Prove that the function f(x) = 3sin(2x) - cos(2x) + 4 is one-one in the interval ![[-\frac{\pi }{6} , \frac{\pi }{3} ]](https://tex.z-dn.net/?f=%5B-%5Cfrac%7B%5Cpi+%7D%7B6%7D+%2C+%5Cfrac%7B%5Cpi+%7D%7B3%7D+%5D "[-\frac{\pi }{6} , \frac{\pi }{3} ]")
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Answer:
trigonometric function inequalities Sine integral Catalan ... \frac{\sin t }{t}\geq\frac{\pi^{2}-t^{2}}{\pi^{2}+t^{2}}, \quad t\ in ( 0 ... 2^{4n} ( n-1 ) ( n+1 ) }>0\quad\text{for } n\geq4, ...
Answered by
27
[
6
−π
,
3
π
]
f:X→[2,6]
f(x)=
3
sin2x−cos2x+4
f(x)=2×
2
3
sin2x−2×
2
1
cos2x+4
f(x)=2cos
6
π
sin2x−2sin
6
π
cos2x+4
f(x)=2(sin2xcos
6
π
−sin
6
π
cos2x)+4
f(x)=2[sin(2x−
6
π
)]+4
∵f(x)is one-one and onto function.
if f(x)=2
∴2[sin(2x−
6
π
)]+4=2
2[sin(2x−
6
π
)]=−2
sin(2x−
6
π
)=−1
(2x−
6
π
)=sin
−1
(−1)
2x−
6
π
=
2
−π
2x=
2
−π
+
3
π
2x=
3
−π
⇒x=
6
−π
if f(x)=6
2[sin(2x−
6
π
)]+4=6
sin(2x−
6
π
)=1
2x−
6
π
=
2
π
x=
3
π
∴xϵ[
6
−π
,
3
π
]
hope it helps you
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