JEE Advanced Model Paper 16 with solutions for online practice

a = 20 - 6π , b = 10π - 30 , c = sin−1 sin (4π - 10) = 10 - 3π
So a + b + c = π
f(x) = sin (2 (a + b + c) x) + cos2 ((a + b + c)x)
= sin 2πx + cos2 πx
= sin 2πx +

1+cos2πx

2

⇒ maximum value of f(x) =

√5+1

2