VITEEE 2017 Paper

f(x) = 2sinx + sin2x
f ’ (x) = 2 cos x + 2 cos 2x = 2 (cos x + cos 2x)
∴ f'(x) = 0 ⇒ 2cos2x + cosx - 1 = 0
cos x =

−1±3

4

= - 1 ,

1

2

∴ x = π,

π

3

Now , f (0) = 0 , f (

3π

2

) = - 2
f (π) = 0 , f (

π

3

) = 2

√3

2

+

√3

2

=

3√3

2

∴ difference between greatest value and least value
=

3√3

2

+ 2