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CBSE Questions for Class 12 Commerce Maths Relations And Functions Quiz 8 - MCQExams.com
CBSE
Class 12 Commerce Maths
Relations And Functions
Quiz 8
Let
f
(
x
)
=
x
2
and
g
(
x
)
=
2
x
. Then the solution of the equation
f
o
g
(
x
)
=
g
o
f
(
x
)
is
Report Question
0%
R
0%
{
0
}
0%
{
0
,
2
}
0%
none
Explanation
f
(
x
)
=
x
2
f
g
(
x
)
=
f
(
2
x
)
=
(
2
x
)
2
=
2
2
x
g
(
x
)
=
2
x
g
f
(
x
)
=
g
(
x
2
)
=
2
x
2
Given:
f
g
(
x
)
=
g
f
(
x
)
⇒
2
2
x
=
2
x
2
Since bases are same we can equate the powers
⇒
2
x
=
x
2
⇒
x
2
−
2
x
=
0
⇒
x
(
x
−
2
)
=
0
⇒
x
=
0
,
2
When
x
=
0
,
f
g
(
x
)
=
g
f
(
x
)
⇒
2
0
=
2
0
=
1
When
x
=
2
,
f
g
(
x
)
=
g
f
(
x
)
⇒
2
2
×
2
=
2
2
2
⇒
16
=
16
Hence
x
=
{
0
,
2
}
Let
g
(
x
)
=
1
+
x
−
[
x
]
and
f
(
x
)
=
{
−
1
i
f
x
<
0
0
i
f
x
=
0
1
i
f
x
>
0
, then
∀
x
,
f
o
g
(
x
)
equals
Report Question
0%
x
0%
1
0%
f
(
x
)
0%
g
(
x
)
Explanation
given g(x)=1+x=[x] 1+{x} {x} donates the fractional part of x so,
g(x) is always positive
so, input to f(g(x)) is always positive and
hence, fog(x)=1
If
f
(
x
)
=
(
a
−
x
n
)
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
(
n
)
=
(
a
−
x
n
)
1
/
n
(
f
o
f
)
(
n
)
=
(
a
−
(
(
a
−
x
n
)
1
/
n
)
n
)
1
/
n
=
(
a
−
a
+
x
n
)
1
/
n
f
o
f
(
n
)
=
x
The inverse of the function
f
(
x
)
=
e
x
−
e
−
x
e
x
+
e
−
x
+
2
is given by
Report Question
0%
log
e
(
x
−
1
x
+
1
)
−
2
0%
log
e
(
x
−
2
x
+
1
)
1
/
2
0%
log
e
(
x
2
−
x
)
1
/
2
0%
log
e
(
x
−
1
3
−
x
)
1
/
2
f
:
R
→
R
such that
f
(
x
)
=
ℓ
n
(
x
+
√
x
2
+
1
)
. Another function
g
(
x
)
is defined such that
g
o
f
(
x
)
=
x
∀
x
∈
R
. Then
g
(
2
)
is -
Report Question
0%
e
2
+
e
−
2
2
0%
e
2
0%
e
2
−
e
−
2
2
0%
e
−
2
Explanation
f
(
x
)
=
l
n
(
x
+
√
x
2
+
1
)
g
(
x
)
=
g
(
f
(
x
)
)
g
(
2
)
=
?
ln
(
x
+
√
x
2
+
1
)
=
y
√
x
2
+
1
=
e
y
−
x
x
2
+
1
=
e
2
y
−
2
e
y
x
+
x
2
e
2
y
−
2
e
y
x
−
1
=
o
⇒
x
=
e
2
y
−
1
2
e
y
=
(
e
y
−
e
−
y
)
2
∴
g
(
x
)
=
e
y
−
e
−
y
2
⇒
g
(
2
)
=
e
2
−
e
−
2
2
Let
f
:
R
→
R
is a function satisfying
f
(
2
−
x
)
=
f
(
2
+
x
)
and
f
(
20
−
x
)
=
f
(
x
)
∀
x
∈
R
If
f
(
0
)
=
5
then the minimum possible no. of values of
x
satisfying
f
(
x
)
=
5
for
x
=
[
0.
,
70
]
, is
Report Question
0%
21
0%
12
0%
11
0%
22
Explanation
Given,
f
(
2
−
x
)
=
f
(
2
+
x
)
.
.
(
i
)
∴
f
(
x
)
is symmetric about
x
=
2
f
(
20
−
x
)
=
f
(
x
)
.
.
(
i
i
)
x
→
x
+
10
f
(
10
−
x
)
=
f
(
10
+
x
)
f
(
x
)
is symmetric about
x
=
10
x
→
x
+
2
f
(
18
−
x
)
=
f
(
x
+
2
)
.
.
(
i
i
)
f
(
2
−
x
)
=
f
(
18
−
x
)
x
→
−
x
f
(
2
+
x
)
=
f
(
18
+
x
)
∴
x
+
2
→
x
∴
f
(
x
)
=
f
(
x
+
16
)
f
(
x
)
has period =
16
f
(
x
)
=
5
,
x
∈
[
0
,
170
]
put
x
=
2
in (i)
f
(
0
)
=
f
(
4
)
put
x
=
4
in (ii)
f
(
4
)
=
f
(
16
)
when
x
∈
[
0
,
16
]
→
f
(
x
)
has two solution
when
x
∈
[
0
,
160
]
has
20
solution.
When
x
i
n
[
0
,
170
]
has
′
21
′
solution
All values of a for which f : R
→
R
defined by f(x)=
x
3
+
a
x
2
+
3
x
+
100
is a one one functions, are
Report Question
0%
(
−
∞
,
−
2
)
0%
(
−
∞
,
4
)
0%
(
4
,
−
4
)
0%
(
−
3
,
3
)
Explanation
f
′
(
x
)
=
3
x
2
+
2
a
x
+
3
f
′
(
x
)
=
0
and f ( x ) > 0
f
′
(
x
)
<
0
3
x
2
+
2
a
x
+
3
=
0
x
=
−
2
a
±
√
4
a
2
−
36
6
4
a
2
<
36
a
∈
(
−
3
,
3
)
Let
A
=
{
1
,
2
,
3
}
. Which of the following functions on
A is invertible?
Report Question
0%
f
=
{
(
1
,
1
)
,
(
2
,
1
)
,
(
3
,
1
)
}
0%
f
=
{
(
1
,
2
)
,
(
2
,
3
)
,
(
3
,
1
)
}
0%
f
=
{
(
1
,
2
)
,
(
2
,
3
)
,
(
3
,
2
)
}
0%
f
=
{
(
1
,
1
)
,
(
2
,
2
)
,
(
3
,
1
)
}
Explanation
A
=
{
1
,
2
,
3
}
∴
A
is invertible when obviously
f
=
{
(
1
,
1
)
,
(
2
,
1
)
,
(
3
,
1
)
}
If
f
(
x
)
=
sin
−
1
(
sin
x
)
+
cos
−
1
(
sin
x
)
and
ϕ
(
x
)
=
f
(
f
(
f
(
x
)
)
)
then
ϕ
′
(
x
)
Report Question
0%
1
0%
sin
x
0%
0
0%
none of these
Explanation
f
(
x
)
=
sin
−
1
(
sin
x
)
+
cos
−
1
(
sin
x
)
=
π
2
[
∵
sin
−
1
θ
+
cos
−
1
θ
=
π
2
]
f
(
f
(
x
)
)
=
f
(
π
2
)
=
π
2
ϕ
(
x
)
=
f
(
f
(
f
(
x
)
)
)
=
f
(
π
2
)
=
π
2
ϕ
(
x
)
=
π
2
ϕ
′
(
x
)
=
0
if
f
(
x
)
=
3
x
+
2
,
g
(
x
)
=
x
2
+
1
,then the values of
(
f
o
g
)
(
x
2
−
1
)
Report Question
0%
3
x
4
−
6
x
2
+
8
0%
3
x
4
+
3
x
+
4
0%
6
x
4
+
3
x
2
−
2
0%
6
x
4
+
3
x
2
+
2
Explanation
f
(
x
)
=
3
x
+
2
g
(
x
)
=
x
2
+
1
f
.
g
(
x
)
=
f
(
x
2
+
1
)
f
.
g
(
x
)
=
3
(
x
2
+
1
)
+
2
=
3
x
2
+
5
f
.
g
(
x
)
=
3
x
2
+
5
f
.
g
(
x
2
−
1
)
=
3
(
x
2
−
1
)
2
+
5
=
3
(
x
4
−
2
x
2
+
1
)
+
5
=
3
x
4
−
6
x
2
+
3
+
5
=
3
x
4
−
6
x
2
+
8
Let A = {1,2,3,4,5} and B={1,2,3,4,5}. If
f
:
A
→
B
is an one-one function and
f
(
x
)
=
x
holds only for one value of
x
ϵ
{
1
,
2
,
3
,
4
,
5
}
,
then the number of such possible function is
Report Question
0%
120
0%
36
0%
45
0%
44
Difference between the greatest and the least values of the function
f
(
x
)
=
x
(
l
n
x
−
2
)
on
[
1
,
e
2
]
is
Report Question
0%
2
0%
e
0%
e
2
0%
1
The function
f
:
[
−
1
2
,
1
2
]
→
[
−
π
2
,
π
2
]
defined by
f
(
x
)
=
sin
−
1
(
3
x
−
4
x
3
)
is
Report Question
0%
both one-one and onto
0%
onto but not one-one
0%
one-one but not onto
0%
neither one-one nor onto
Explanation
f
:
[
−
1
2
,
1
2
]
→
[
−
π
2
,
π
2
]
f
(
x
)
=
sin
−
1
(
3
x
−
4
x
3
)
Range of
sin
−
1
(
3
x
−
4
x
3
)
is $$
[-\cfrac{\pi}{2},\cfrac{\pi}{2}]$$
And
−
π
2
≤
sin
−
1
(
3
x
−
4
x
3
)
≤
π
2
⇒
sin
(
−
π
2
)
≤
3
x
−
4
x
3
≤
sin
π
2
⇒
−
1
≤
3
x
−
4
x
3
≤
1
3
x
−
4
x
3
≤
−
1
⇒
4
x
3
−
3
x
−
1
≤
0
⇒
x
≤
1
2
and
3
x
−
4
x
3
≤
1
⇒
4
x
3
−
3
x
+
1
≤
0
⇒
x
≤
−
1
2
∴
x
ϵ
[
−
1
2
,
1
2
]
f
x
=
sin
−
1
(
3
x
−
4
x
3
)
⇒
f
1
(
x
)
=
1
√
1
−
(
3
x
−
4
x
3
)
2
×
(
3
−
12
x
2
)
⇒
f
1
(
x
)
=
3
−
12
x
2
=
−
(
12
x
2
−
3
)
x
2
≤
0
⇒
12
x
2
≤
0
⇒
12
x
2
−
3
≤
−
3
−
3
(
12
x
2
−
3
)
≤
3
∴
f
1
(
x
)
≤
0
∴
f
1
(
x
)
is a decreasing function.
Hence
f
(
x
)
is one-one and range of function
=
its co-domain.
Hence it is both one-one and onto.
If
f
(
x
)
=
x
−
1
x
+
1
, then
f
−
1
(
x
)
is
Report Question
0%
f
(
x
)
+
1
f
(
x
)
+
3
0%
3
f
(
x
)
+
1
f
(
x
)
+
3
0%
f
(
x
)
+
3
f
(
x
)
+
1
0%
f
(
x
)
+
3
3
f
(
x
)
+
1
Let g be the inverse function of differentiable function f and
G
(
x
)
=
1
g
(
x
)
i
f
f
(
4
=
2
)
and
f
′
(
4
)
=
1
16
, then the value of
(
G
′
(
2
)
)
2
equals to:
Report Question
0%
1
0%
4
0%
16
0%
64
If
f
:
(
−
1
,
1
)
→
B
, is a function defined by
f
(
x
)
=
tan
−
1
2
x
1
−
x
2
, then find
B
when
f
(
x
)
is both one-one and onto function.
Report Question
0%
[
−
π
2
,
π
2
]
0%
(
−
π
2
,
π
2
)
0%
(
0
,
π
2
)
0%
[
0
,
π
2
)
Explanation
For
x
ϵ
(
−
1
,
1
)
, we have
f
(
x
)
=
tan
−
1
[
2
x
1
−
x
2
]
Substituting
x
=
tan
θ
in above equation.
Therefore,
f
(
tan
θ
)
=
tan
−
1
[
2
tan
θ
1
−
tan
2
θ
]
=
tan
−
1
tan
(
2
θ
)
=
2
θ
=
2
tan
−
1
x
Thus
−
π
2
<
tan
−
1
[
2
x
1
−
x
2
]
<
π
2
Thus option B is correct.
If
f
(
x
)
=
x
3
+
x
2
f
′
(
1
)
+
x
f
″
, then
f(x)
is
Report Question
0%
one-one and onto
0%
one-one and into
0%
many-one and onto
0%
non-invertible
The multiplicative inverse of the product of the additive inverse of x+1 is ________________.
Report Question
0%
x-1
0%
\dfrac { 1 }{ 1-x }
0%
{ x }^{ 2 }-1
0%
\dfrac { 1 }{ 1-{ x }^{ 2 } }
Let S be a non-empty set and P(S) be the power set of set S. Find the identity element for the union
(\cup)
as a binary operation on
P(S)
.
Report Question
0%
\phi
0%
1
0%
0
0%
None of these
Explanation
We observe that
A \cup \phi = A = \phi \cup A
for every subset A of set S.
A \cup \phi = A =\phi \cup A
for all
A \in P(S)
\phi
is the identity element for union
(\cup)
on
P(S)
.
If
\begin{bmatrix} \sin { \left( \dfrac { \pi }{ 2 } \right) } & \cos { \left( \dfrac { \pi }{ 3 } \right) } \\ 2\tan { \left( \dfrac { \pi }{ 4 } \right) } & 2k \end{bmatrix}
is not invertible, then
k=
Report Question
0%
2
0%
\dfrac{1}{2}
0%
1
0%
3
Explanation
We have,
\begin{bmatrix} \sin { \left( \dfrac { \pi }{ 2 } \right) } & \cos { \left( \dfrac { \pi }{ 3 } \right) } \\ 2\tan { \left( \dfrac { \pi }{ 4 } \right) } & 2k \end{bmatrix}
Since,
|A| = 0
if it is not invertible,
1(2k) - ((\cos 60) ( 2 \tan 45 )) = 0
2k - 1 = 0
k = \dfrac{1}{2}
Hence, this is the answer.
Let
N
be the set of natural numbers and two functions
f
and
g
be defined as
f,g : N\to N
such that :
f (n)= \begin{cases}\dfrac{n+1}{2}& \text{if n is odd}\\ \dfrac{n}{2} & \text{in n is even} \end{cases}
and
g(n) = n - (-1)^n
. The fog is:
Report Question
0%
Both one-one and onto
0%
One-one but not onto
0%
Neither one-one nor onto
0%
onto but not one-one
Explanation
fx = \begin{cases} \dfrac{n+1}{2}& \text{n is odd} \\\dfrac{n}{2}& \text{n is even}\end{cases}
g(x) = n -(-1)^n \begin{cases}n+1; \text{n is odd} \\n-1; \text{n is even} \end{cases}
f(g(n)) = \begin{cases}\dfrac{n}{2}; & \text{n is even}\\ \dfrac{n+1}{2}; &\text{n is odd} \end{cases}
\therefore
onto but not one-one
The numbers system which uses alphabets as well as numbers is-
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0%
Binary numbers system
0%
Octal numbers system
0%
Decimal numbers system
0%
Hexadecimal numbers system
Explanation
HEXADECIMAL NUMBERS SYSTEM
f : R \rightarrow R , f ( x ) = e ^ { | x | } - e ^ { - x }
is many-one into function.
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0%
True
0%
False
Explanation
f(x)=e^{(x)}-e^{-x}
For every
x
, there will be different
f(x)
\therefore
It is a one - one function
\therefore False
Number of one-one functions from A to B where
n(A)=4, n(B)=5
.
Report Question
0%
4
0%
5
0%
120
0%
90
Explanation
n(A)=4
and
n(B)=5
For one-one mapping
4 elements can be selected out of 5 elements of set B in
{}^5C_4
ways
and then those 4 selected elements can be mapped with 4 elements of set A in
4!
ways.
Number of one-one mapping from
A
to
B
={}^5C_4\times4!={}^5P_4=\dfrac{5!}{(5-4)!}=5!=120
Let
f(x)=x^ {135}+x^ {125}-x^ {115}+x^ {5}+1
. If
f(x)
divided by
x^ {3}-x
, then the remainder is some function of
x
say
g(x)
. Then
g(x)
is an:-
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one-one function
0%
many one function
0%
into function
0%
onto function
Explanation
f : R \rightarrow R , f ( x ) = 2 x + | \sin x |
is one-one onto.
Report Question
0%
True
0%
False
Explanation
f(x) = 2x + \left | sin x \right |
for query x,there will be a defferent f(x)
\therefore one-one
& domain
\rightarrow
R
\therefore onto
\therefore True.
If
f : R \rightarrow R
be given by
f(x)=\left(3-x^{3}\right)^{\dfrac{1}{3}},
then
fof(x)
is
Report Question
0%
x^{\dfrac{1}{3}}
0%
1^{3}
0%
x
0%
\left(3-x^{3}\right)
Explanation
Given,
f(x)=\left(3-x^{3}\right)^{\dfrac{1}{3}}
.
Now,
fof(x)
=f[f(x)]
=\left(3-[f(x)]^{3}\right)^{\dfrac{1}{3}},
=\left(3-[(3-x^3)^{\dfrac{1}{3}}]^{3}\right)^{\dfrac{1}{3}},
=\left(3-[(3-x^3)]\right)^{\dfrac{1}{3}},
=[x^3]^{\dfrac{1}{3}}
=x
.
Let :
R\rightarrow R
defined as
f\left( x \right) =\dfrac { x\left( x+1 \right) \left( { x }^{ 4 }+1 \right) +{ 2x }^{ 4 }+{ x }^{ 2 }+2 }{ { x }^{ 2 }+x+1 }
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odd and one-one
0%
even and one-one
0%
many to one and even
0%
many to one and neither even nor odd
If a binary operation is defined
a\star b=a^b
then 2
\star 2
is equal to:
Report Question
0%
4
0%
2
0%
9
0%
8
Explanation
a \star b = {a}^{b} \quad \left( \text{Given} \right)
Therefore,
2 \star 2 = {2}^{2} = 4
Hence the correct answer is
4
.
If is a binary operation such that
a * b = a^2 + b^2
then
3 * 5
is
Report Question
0%
34
0%
9
0%
8
0%
25
Explanation
a \star b = {a}^{2} + {b}^{2} \quad \left( \text{Given} \right)
Therefore,
3 \star 5 = {3}^{2} + {5}^{2} = 9 + 25 = 34
Hence the correct answer is
34
.
Consider
f(x) = \dfrac{x^2}{1 + x^3}
;
g(t) = \displaystyle \int f(t) dt
. If
g(1) = 0
then
g(x)
equals
Report Question
0%
\dfrac{1}{3} ln(1 + x^3)
0%
\dfrac{1}{3} ln\left ( \dfrac{1 + x^3}{2} \right )
0%
\dfrac{1}{2} ln\left ( \dfrac{1 + x^3}{3} \right )
0%
\dfrac{1}{3}l n\left ( \dfrac{1 + x^3}{3} \right )
Let f :
R\rightarrow R
be a function defined by f(x) =
{ x }^{ 3 }+{ x }^{ 2 }+3x+sin\times .
Then f is.
Report Question
0%
one-one & onto
0%
one-one & into
0%
many one & onto
0%
many one & into
Explanation
f(x)=x^{3}+x^{2}+3 x+\sin x
Suppose
f(x_{1})=f(x_{2})
x_{1}^{3}+x_{1}^{2}+3 x_{1}+\sin x_{1}=x_{2}+x_{2}+\sin x-\sin x_{2}
Equating\;above\;equations\;to \;0
x_{1}^{3}+x_{1}^{2}+3 x_{1}+\sin x_{1}=x_{2}+x_{2}+\sin x-\sin x_{2}=0
Solving\;them
It\; will\; not\; prove\; that
x_{1}=x_{2}
So,\;f(x)\;is\;one-one
Range\;f(x)=Real
Co-domain\;f(x)=Real
Range=Co-domain
Hence,\; f(x) \;is \;onto
So,\;option\; C \;is\;correct.
A function
f
from the set of natural numbers to integers defined by
f(n)=\begin{cases} \cfrac { n-1 }{ 2 } ,\quad \text{when n is odd} \\ -\cfrac { n }{ 2 } ,\quad \text{when n is even} \end{cases}
is
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0%
neither one-one nor onto
0%
one-one but not onto
0%
onto but not one-one
0%
one-one and onto both
Explanation
one-one
test of
f:
Let
x_1
and
x_2
be any two elements in the domain
(N).
Case\,I:
When both
x_1
and
x_2
are even.
Let
f(x_1)=f(x_2)
\Rightarrow
\dfrac{-x_1}{2}=\dfrac{x_2}{2}
\Rightarrow
-x_1=-x_2
\Rightarrow
x_1=x_2
Case\,II:
When both
x_1
and
x_2
are odd.
Let
f(x_1)=f(x_2)
\Rightarrow
\dfrac{x_1-1}{2}=\dfrac{x_2-1}{2}
\Rightarrow
x_1-1=x_2-1
\Rightarrow
x_1=x_2
Case\,III:
When
x_1
be even and
x_2
be odd.
Then,
f(x_1)=\dfrac{-x_1}{2}
and
f(x_2)=\dfrac{x_2-1}{2}
Then clearly,
\Rightarrow
x_1\ne x_2
\Rightarrow
f(x_1)\ne f(x_2)
From, all the cases, we can say that,
f
is one-one.
onto
test of
f:
Co-domain of
f=Z=\{....,-3,-2,-1,0,1,2,3,...\}
Range of
f=\left\{...,\dfrac{-2-1}{2},\dfrac{-(-2)}{2},\dfrac{-1-1}{2},\dfrac{0}{2},\dfrac{1-1}{2},\dfrac{-2}{2},\dfrac{3-1}{2},...\right\}
Range of
f=\{...,-2,1,-1,0,0,-1,1,..\}
\Rightarrow
Co-domain of
f=
Range of
f
\therefore
f
is onto.
Let
f:[2,\infty )\rightarrow X
be defined by
f(x)=4x-{x}^{2}
. Then,
f
is invertible, if
X=
Report Question
0%
[2,\infty)
0%
(-\infty,2]
0%
(-\infty,4]
0%
[4,\infty)
Explanation
Since,
f
is invertible, range of
f=
Co-domain of
f=X
So, we need to find the range of
f
to find
X.
Foe finding the range, let
f(x)=y
\Rightarrow
4x-x^2=y
\Rightarrow
x^2-4x=-y
\Rightarrow
x^2-4x+4=4-y
[ Adding
4
on both sides ]
\Rightarrow
(x-2)^2=4-y
\Rightarrow
x-2=\pm\sqrt{4-y}
\Rightarrow
x=2\pm\sqrt{4-y}
This is defined only when,
4-y\ge 0
\Rightarrow
y\le 4
X=
Range of
f=(-\infty , 4]
If
g(x)={ x }^{ 2 }+x-2
and
\cfrac { 1 }{ 2 } (g\circ f(x))=2{ x }^{ 2 }-5x+2
, then
f(x)
is equal to
Report Question
0%
2x-3
0%
2x+3
0%
2{x}^{2}+3x+1
0%
2{x}^{2}-3x-1
Explanation
We will solve this problem by the trial and error method.
Let us check option
A
first.
If
f(x)=2x-3
g(x)=x^2+x-2
[ Given ]
\Rightarrow
\dfrac{1}{2}(g\circ f)(x)=g[f(x)]
=\dfrac{1}{2}g(2x-3)
=\dfrac{1}{2}[(2x-3)^2+(2x-3)-2]
=\dfrac{1}{2}[4x^2+9-12x+2x-3-2]
=\dfrac{1}{2}[4x^2-10x+4]
=2x^2-5x+2
The given condition is satisfied by
A.
If
g(x)=x^2+x-1
and
(gof)(x)=4x^2-10x+5
, then
f\left(\dfrac{5}{4}\right)
is equal to:
Report Question
0%
\dfrac{3}{2}
0%
\dfrac{1}{2}
0%
-\dfrac{3}{2}
0%
-\dfrac{1}{2}
Explanation
g(x)=x^2+x-1
g\left(f\left(\dfrac{5}{4}\right)\right)=4\left(\dfrac{5}{4}\right)^2-10\dfrac{5}{4}+5=-\dfrac{5}{4}
g\left(f\left(\dfrac{5}{4}\right)\right)=f^2\left(\dfrac{5}{4}\right)+f\left(\dfrac{5}{4}\right)-1
-\dfrac{5}{4}=f^2\left(\dfrac{5}{4}\right)+f\left(\dfrac{5}{4}\right)-1
f^2\left(\dfrac{5}{4}\right)+f\left(\dfrac{5}{4}\right)+\dfrac{1}{4}=0
\left(f\left(\dfrac{5}{4}\right)+\dfrac{1}{2}\right)^2=0
\boxed{f\left(\dfrac{5}{4}\right)=\dfrac{-1}{2}}
The inverse function of
f(x) = \dfrac{8^{2x}-{8^{-2x}}} {8^{2x}+8^{-2x}}\in (-1, 1)
, is ________.
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0%
\dfrac{1}{4} \log_e\left(\dfrac{1-x}{1+x}\right)
0%
\dfrac{1}{4} \log_e\left(\dfrac{1+x}{1-x}\right)
0%
\dfrac{1}{4} (\log_e)\log_e\left(\dfrac{1-x}{1+x}\right)
0%
\dfrac{1}{4} (\log_e)\log_e\left(\dfrac{1+x}{1-x}\right)
Explanation
f(x) = \dfrac{8^{2x} - 8^{-2x}}{8^{2x}+8^{-2x}}
y =\dfrac{8^{2x} - 8^{-2x}}{8^{2x}+8^{-2x}}
\dfrac{1+y}{1-y} = \dfrac{8^{2x}}{8^{-2x}}
8^{4x} = \dfrac{1+y}{1-y}
4x = \log_8\left(\dfrac{1+y}{1-y}\right)
x = \dfrac{1}{4} \log_8\left(\dfrac{1+y}{1-y}\right)
f^{-1}(y) = \dfrac{1}{4} \log_8\left(\dfrac{1+y}{1-y}\right)
\therefore \boxed {f^{-1}(x) = \dfrac{1}{4} \log_8 \left(\dfrac{1+x}{1-x}\right)}....Answer
Hence option
'B'
is the answer.
If
f(x) = \dfrac{x+1}{x-1}
, then the valueof
f(f(x))
is equal to
Report Question
0%
x
0%
0
0%
-x
0%
1
Explanation
f(x)=\dfrac{x+1}{x-1}
\therefore f(f(x))=f\left(\dfrac{x+1}{x-1}\right)
\dfrac{\dfrac{x+1}{x-1}+1}{\dfrac{x+1}{x-1}-1}
=\dfrac{x+1+x-1}{x+1-x+1}
=\dfrac{2x}{2}
=x
Let
f : x \rightarrow y
be such that
f(1) = 2
and
f(x + y) = f(x) f(y)
for all natural numbers x and y. If
\displaystyle \sum_{k= 1}^n f(a + k) = 16 (2^n - 1)
, then a is equal to
Report Question
0%
3
0%
4
0%
5
0%
6
0%
7
Explanation
We have,
f(1)=2
and
f(x+y)=f(x).f(y)
Now,
f(2)=f(1+1)=f(1).f(1)=2.2=2^2
f(3)=f(2+1)+f(2).f(1)=2^2.2=2^3
and so on
\therefore f(x)=2^n
......(i)
Now, we have
\displaystyle \sum^n_{k=1}f(a+k)=16(2^n-1)
\Rightarrow f(a+1)+f(a+2)+----f(a+n)=16(2^n-1)
\Rightarrow f(a).f(1)+f(a).f(2)+-----f(a).f(n)=16(2^n-1)
\Rightarrow f(a)=[f(1)+f(2)+---f(n)]=16(2^n-1)
\Rightarrow f(a)[2+2^2+----+2^n]=16(2^n-1)
\Rightarrow f(a).[2\dfrac{(2^n-1)}{2-1}]=16(2^n-1)
\Rightarrow 2f(a).(2^n-1)=16.(2^n-1)\Rightarrow f(a)=8
\Rightarrow 2^a=8[\because f(x)=2^n\Rightarrow f(a)=2^a]
\Rightarrow 2^a=2^3=a=3
If
f(x)=\dfrac{(4x+3)}{(6x-4)}, x\neq \dfrac{2}{3}
then
(f o f)(x)=?
Report Question
0%
x
0%
(2x-3)
0%
\dfrac{4x-6}{3x+4}
0%
None of these
If
f(x)=\sqrt[3]{3-x^3}
then
(f o f)(x)=?
Report Question
0%
x^{1/3}
0%
x
0%
(1-x^{1/3})
0%
None of these
Let
f : R \rightarrow R : f(x) =x +1
and
g : R \rightarrow R : g(x) = 2x -3
.
Find
(f +g) (x)
.
Report Question
0%
3x -2
0%
4x -5
0%
3x -4
0%
2x -3
Explanation
Given ,
f(x)=x+1,g(x)=2x-3
\implies (f+g)x=f(x)+g(x)=x+1+2x-3=3x-2
If
\displaystyle f(x) = | x - 2 |
and
g(x) = fof\,(x)
, then for
x > 20 , {g}\,'(x) =
Report Question
0%
2
0%
1
0%
3
0%
None of these
If
\displaystyle {f}'(x) = g\,(x)
and
\displaystyle {g}'(x) = - f\,(x)
for all
x
and
f\,(2) = 4 = {f}'(2)
then
\displaystyle f^{2}\,(19) + g^{2} \,(19)
is
Report Question
0%
16
0%
32
0%
64
0%
None of these
The value of f(0), so that the function
f(x) =
\dfrac{2x-sin^{-1}x}{2x+tan^{-1}x}
is continuous at each point in its domain, is equal to
Report Question
0%
2
0%
1/3
0%
2/3
0%
-1/3
Explanation
The function f is clearly continuous at each point in its domain except possibly at x=0 Given that f(x) is continuous at x=0
Therfore,f (0) =
\underset{x\rightarrow 0}{lim}f(x)
=\underset{x\rightarrow 0}{lim}\dfrac{2x-sin^{-1}x}{2x+tan^{-1}x}
= \underset{x\rightarrow 0}{lim}\dfrac{2-\frac{(sin^{-1}x)}x}{2+\frac{(tan^{-1}x)}x}
\dfrac{2-1}{2+1}=\dfrac13
let
f(x) = sin^2 x/2 + cos ^2 x/2
and
g(x) = sec^2 x - tan ^2 x.
The two functions are equal over the set
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0%
\phi
0%
R
0%
R-{ x:x (2n+1) \frac{\pi}{2}, n\in1}
0%
None of these
Let
f(n)
denote the number of different ways in which the positive integer
n
can be expressed as the sum of
1s
and
2s
. For example,
f(4) = 5
, since
4 = 2 + 2 = 2 + 1 + 1 = 1 + 2 + 1 = 1 + 1 + 2 = 1 + 1 + 1 + 1
. Note that order of
1s
and
2s
is important.
f : N\rightarrow N
is
Report Question
0%
One-one and onto
0%
One-one and into
0%
Many-one and onto
0%
Many-one and into
Explanation
6 = 0(2) + 6(1) = 1(2) + 4(1) = 2(2) + 2(1) = 3(2) + 0(1)
Number of
2s
Number of
1s
Number of permutations
0
6
1
1
4
\dfrac {5!}{4!} = 5
2
2
\dfrac {4!}{2!2!} = 6
3
0
\dfrac {3!}{3!} = 1
Total = 13
\therefore f(6) = 13
Now,
f(f(6)) = f(13)
Number of
1s
Number of
2s
Number of permutations
13
0
1
11
1
\dfrac {12!}{11!} = 12
9
2
\dfrac {11!}{9!2!} = 55
7
3
\dfrac {10!}{7!3!} = 120
5
4
\dfrac {9!}{5!4!} = 126
3
5
\dfrac {8!}{3!5!} = 56
1
6
\dfrac {7!}{6!} = 7
Total = 377
\therefore f(f(6)) = f(13) = 377
f(1) = 1(1)
f(2) = 2 (1, 1
or
2)
f(3) = 3(1, 1, 1\ or\ 2, 1\ or\ 1, 2)
f(4) = 5
(explained in the paragraph)
By taking higher value of
n
in
f(n)
, we always get more value of
f(n)
. Hence,
f(x)
is one-one. Clearly,
f(x)
is into.
The function
f(x)= \dfrac{(3^{x}-1^{})^2}{\sin x. \ln(1+x)}, x\neq 0
, is continuous at
x=0
. Then the value of
f(0)
is
Report Question
0%
2log 3
0%
(\log_{e}3)^{2}
0%
\log_{e} 6
0%
None of these
Explanation
Given f(x) is continuous at
x=0
\Rightarrow \underset{x\rightarrow 0}{\lim}f(x)=f(0)
\Rightarrow \underset{x\rightarrow 0}{\lim}\dfrac{(3^{x}-1)^{2}}{\sin x\ln(1+x)}=f(0)
\Rightarrow f(0)=\underset{x\rightarrow 0}{\lim} \dfrac{\bigg({\dfrac{3^x-1}{x}}\bigg)^2}{\dfrac{\sin x}{x}\dfrac{\ln(1+x)}{x}}=
(\log_e3)^{2}
Let f(x)= max { 1+sinx, 1, 1 -cosx},
x \epsilon [0, 2 \pi]
and g(x)= max {1, |x-1|}
x \epsilon R
, then
Report Question
0%
g(f(0))=1
0%
g(f(1))=1
0%
f(f(1))=1
0%
f(g(0))=1+sin1
If f:
R\rightarrow R
be given by
f(x) = 3 + 4x
and
a_n = A + Bx
, then which of the following is not true?
Report Question
0%
A + B + 1 =
2^{2n + 1}
0%
| A - B| = 1`
0%
lim_{n \to \infty} \dfrac{A}{B} = -1
0%
None of these
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Practice Class 12 Commerce Maths Quiz Questions and Answers
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