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Byju's Answer
Standard XII
Mathematics
Domain
Domain of s...
Question
Domain of
sin
(
cos
θ
)
is ......
A
[
−
π
2
,
π
2
]
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B
R
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C
[
0.
π
]
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D
[
−
1
,
1
]
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Solution
The correct option is
C
R
The domain of
sin
(
cos
θ
)
The domain of
cos
is
R
The range of
cos
is
[
−
1
,
1
]
⟹
The domain of
sin
(
cos
θ
)
is
R
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0
Similar questions
Q.
Assertion :The area bounded by the curve
y
=
sin
−
1
x
& the line x=0 &
|
y
|
=
π
2
is
√
2
square units. Reason: The domain & principal value branch of
y
=
sin
−
1
x
are [-1,1] &
[
−
π
2
,
π
2
]
respectively
Q.
Let
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:
(
−
1
,
1
)
→
B
, be a function defined by
f
(
x
)
=
t
a
n
−
1
2
x
1
−
x
2
, then f is both one - one and onto when B is the interval
Q.
The equation
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=
|
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|
will have atleast one solution if
a
∈
Q.
Verify Rolle's theorem for each of the following functions on the indicated intervals
(i) f(x) = cos 2 (x − π/4) on [0, π/2]
(ii) f(x) = sin 2x on [0, π/2]
(iii) f(x) = cos 2x on [−π/4, π/4]
(iv) f(x) = e
x
sin x on [0, π]
(v) f(x) = e
x
cos x on [−π/2, π/2]
(vi) f(x) = cos 2x on [0, π]
(vii) f(x) =
sin
x
e
x
on 0 ≤ x ≤ π
(viii) f(x) = sin 3x on [0, π]
(ix) f(x) =
e
1
-
x
2
on [−1, 1]
(x) f(x) = log (x
2
+ 2) − log 3 on [−1, 1]
(xi) f(x) = sin x + cos x on [0, π/2]
(xii) f(x) = 2 sin x + sin 2x on [0, π]
(xiii)
f
x
=
x
2
-
sin
π
x
6
on
[
-
1
,
0
]
(xiv)
f
x
=
6
x
π
-
4
sin
2
x
on
[
0
,
π
/
6
]
(xv) f(x) = 4
sin
x
on [0, π]
(xvi) f(x) = x
2
− 5x + 4 on [1, 4]
(xvii) f(x) = sin
4
x + cos
4
x on
0
,
π
2
(xviii) f(x) = sin x − sin 2x on [0, π]