Prove that the function f(x) = 3sin(2x) – cos(2x) = 4 is one one in the interval [tex][-\frac{\pi }{6} , \frac{\pi }{3} ][/tex] About the author Brielle
Answer: Correct option is C [ 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 ] Reply
Answer:
Correct option is
C
[
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
]