Question

Prove that the function f(x) = 3sin(2x) - cos(2x) = 4 is one one in the interval $[-\frac{\pi }{6} , \frac{\pi }{3} ]$

Answer 1

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

]