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Byju's Answer
Standard XII
Mathematics
Integration by Substitution
A=sin8θ+cos14...
Question
A
=
sin
8
θ
+
cos
14
θ
, then for all values of
θ
A
A
≤
1
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B
1
<
2
A
≤
3
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C
A
≥
1
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D
0
≤
A
≤
1
2
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Solution
The correct option is
A
A
≤
1
s
i
n
8
θ
≤
s
i
n
2
θ
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
(
1
)
c
o
s
14
θ
≤
c
o
s
2
θ
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
(
2
)
Adding equation (1) and (2):
A
≤
1
Suggest Corrections
0
Similar questions
Q.
If
A
=
s
i
n
8
θ
+
c
o
s
14
θ
,
then for all real values of
θ
Q.
Let
A
=
sin
8
θ
+
cos
14
θ
;
then
∀
real
θ
Q.
If
A
=
sin
8
θ
+
cos
14
θ
, then for all values of
θ
Q.
Show that for all real values of
θ
, the expression a
sin
2
θ
+
b
sin
θ
cos
θ
+
c
cos
2
θ
lies between
1
2
(
a
+
c
)
−
1
2
√
b
2
+
(
a
−
c
)
2
1
2
(
a
+
c
)
+
1
2
√
b
2
+
(
a
−
c
)
2
Q.
Show that for all real values of
θ
, the expansion
a
sin
2
θ
+
b
sin
θ
cos
θ
+
c
cos
2
θ
lies between
1
2
(
a
+
c
)
−
1
2
√
b
2
+
(
a
−
c
)
2
and
1
2
(
a
+
c
)
+
1
2
√
b
2
+
(
a
−
c
)
2
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