3
)
=
g
[
f
−
1
(
3
)
]
=
g
(
3
+
3
5
)
=
g
(
6
5
)
=
[
(
6
5
)
2
+
3
]
=
36
25
+
3
=
111
25
Hence,
111
25
is the correct answer.
The inverse of the function
y
=
5
ln
x
is
Report Question
0%
x
=
y
1
ln
5
,
y
>
0
0%
x
=
y
ln
5
,
y
>
0
0%
x
=
y
1
ln
5
,
y
<
0
0%
x
=
5
ln
y
,
y
>
0
Explanation
Given that
y
=
5
ln
x
ln
y
=
ln
(
5
ln
x
)
=
(
ln
x
)
(
ln
5
)
∴
ln
x
=
ln
y
ln
5
x
=
e
ln
y
ln
5
=
(
e
ln
y
)
1
ln
5
=
y
1
ln
5
Since the domain of the natural log function ln() is the set of positive real numbers, it is required here that
y
>
0
.
The inverse function:
x
=
y
1
ln
5
,
y
>
0
If
f
(
x
)
=
2
x
3
+
7
x
−
5
then
f
−
1
(
4
)
is
Report Question
0%
Equal to
1
0%
Equal to
2
0%
Equal to
1
/
3
0%
Non existent
Explanation
f
(
x
)
=
2
x
3
+
7
x
−
5
x
=
2
(
f
−
1
(
x
)
3
)
+
7
f
−
1
(
x
)
−
5
f
−
1
(
4
)
=
?
4
=
2
(
f
−
1
(
4
)
3
)
+
7
f
−
1
(
4
)
−
5
2
(
f
−
1
(
4
)
3
)
+
7
f
−
1
(
4
)
−
9
=
0
Let
f
−
1
(
4
)
=
t
2
t
3
+
7
t
−
9
=
0
2
t
3
−
2
+
7
t
−
7
=
0
2
(
t
3
−
1
)
+
7
(
t
−
1
)
=
0
2
(
t
−
1
)
(
t
2
+
t
+
1
)
+
7
t
−
1
=
0
(
t
−
1
)
(
2
t
2
+
2
t
+
2
+
7
)
=
0
t
=
1
2
t
2
+
2
t
+
9
=
0
So,
f
−
1
(
4
)
=
1
Answer A
f
,
g
:
R
→
R
are functions such that
f
(
x
)
=
3
x
−
sin
(
π
x
2
)
,
g
(
x
)
=
x
3
+
2
x
−
sin
(
π
x
2
)
The value of
d
d
x
f
−
1
(
g
−
1
(
x
)
)
x
=
12
is equal to
Report Question
0%
2
30
+
x
0%
2
30
−
x
0%
2
3
(
28
−
π
)
0%
2
3
(
28
+
π
)
Explanation
Given that,
g
:
R
→
R
f
(
x
)
=
3
x
−
sin
(
π
x
2
)
g
(
x
)
=
x
3
+
2
x
−
sin
(
π
x
2
)
∴
d
f
−
1
d
x
(
g
′
(
x
)
)
x
=
12
=
2
30
+
x
Hence, the option
(
A
)
is correct.
If
f
:
R
→
R
and
g
:
R
→
R
are defined
f
(
x
)
=
x
−
[
x
]
and
g
(
x
)
=
[
x
]
∀
x
ϵ
R
,
f
(
g
(
x
)
)
.
Report Question
0%
x
0%
0
0%
f
(
x
)
0%
g
(
x
)
Explanation
We have,
f
(
x
)
=
x
−
[
x
]
=
{
x
}
…
(
1
)
where
{
x
}
is the fractional part of
x
If
g
(
x
)
=
[
x
]
, where
[
x
]
is the greatest integer part of
x
, then
The value of
g
(
x
)
will always be an integer.
And
∴
f
(
g
(
x
)
)
=
f
(
[
x
]
)
=
0
∵
the fractional part of
g
(
x
)
i.e
[
x
]
is always
0
.
Let
f
:
A
→
B
be a function defined as
f
(
x
)
=
x
−
1
x
−
2
, where
A
=
R
−
{
2
}
and
B
=
R
−
{
1
}
. Then
f
is :
Report Question
0%
invertible and
f
−
1
(
y
)
=
2
y
+
1
y
−
1
0%
invertible and
f
−
1
(
y
)
=
3
y
−
1
y
−
1
0%
not invertible
0%
invertible and
f
−
1
(
y
)
=
2
y
−
1
y
−
1
Explanation
Let
y
=
f
(
x
)
⇒
y
=
x
−
1
x
−
2
⇒
y
x
−
2
y
=
x
−
1
⇒
(
y
−
1
)
x
=
2
y
−
1
⇒
x
=
f
−
1
(
y
)
=
2
y
−
1
y
−
1
So on the given domain the function is invertible and its inverse can be computed as shown above.
So, option D is the correct answer.
If
f
(
x
)
is a real valued function, then which of the following is one-one function?
Report Question
0%
f
(
x
)
=
e
|
x
|
0%
f
(
x
)
=
|
e
x
|
0%
f
(
x
)
=
sin
x
0%
f
(
x
)
=
|
sin
x
|
Explanation
(
1
)
f
(
x
)
=
e
|
x
|
Since
f
(
x
)
=
f
(
−
x
)
So
f
(
x
)
is many one.
(
2
)
f
(
x
)
=
|
e
x
|
so,
f
(
x
)
=
e
x
and as
e
x
is one-one
So
f
(
x
)
is one-one.
(
3
)
f
(
x
)
=
sin
x
Since
sin
x
is many one
So
f
(
x
)
is also many-one.
(
4
)
f
(
x
)
=
|
sin
x
|
Since
sin
x
is many one
So
f
(
x
)
is also many-one.
If
A
=
{
1
,
2
,
3
}
and
B
=
{
4
,
5
}
then the number of function
f
:
A
→
B
which is not onto is ______
Report Question
0%
2
0%
6
0%
8
0%
4
Explanation
Total number of function
f
:
A
→
B
is the number of relations from
A
to
B
such that all the elements of
A
are in the first elements of function
f
.
Hence, total number of functions are
=
2
×
2
×
2
=
2
3
=
8
There are two functions for which the function is
{
(
1
,
4
)
,
(
2
,
4
)
,
(
3
,
4
)
}
and
{
(
1
,
5
)
,
(
2
,
5
)
,
(
3
,
5
)
}
.
Hence there are two non-onto functions.
If
f
:
R
→
R
,
g
:
R
→
R
are defined by
f
(
x
)
=
5
x
−
3
,
g
(
x
)
=
x
2
+
3
, then,
(
g
o
f
−
1
)
(
3
)
=
Report Question
0%
25
3
0%
111
25
0%
9
25
0%
25
111
Explanation
f
(
x
)
=
5
x
−
3
,
g
(
x
)
=
x
2
+
3
Let
f
(
x
)
=
y
⇒
5
x
−
3
=
y
⇒
x
=
y
+
3
5
----( 1 )
Now,
f
(
x
)
=
y
f
−
1
(
y
)
=
x
f
−
1
(
y
)
=
y
+
3
5
[ From ( 1 ) ]
f
−
1
(
x
)
=
x
+
3
5
Next,
(
g
o
f
−
1
)
(
3
)
⇒
g
[
f
−
1
(
3
)
]
⇒
g
[
3
+
3
5
]
⇒
g
[
6
5
]
⇒
(
6
5
)
2
+
3
[ Since,
g
(
x
)
=
x
2
+
3
]
⇒
36
25
+
3
⇒
36
+
75
25
⇒
111
25
Let
f
:
A
→
b
be a function defined by f(x) =
√
1
−
x
2
Report Question
0%
f(x) is one-one if A =[0,1]
0%
f(x) is onto if B = [0,1]
0%
f(x) is one-one if A =[-1 , 0]
0%
f(x) is onto if B = [-1,1]
Explanation
f
:
A
→
B
f
(
x
)
=
√
1
−
x
2
When
x
=
0
f
(
0
)
=
√
1
−
0
=
1
f
(
1
)
=
√
1
−
1
2
=
0
f
(
x
)
is one - one function when
A
=
[
0
,
1
]
If
f
:
R
→
R
,
f
(
x
)
=
{
1
x
>
0
0
x
=
0
−
1
x
<
0
and
g
:
R
→
R
,
g
(
x
)
=
[
x
]
, then
(
f
∘
g
)
(
π
)
is:
Report Question
0%
π
0%
0
0%
1
0%
−
1
Explanation
Given that
g
(
x
)
=
[
x
]
⇒
g
(
π
)
=
g
(
3.141....
)
i.e
g
(
π
)
=
3
(
f
∘
g
)
(
x
)
=
f
(
g
(
x
)
)
⇒
(
f
∘
g
)
(
π
)
)
=
f
(
g
(
π
)
)
But
g
(
π
)
=
3
⇒
f
(
3
)
=
1
......
[
∵
f
(
x
)
=
1
for
x
>
0
]
The inverse of the function
y
=
e
x
−
e
−
x
e
x
+
e
−
x
is
Report Question
0%
1
2
log
1
+
x
1
−
x
0%
1
2
log
2
+
x
2
−
x
0%
1
2
log
1
−
x
1
+
x
0%
2
log
(
1
+
x
)
Explanation
Given
y
=
e
x
−
e
−
x
e
x
+
e
−
x
or,
1
+
y
1
−
y
=
2
e
x
2
e
−
x
or,
e
2
x
=
1
+
y
1
−
y
or,
x
=
1
2
log
1
+
y
1
−
y
So the inverse is
1
2
log
1
+
x
1
−
x
.
Let
f
(
x
)
=
x
2
and
g
(
x
)
=
√
x
(where
x
>
0
),then
Report Question
0%
f
(
g
(
x
)
)
=
x
0%
g
(
f
(
x
)
)
=
x
0%
The least value of
f
(
g
(
x
)
)
+
1
g
(
f
(
x
)
)
is
2
0%
The least value of
g
(
f
(
x
)
)
+
1
f
(
g
(
x
)
)
is
2
Explanation
f
(
x
)
=
x
2
,
g
(
x
)
=
√
x
f
(
g
(
x
)
)
=
(
√
x
)
2
=
x
g
(
f
(
x
)
)
=
√
x
2
=
x
f
(
g
(
x
)
)
+
1
g
(
f
(
x
)
)
=
g
(
f
(
x
)
)
+
1
f
(
g
(
x
)
)
=
x
+
1
x
≥
2
Hence, all options are correct.
The solution of
(
3
∗
4
)
∗
3
, when
∗
is a binary operation on Z such that:
a
∗
b
=
a
+
b
, is.
Report Question
0%
10
0%
−
10
0%
−
1
6
0%
−
6
Explanation
Given
(
3
∗
4
)
∗
3
(
3
∗
4
)
∗
3
=
(
3
+
4
)
+
3
(as
(
a
∗
b
)
=
(
a
+
b
)
)
=
7
+
3
=
10
If
g
(
x
)
=
x
2
+
x
−
2
and
1
2
g
o
f
(
x
)
=
2
x
2
+
5
x
+
2
, then
f
(
x
)
is
Report Question
0%
2
x
−
3
0%
2
x
+
3
0%
2
x
2
+
3
x
+
1
0%
2
x
2
−
3
x
−
1
Explanation
g
(
x
)
=
x
2
+
x
−
2
1
2
g
(
f
(
x
)
)
=
2
x
2
+
5
x
+
2
⇒
f
2
(
x
)
+
f
(
x
)
−
2
=
4
x
2
−
10
x
+
4
⇒
f
2
(
x
)
+
f
(
x
)
−
(
4
x
2
−
10
x
+
6
)
=
0
f
(
x
)
=
−
1
±
√
1
+
4
(
4
x
2
−
10
x
+
6
)
2
f
(
x
)
=
−
1
±
√
1
+
16
x
2
−
40
x
+
24
2
f
(
x
)
=
−
1
±
(
4
x
−
5
)
2
=
(
2
x
−
3
)
Let f :
R
→
R
and g :
R
→
R
be two one-one and onto functions such that they are the mirror images of each other about the line y =If h(x) = f(x) + g(x), then h(0) equal to
Report Question
0%
2
0%
4
0%
0
0%
1
Explanation
Given
g
(
x
)
and
f
(
x
)
are mirror images about
y
=
2
⟹
g
(
x
)
−
2
=
2
−
f
(
x
)
⟹
f
(
x
)
+
g
(
x
)
=
4
⟹
h
(
x
)
=
4
h
(
0
)
=
4
If
f
:
A
→
A
defined by
f
(
x
)
=
4
x
+
3
6
x
−
4
where
A
=
R
−
2
3
. Find
f
−
1
Report Question
0%
2
x
0%
4
x
+
3
6
x
−
4
0%
x
2
0%
None of these
Explanation
f
(
x
)
=
4
x
+
3
6
x
−
4
l
e
t
y
=
4
x
+
3
6
x
−
4
⇒
6
x
y
−
4
y
=
4
x
+
3
⇒
6
x
y
−
4
x
=
3
+
4
y
⇒
x
(
6
y
−
4
)
=
3
+
4
y
⇒
x
=
3
+
4
y
6
y
−
4
R
e
p
l
a
c
e
x
b
y
y
&
y
b
y
x
.
y
=
3
+
4
x
6
x
−
4
f
′
(
x
)
=
4
x
+
3
6
x
−
4
If the binary operation
∗
is defined on a set of integers as
a
∗
b
=
a
+
3
b
2
, then the value of
2
∗
3
is
Report Question
0%
27
0%
29
0%
2
0%
None of these
Explanation
a
∗
b
=
a
+
3
b
2
⇒
2
∗
3
=
2
+
3
(
3
)
2
=
2
+
27
=
29
Let
f
,
g
:
R
→
R
be two functions defined as
f
(
x
)
=
|
x
|
+
x
,
g
(
x
)
=
|
x
|
−
x
,
∀
x
∈
R
. Then, find
f
o
g
(
x
)
Report Question
0%
|
|
x
|
−
x
|
−
|
x
|
−
x
0%
|
|
x
|
−
x
|
+
|
x
|
−
x
0%
|
|
x
|
−
x
|
−
|
x
|
+
x
0%
None of thesse
Explanation
f
(
x
)
=
|
x
|
+
x
g
(
x
)
=
|
x
|
−
x
f
o
g
(
x
)
=
f
(
g
(
x
)
)
=
f
(
|
x
|
−
x
)
=
|
|
x
|
−
x
|
+
|
x
|
−
x
Consider set
A
=
1
,
2
,
3
,
4
and set
B
=
0
,
2
,
4
,
6
,
8
, then the
number of one-one function from set
A
to set
B
is ?
Report Question
0%
5
0%
24
0%
120
0%
None of these
Explanation
Number of one-one function from
A
(
m
)
to
B
(
n
)
=
{
n
P
m
,
i
f
n
≥
m
0
,
i
f
n
<
m
m
=
4
,
n
=
5
One-one function
=
5
P
4
=
5
!
1
!
=
120
The function
∗
on
N
as
a
∗
b
=
(
a
−
b
)
2
is a binary operator
Report Question
0%
True
0%
False
Explanation
∗
is relation of
N
:
a
∗
b
=
(
a
−
b
)
2
Checking at
a
=
b
,
a
,
b
∈
N
,
⇒
a
∗
b
=
(
a
−
b
)
2
=
0
∉
N
⇒
∗
is not a binary opeation.
If the binary operation
∗
is on set of integers
Z
is defined as
a
∗
b
=
a
+
2
b
2
, then the value of
(
8
∗
3
)
∗
2
Report Question
0%
26
0%
22
0%
32
0%
34
Explanation
a
∗
b
=
a
+
2
b
2
⇒
(
8
∗
3
)
∗
2
=
(
8
+
2
(
3
)
2
)
∗
2
=
(
26
∗
2
)
=
26
+
2
(
2
)
2
=
26
+
8
=
34
If
f
(
x
)
=
2
x
+
5
and
g
(
x
)
=
x
2
+
1
be two real function , then value of
f
o
g
at x=1
Report Question
0%
9
0%
6
0%
5
0%
4
If
g
(
f
(
x
)
)
=
|
sin
x
|
and
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 and g can not be determined
Explanation
g
(
f
(
x
)
)
=
|
s
i
n
x
|
=
√
(
s
i
n
x
)
2
f
(
g
(
x
)
)
=
(
s
i
n
√
x
)
2
=
s
i
n
2
√
x
∴
f
(
x
)
=
s
i
n
2
x
a
n
d
g
(
x
)
=
√
x
Let
f
:
R
→
R
be defined by
f
(
x
)
=
x
2
−
3
x
+
4
for all
x
ϵ
R
, then
f
−
1
(
2
)
is
Report Question
0%
2
0%
1
0%
Not defined
0%
1
2
Explanation
F
:
R
→
R
f
(
x
)
=
x
2
−
3
x
+
4
f
(
1
)
=
1
−
3
+
4
f
(
1
)
=
2
f
−
1
(
2
)
=
1
Let
f
(
x
+
1
x
)
=
x
2
+
1
x
2
(
x
≠
0
)
, then
f
(
x
)
=
Report Question
0%
x
2
0%
x
2
−
1
0%
x
2
−
2
0%
N.O.T
Explanation
We are given
f
(
x
+
1
2
)
=
x
2
+
1
x
2
(where
x
≠
0
)
=
x
2
+
1
x
2
+
2
−
2
.
=
x
2
+
1
x
2
+
2
(
x
2
)
(
1
x
2
)
−
2
.
f\left(x+ \dfrac{1}{x} \right)= \left(x+ \dfrac{1}{x} \right)^{2}-2
So, simply put
x+ \dfrac{1}{x} \rightarrow x
we get
f(x) = x^{2}-2
The set onto which the derivative of the function
f(x)=x(\log x-1)
maps the range
[1,\infty )
is
Report Question
0%
\left[1,\infty \right)
0%
\left( e,\infty \right)
0%
\left[e,\infty \right)
0%
\left( 0,0 \right)
Explanation
Given
f (x) = x [\log x-1] x \in [+, \infty]
f (x) = x \left[ \dfrac{d}{dx} (\log x-1) \right]+ (\log x-1) \dfrac{d}{dx} (x)
=x. \dfrac{1}{x}+ (\log x-1).1
=1 -1 \log x-1
f(x)= \log x
\therefore
The set of
f(x)= \log x
that maps
[1, \infty )
is
[e, \infty )
Let
E=\{1, 2, 3, 4\}
and
F=\{1, 2\}
then the number of onto functions from E to F is
Report Question
0%
14
0%
16
0%
12
0%
8
Explanation
Number of onto function from
E
to
F.
E=\{1,2,3,4\}
F=\{1,2\}
Number of function from
E
to
F=2\times 2\times 2\times 2=16
We have to exclude functions where
f(x)=1
&
f(x)=2
.
\therefore
Total number of onto function
=16-2=14
.
Let
f\left( x \right) ={ x }^{ 2 },g\left( x \right) ={ 2 }^{ x }
, then solution set of
fog\left( x \right) =gof\left( x \right)
is
Report Question
0%
R
0%
\left\{ 0 \right\}
0%
\left\{ 0,2 \right\}
0%
None of these
If
f\left( x \right) =\begin{cases} 2+x,\quad x\ge 0 \\ 2-x,\quad x<0 \end{cases}
then
f\left( f\left( x \right) \right)
is given by
Report Question
0%
f\left( f\left( x \right) \right) =\begin{cases} 2+x,\quad x\ge 0 \\ 4-x,\quad x<0 \end{cases}
0%
f\left( f\left( x \right) \right) =\begin{cases} 2+x,\quad x\ge 0 \\ 2-x,\quad x<0 \end{cases}
0%
f\left( f\left( x \right) \right) =\begin{cases} 4+x,\quad x< 0 \\ x,\quad x\ge 0 \end{cases}
0%
f\left( f\left( x \right) \right) =\begin{cases} 4+x,\quad x\ge 0 \\ x,\quad x<0 \end{cases}
Let
f\left[-1, \dfrac{-1}{2}\right] \to [-1, 1]
is defined by
f(x) = 4x^3 - 3x
, then
f^{-1} (x) =
____ .
Report Question
0%
\cos \left(\dfrac{1}{3}\cos^{-1} x\right)
0%
\cos \left(3\cos^{-1} x\right)
0%
\sin \left(\dfrac{1}{3}\sin^{-1} x\right)
0%
\cos \left(\dfrac{2\pi}{3}+\dfrac{1}{3} \cos^{-1} x\right)
Explanation
f(x)=4x^3-3x
f\left[-1,\cfrac{-1}{2}\right]\longrightarrow \left[-1,1\right]
let,
x=\cos\theta
\because y=\cos 3\theta
f(\cos\theta)=4cos^3\theta-3\cos\theta
=\cos3\theta
\because 3\theta=\cos^{-1}y
\theta=\cfrac{1}{3}\cos^{-1}y
f^{-1}(\cos3\theta)=\cos\theta
\because y=f(x)
f^{-1}(y)=(x)
\because f^{-1}(\cos3\theta)=\cos\theta
f^{-1}(y)=\cos\left(\cfrac{1}{3}\cos^{-1}y\right)
f^{-1}(x)=\cos\left(\cfrac{1}{3}\cos^{-1}x\right)
If :
f(x) = 5 {x}^{2}
,
g(x) = 3x^{4}
, then :
(fog) (-1) =
Report Question
0%
45
0%
-54
0%
-32
0%
-64
Let
f:X \to \left[ {1,\,27} \right]
be a function by
f\left( x \right) = 5\sin x + 12\cos x + 14
. The set
X
so that
f
is one-one and onto is
Report Question
0%
\left[ { - \pi /2,\pi /2\,} \right]
0%
\left[ {0,\,\pi } \right]
0%
\left[ {0,\,\pi /2} \right]
0%
non of these
For
a,\ b\ \in \ R-\left\{ 0 \right\}
, let
f(x)=ax^{2}+bx+a
satisfies
f\left(x+\dfrac{7}{4}\right)=f\left(\dfrac{7}{4}-x\right) \forall \ x\ \in\ R
.
Also the equation
f(x)=7x+a
has only one real distinct solution. The minimum value of
f(x)
in
\left[0,\dfrac{3}{2}\right]
is equal to
Report Question
0%
\dfrac{-33}{8}
0%
0
0%
4
0%
-2
If
f\left( x \right) = (1 - x)
,
x \in \left[ { - 3,3} \right]
, then the domain of
f\left( {f\left( x \right)} \right)
is
Report Question
0%
\left[ { - 2,3} \right]
0%
\left( { - 2,3} \right)
0%
\left[ { - 2,3]} \right.
0%
( - 2,3]
Explanation
f\left(x\right)=1-x
f\left(f\left(x\right)\right)=f\left(1-x\right)=1-1+x=x
And
x\in\left[-3,3\right]
Domain of
f\left(x\right)
is
\left[-3,3\right]
Domain of
f\left(f\left(x\right)\right)=
Domain of
f\left(1-x\right)
Domain of
f\left(f\left(x\right)\right)=\left[1-3,3\right]=\left[-2,3\right]
If f(g(x))=5x+2 and g(x)=8x then f(x)=
Report Question
0%
\frac{5}{8}x+2
.
0%
\frac{8}{5}x+2
.
0%
\frac{5}{8}x-2
.
0%
8x-2
0%
5x-2
Let
g\left( x \right) =1+x-\left[ x \right]
and
f\left( x \right) =\begin{cases} -1,x<0 \\ 0,x=0 \\ 1,x>0 \end{cases}
Then for all
x,f\left( g\left( x \right) \right)
is equal to (where
\left[ . \right]
represents the greatest integer function)
Report Question
0%
x
0%
1
0%
f\left( x \right)
0%
g\left( x \right)
Let
f:(2,3) \rightarrow (0,1)
be defined by
f(x)=x-[x]
then
f^{-1}(x)
equals
Report Question
0%
x-2
0%
x+1
0%
x-1
0%
x+2
If
f(x)=\frac{x}{\sqrt{1-x^{2}}}
and g(x) =
f(x)=\frac{x}{\sqrt{1+x^{2}}}
, then (fog)(x) =
Report Question
0%
f(x)=\frac{x}{\sqrt{1-x^{2}}}
0%
f(x)=\frac{x}{\sqrt{1+x^{2}}}
0%
x^{2}
0%
x
Explanation
Given:-
f{\left( x \right)} = \cfrac{x}{\sqrt{1 - {x}^{2}}}
g{\left( x \right)} = \cfrac{x}{\sqrt{1 + {x}^{2}}}
To find:-
fog{\left( x \right)} = ?
fog{\left( x \right)}
= f{\left( g{\left( x \right)} \right)}
= f{\left( \cfrac{x}{\sqrt{1 + {x}^{2}}} \right)}
= \cfrac{\left( \cfrac{x}{\sqrt{1 + {x}^{2}}} \right)}{\sqrt{1 - {\left( \cfrac{x}{\sqrt{1 + {x}^{2}}} \right)}^{2}}}
= \cfrac{\left( \cfrac{x}{\sqrt{1 + {x}^{2}}} \right)}{\sqrt{1 - \left( \cfrac{{x}^{2}}{1 + {x}^{2}} \right)}}
= \cfrac{\left( \cfrac{x}{\sqrt{1 + {x}^{2}}} \right)}{\sqrt{\left( \cfrac{1 + {x}^{2} - {x}^{2}}{1 + {x}^{2}} \right)}}
= \cfrac{\left( \cfrac{x}{\sqrt{1 + {x}^{2}}} \right)}{\left( \cfrac{1}{1 + {x}^{2}} \right)}
= \cfrac{x}{\sqrt{1 + {x}^{2}}}
\Rightarrow fog{\left( x \right)} = \cfrac{x}{\sqrt{1 + {x}^{2}}}
Hence the correct answer is
\cfrac{x}{\sqrt{1 + {x}^{2}}}
.
Let
f:X\rightarrow Y
be an invertible function. Then f has unique inverse.
Report Question
0%
True
0%
False
If
f:R \rightarrow R, f(x)=2x-1
and
g; R \rightarrow R, g(x)=x^{2}+2
, then
(gof)(x)
equals-
Report Question
0%
2x^{2}-1
0%
(2x-1)^{2}
0%
2x^{2}+3
0%
4x^{2}-4x+3
Explanation
f(x)=2x-1
g(x)=x^2+2
g(f(x))=(2x-1)^2+2=4x^2-4x+3
.
Let
f(x)=\dfrac{1}{x^{2}}
for
x\ge 1
, and
g(x)
is its reflection in the line mirror
y=x
, then function
h(x)=\begin{cases} f\left( x \right) & x\ge 1 \\ g\left( x \right) & 0<x<1 \end{cases}
, is
Report Question
0%
derivable at
x=1
0%
continuous at
x=1
0%
not derivable at
x=1
0%
not continuous at
x=1
Explanation
Let
\left(t, \dfrac{1}{t^2} \right)
be a point on f(x)
\therefore
reflection in line mirror y = x
\therefore t = \gamma
\therefore \dfrac{\gamma - \dfrac{1}{t^2}}{1} = \dfrac{x - t}{-1} = -2 \dfrac{(-t + \dfrac{1}{t^2})}{2}
\therefore \gamma = \dfrac{1}{t^2} + t - \dfrac{1}{t^2}
\therefore \gamma = t
\therefore x = \dfrac{1}{t^2}
\therefore x = \dfrac{1}{y^2}
\therefore xy^2 = 1
=9(x)
\therefore h'(x)
at
x = 1
for
x \ge 1 = f'(x)
= \dfrac{-2}{x^3} = \dfrac{-2}{1} = -2
\therefore h'(x)
at x = 1 for
x \le 1 = 9'(x)
= \dfrac{-2}{13} = -2
derivable at
x = 1
.
If
f(x)=\begin{cases} x+1 & x\epsilon \left[ -1,0 \right] \\ { x }^{ 2 }+1 & x\epsilon \left( 0,1 \right) \end{cases}
, then the value of
\dfrac{f^{-1}(0)+f^{-1}(1)+f^{-1}(2)}{f(-1)+f(0)+f(1)}
is-
Report Question
0%
0
0%
1
0%
2
0%
\dfrac{1}{3}
Explanation
f(x) = \begin{cases} x +1 & x \in (-1, 0) \\ x^2 + 1 & x \in (0,1) \end{cases}
\therefore x = \begin{cases} f^{-1} (x + 1) & x \in [-1, 0] \\ f^{-1} (x^2 + 1) & x \in (0, 1)\end{cases}
f^{-1} (0)
:- put
x = -1
\therefore f^{-1} (0) = x = -1
\therefore f^{-1} (1)
put
x = 0
f^{-1} = 0
f^{-1} (2)
:- put
x = 1
f^{-1} (2) = 1
\therefore f(0) = 1
\therefore f(1) = 1 + 1 = 2
\therefore f(-1) = -1 + 1 = 0
\therefore \dfrac{f^{-1} (0) + f^{-1} (1) + f^{-1} (2)}{f(-1) + f(0) + f(1)} = \dfrac{-1 + 0 + 1}{3}
= \dfrac{0}{3}
= 0
.
The last three digits, if
(12345956)_{10}
is expressed in binary system.
Report Question
0%
110
0%
210
0%
100
0%
010
Explanation
undefined
If
f ( x ) = \left( a - x ^ { n } \right) ^ { 1 / n }
where
a > 0
and }
n
is a positive integer then
( f o f ) ( x )
is
Report Question
0%
f ( x )
0%
x
0%
0
0%
1
Explanation
f(x)=(a-x^n)^{1/n}
\therefore f(f(x))=(a-(f(x))^n))^{1/n}
\therefore f(f(x))=(a-((a-x^n)^{1/n)^n)^{1/n}}
\therefore f(f(x))=(a-((a-x^n)))^{1/n}
\therefore f(f(x))=(a-a+x^n)^{1/n}
\therefore f(f(x))=(x^n)^{1/n}
\therefore(f(x))=x
if
f\left( x \right) = \log \left( {\dfrac{{1 +x}}{{1 - x}}} \right)
and
g\left( x \right) = \dfrac{{3x + {x^3}}}{{1 + 3{x^2}}}
then
\left( {f(g(x)))} \right)
is equal to
Report Question
0%
- f\left( x \right)
0%
3f\left( x \right)
0%
{\left( {f\left( x \right)} \right)^3}
0%
f\left( {3x} \right)
Explanation
f(x)=\log\bigg(\dfrac{1+x}{1-x}\bigg)\ and\ g(x)=\dfrac{3x+x^{3}}{1+3x^{2}}
f(g(x))=\log\bigg(\dfrac{1+\dfrac{3x+x^{3}}{1+3x^{2}}}{1-\dfrac{3x+x^{3}}{1+3x^{2}}}\bigg)
f(g(x))=\log\bigg(\dfrac{\dfrac{1+3x^{2}+3x+x^{3}}{1+3x^{2}}}{\dfrac{1+3x^{2}-3x-x^{3}}{1+3x^{2}}}\bigg)
f(g(x))=\log\bigg(\dfrac{1+3x^{2}+3x+x^{3}}{1+3x^{2}-3x-x^{3}}\bigg)
f(g(x))=\log\bigg(\dfrac{(1+x)^3}{(1-x)^{3}}\bigg)
f(g(x))=\log\bigg(\dfrac{1+x}{1-x}\bigg)^{3}
f(g(x))=3\log\bigg(\dfrac{1+x}{1-x}\bigg)
f(g(x))=3f(x)
0:0:1
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
0
Answered
1
Not Answered
49
Not Visited
Correct : 0
Incorrect : 0
Report Question
×
What's an issue?
Question is wrong
Answer is wrong
Other Reason
Want to elaborate a bit more? (optional)
Practice Class 12 Commerce Maths Quiz Questions and Answers
<
>
Support mcqexams.com by disabling your adblocker.
×
Please disable the adBlock and continue.
Thank you.
Reload page