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Byju's Answer
Standard IX
Mathematics
Inequality
For real a, b...
Question
For real a, b and x
−
√
(
a
2
+
b
2
)
≤
a
s
i
n
x
+
b
c
o
s
x
≤
√
(
a
2
+
b
2
)
A
True
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B
False
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Solution
Put
a
=
r
c
o
s
α
,
b
=
r
s
i
n
α
where
r
=
√
(
a
2
+
b
2
)
∴
a
s
i
n
x
+
b
c
o
s
x
=
r
s
i
n
(
x
+
α
)
since
−
1
≤
s
i
n
θ
≤
1
therefore
−
r
≤
s
i
n
(
x
+
α
)
≤
r
etc
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