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CBSE Questions for Class 12 Commerce Maths Inverse Trigonometric Functions Quiz 10 - MCQExams.com
CBSE
Class 12 Commerce Maths
Inverse Trigonometric Functions
Quiz 10
If for
x
<
−
1
,
c
o
s
−
1
(
x
2
−
1
x
2
+
1
)
+
s
i
n
−
1
(
2
x
1
+
x
2
)
−
t
a
n
−
1
(
2
x
x
2
−
1
)
=
π
3
, then
x
=
Report Question
0%
−
√
3
0%
−
√
2
0%
−
2
0%
−
2
√
3
The value of a for which
a
x
+
s
e
c
−
1
√
2
x
2
−
x
4
+
c
o
s
e
−
1
√
2
x
2
−
x
4
=
0
Report Question
0%
π
2
0%
−
π
2
0%
2
π
0%
−
2
π
The sum of the solution of the equation
2
sin
−
1
√
x
2
+
x
+
1
+
cos
−
1
√
x
2
+
x
3
π
2
is
Report Question
0%
0
0%
−
1
0%
3
0%
2
If
tan
−
1
(
3
a
2
x
3
a
3
−
3
a
x
2
)
=
k
tan
−
1
(
x
a
)
, then
k
=
Report Question
0%
2
0%
3
0%
−
2
0%
4
If
c
o
s
−
1
√
p
+
c
o
s
−
1
√
1
−
p
+
c
o
s
−
1
√
1
−
q
=
3
π
4
then the value of q is equal to:
Report Question
0%
1
0%
1
√
2
0%
1
3
0%
1
2
The value of
sin
−
1
[
cot
(
sin
−
1
√
(
2
−
√
3
4
)
+
cos
−
1
√
12
4
)
+
sec
−
1
√
2
]
Report Question
0%
0
0%
π
4
0%
π
6
0%
π
2
If
t
a
n
−
1
(
s
i
n
2
θ
+
2
s
i
n
+
2
)
+
c
o
t
−
1
(
4
s
e
c
2
+
1
)
=
π
2
has solution for some
θ
and
ϕ
then
Report Question
0%
s
i
n
θ
=
−
1
0%
c
o
s
ϕ
=
−
1
0%
c
o
s
ϕ
=
1
0%
s
i
n
θ
=
1
The sum of roots of the equation
tan
−
1
1
1
+
2
x
+
tan
−
1
1
1
+
4
x
=
tan
−
1
2
x
2
is
Report Question
0%
7
3
0%
3
0%
4
0%
5
If
sin
−
1
x
+
sin
−
1
y
=
π
3
then the value of
cos
−
1
x
+
cos
−
1
y
is equal to which of the following :
Report Question
0%
π
6
0%
π
3
0%
2
π
3
0%
π
tan
−
1
(
x
y
)
−
tan
−
1
(
x
−
y
x
+
y
)
=
…
Report Question
0%
π
2
0%
π
3
0%
π
2
0%
π
4
or
3
π
4
If
sin
−
1
x
+
cos
−
1
y
=
2
π
5
,
then
cos
−
1
x
+
sin
−
1
y
is
Report Question
0%
2
π
5
0%
3
π
5
0%
4
π
5
0%
3
π
10
The value of x satisfying
s
i
n
−
1
x
+
s
i
n
−
1
(
1
−
x
)
=
c
o
s
−
1
x
are
Report Question
0%
0
0%
1/2
0%
1
0%
2
Solve
tan
−
1
(
cos
x
1
+
sin
x
)
.
Report Question
0%
π
4
+
x
2
0%
π
4
−
x
2
0%
π
4
+
x
0%
π
4
−
x
If
x
=
sin
−
1
(
sin
10
)
a
n
d
y
=
cos
−
1
(
cos
10
)
,
t
h
e
n
y
−
x
is equal to:
Report Question
0%
π
0%
0
0%
7
π
0%
10
If
x
=
sin
−
1
(
sin
10
)
a
n
d
y
=
cos
−
1
(
cos
10
)
Report Question
0%
0
0%
10
0%
7
π
0%
π
If
θ
ϵ
[
4
π
,
5
π
]
, then
c
o
s
−
1
(
c
o
s
θ
)
equals
Report Question
0%
−
4
π
+
θ
0%
5
π
−
θ
0%
4
π
−
θ
0%
θ
−
5
π
The solution set of the equation
s
i
n
−
1
√
1
−
x
2
+
c
o
s
−
1
x
=
c
o
t
−
1
(
√
1
−
x
2
x
)
−
s
i
n
−
1
x
is
Report Question
0%
[-1,1] - {0}
0%
(0,1] U {-1]
0%
[-1,0) U {1}
0%
[-1,1]
If
[
sin
−
1
(
cos
−
1
(
sin
−
1
(
tan
−
1
x
)
)
)
]
=
1
, where
[
∙
]
denotes the greatest integer function, then
x
∈
Report Question
0%
[
tan
sin
cos
1
,
tan
sin
cos
sin
1
]
0%
(
tan
sin
cos
1
,
tan
sin
cos
sin
1
)
0%
[-1, 1]
0%
None of these
Number of solutions of the equation
t
a
n
−
1
(
1
a
−
1
)
=
t
a
n
−
1
(
1
x
)
+
t
a
n
−
1
(
1
a
2
x
+
1
)
Report Question
0%
o
n
e
0%
T
w
o
0%
T
h
r
e
e
0%
Z
e
r
o
Solution of
tan
−
1
2
x
1
−
x
2
+
cot
−
1
1
−
x
2
2
x
=
2
π
3
are
Report Question
0%
1
√
3
0%
−
√
3
0%
√
3
+
2
0%
√
3
−
2
cos
−
1
(
cos
x
)
=
[
x
]
,
[
.
]
d
e
n
o
t
e
s
t
h
e
g
r
e
a
t
e
s
t
int
e
g
e
r
f
u
n
c
t
i
o
n
,
i
s
Report Question
0%
2
π
+
3
0%
π
+
3
0%
π
−
3
0%
2
π
−
3
If
α
is the only real root of the equation
x
3
+
b
x
2
+
c
=
0
(
b
<
c
) then the value of
T
a
n
1
α
+
T
a
n
1
(
1
α
)
=
Report Question
0%
π
2
0%
π
2
0%
0
0%
π
The number of solution for the equation
cos
−
1
(
1
−
x
)
+
m
cos
−
1
x
=
n
π
2
where
m
>
0
,
n
≤
0
is
Report Question
0%
0
0%
1
0%
2
0%
i
n
f
i
n
i
t
e
The value of
s
i
n
−
1
[
c
o
t
(
s
i
n
−
1
√
(
2
−
√
3
4
)
+
c
o
s
−
1
(
√
12
4
)
+
s
e
c
−
1
√
2
)
]
is :
Report Question
0%
0
0%
π
4
0%
π
6
0%
π
2
If
x
=
T
a
n
1
(
1
)
+
C
o
s
1
(
1
2
)
+
S
i
n
1
(
1
2
)
and
y
=
c
o
s
[
1
2
C
o
s
1
(
1
/
8
)
]
then
Report Question
0%
x
=
2
π
y
0%
y
=
3
π
x
0%
x
=
π
y
0%
y
=
π
x
Solve the equation :
4
tan
−
1
1
5
−
tan
−
1
1
239
=
?
Report Question
0%
π
0%
π
/
2
0%
π
/
3
0%
π
/
4
Let
f
(
x
)
=
sin
−
1
x
+
cos
−
1
x
⋅
Then
π
2
is equal to:
Report Question
0%
f(-2)
0%
f
(
k
2
−
2
k
+
3
)
,
k
∈
R
0%
f
(
1
1
+
k
2
)
,
k
∈
R
0%
none
If
s
i
n
−
1
x
+
s
i
n
−
1
y
=
2
π
3
, then
c
o
s
−
1
x
+
c
o
s
−
1
y
=
Report Question
0%
2
π
3
0%
π
3
0%
π
6
0%
π
If
t
a
n
−
1
2
x
+
t
a
n
−
1
3
x
=
π
4
then
x
=
Report Question
0%
−
1
0%
1
6
0%
−
1
,
1
6
0%
None of these
If
A
=
tan
−
1
(
x
√
3
2
k
−
x
)
and
B
=
tan
−
1
(
2
x
−
k
k
√
3
)
, then the value of A-B is
Report Question
0%
0
o
0%
45
o
0%
60
o
0%
30
o
If
S
i
n
1
(
x
−
x
2
2
+
x
3
4
−
.
.
.
.
.
.
.
)
+
C
o
s
1
(
x
2
−
x
4
2
+
x
6
4
−
.
.
.
.
.
.
)
=
π
2
for 0 <
|
x
|
<
√
2
then
x
=
Report Question
0%
1/2
0%
1
0%
1/2
0%
1
sin
cot
−
1
tan
cos
−
1
x
is always equal to
Report Question
0%
x
0%
√
1
−
x
2
0%
1
x
0%
None of these
3
t
a
n
−
1
x
=
t
a
n
−
1
{
3
x
−
x
3
1
−
3
x
2
}
,
t
h
e
n
x
b
e
l
o
n
g
t
o
Report Question
0%
[-1,1]
0%
(-1,1)
0%
{
−
1
√
3
,
1
√
3
}
0%
None of these
If
A
=
tan
−
1
(
x
√
3
2
k
−
x
)
and
B
=
tan
−
1
(
2
x
−
k
k
√
3
)
, then the value of
A
−
B
is
Report Question
0%
0
o
0%
45
o
0%
60
o
0%
30
o
Value of
sin
−
1
3
√
13
+
cos
−
1
11
√
146
+
cot
−
1
√
3
is
Report Question
0%
π
0%
π
/2
0%
5
π
/12
0%
π
/3
If
A
=
2
tan
−
1
(
2
√
2
−
1
)
and
B
=
3
sin
−
1
(
1
/
3
)
+
s
i
n
−
1
(
3
/
5
)
, then
Report Question
0%
A
=
B
0%
A
<
B
0%
A
>
B
0%
n
o
n
e
\of
\these
The least positive integer n for which
(
1
+
i
1
−
i
)
n
=
2
π
(
sec
−
1
1
x
+
sin
−
1
x
)
(
w
h
e
r
e
,
x
≠
0
,
−
1
≤
x
≥
1
a
n
d
i
=
√
−
1
)
, is
Report Question
0%
2
0%
4
0%
6
0%
8
If
2
s
i
n
−
1
(
3
5
)
−
c
o
s
−
1
(
5
13
)
=
c
o
s
−
1
(
λ
)
, then
λ
is eual to
Report Question
0%
323/225
0%
223/325
0%
323/325
0%
123/125
the number o real soluttion of
t
a
n
−
1
√
x
(
x
+
1
)
+
s
i
n
−
1
√
x
2
+
x
+
1
=
π
2
i
s
Report Question
0%
zero
0%
one
0%
two
0%
infinit
The numerical value of
c
o
t
(
2
s
i
n
−
1
3
5
+
c
o
s
−
1
3
5
)
is
Report Question
0%
−
4
3
0%
−
3
4
0%
3
4
0%
4
3
The value of
sin
−
1
(
sin
12
)
+
sin
−
1
(
cos
12
)
=
Report Question
0%
0
0%
24
−
2
π
0%
4
π
−
24
0%
8
π
cot
−
1
3
+
cot
−
1
7
+
cot
−
1
13
+
n
t
e
r
m
s
=
Report Question
0%
tan
−
1
(
n
n
+
2
)
0%
tan
−
1
(
n
+
2
n
)
0%
tan
−
1
(
3
n
4
n
+
1
)
0%
tan
−
1
(
4
n
+
1
3
n
)
If
θ
=
cot
−
1
√
cos
x
−
tan
−
1
√
cos
x
, then
sin
θ
=
Report Question
0%
tan
1
2
x
0%
tan
2
(
x
2
)
0%
1
2
tan
−
1
(
x
2
)
0%
none of these
If
sin
−
1
(
2
a
1
+
a
2
)
+
cos
−
1
(
1
−
a
2
1
+
a
2
)
=
tan
−
1
(
2
x
1
x
2
)
where,
a
,
x
ϵ
(
0
,
1
)
then the value of
x
is
Report Question
0%
0
0%
a
2
0%
a
0%
2
a
1
a
2
If
tan
−
1
(
√
1
+
x
2
−
√
1
−
x
2
√
1
+
x
2
+
√
1
−
x
2
)
=
α
,
(
α
∈
[
0
,
π
4
)
]
yhen
x
2
is equal to
Report Question
0%
cos
2
α
0%
tan
2
α
0%
sin
2
α
0%
None of these
If If
x
2
+
y
2
+
z
2
=
r
2
,
then
t
a
n
−
1
(
x
y
z
r
)
+
tan
−
1
(
y
z
x
r
)
+
tan
−
1
(
x
z
y
r
)
=
Report Question
0%
π
0%
π
2
0%
0
0%
none of these.
If
c
o
s
−
1
x
−
c
o
s
−
1
y
2
=
α
, then
4
x
2
−
4
x
y
c
o
s
α
+
y
2
is equal to -
Report Question
0%
2
s
i
n
2
α
0%
4
0%
4
s
i
n
2
α
0%
−
4
s
i
n
2
α
If a is a real of the equation
x
3
+
3
x
−
t
a
n
2
=
0
then
c
o
t
−
1
a
+
c
o
t
−
1
1
a
−
x
2
can be equal to
Report Question
0%
0
0%
π
2
0%
π
0%
3
π
2
If
cos
−
1
(
x
/
a
)
+
cos
−
1
(
y
/
b
)
=
α
,
Then
x
2
/
a
2
+
y
2
/
b
2
is equal to:
Report Question
0%
(
2
x
y
/
a
b
)
cos
α
+
sin
2
α
0%
(
2
x
y
/
a
b
)
sin
α
+
cos
2
α
0%
(
2
x
y
/
a
b
)
cos
2
α
+
sin
α
0%
(
2
x
y
/
a
b
)
sin
2
α
+
cos
α
If
s
i
n
−
1
(
√
x
2
)
+
s
i
n
−
1
(
√
1
−
x
4
)
+
t
a
n
−
1
y
=
2
π
3
then
Report Question
0%
maximum value of
x
2
+
y
2
is
49
3
0%
maximum value of
x
2
+
y
2
is 4
0%
maximum value of
x
2
+
y
2
is
1
2
0%
maximum value of
x
2
+
y
2
is 3
0:0:2
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Practice Class 12 Commerce Maths Quiz Questions and Answers
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