JEE Mains Model Paper 2

We have tan2x - tan4x + tan6x - tan8x + .....
=

tan2x

1−(−tan2x)

=

tan2x

secx

= sin2x
∴ y = exp {(tan2x - tan4x + tan6x - tan8x + .....) loge16}
= exp {(sin2x) loge16} = exp {log3(16sin2x)}
= 16sin2x
As y satisfies x3 -3x + 2 = 0 we get
y = 1 or y = 2
⇒ 16sinx = 1 or 16sin2x = 2
Since 0 < x < π/4 , 0 < sinx < 1/2
⇒ 0 < sin2x < 1/2
∴ 16sin2x = 1 is not possible
Thus, 16sin2x = 2
⇒ sin2x = 1/4
Thus cos2x+cos4x = (1 - sin2x) + (1−sin2x)2 = 21/16