JEE Main 12 Jan 2019 Shift 2 Solved Paper

A.M. ≥ G.M.

sin4α+4cos4β+1+1

4

≥ (sin4α.4cos4β.1,1)

1

4

sin4α+4cos2β + 2 ≥ 4√2 sin α cos β
given the sin4α+4cos4β + 2 = 4√2 sin α cos β
⇒ A.M. = G.M. ⇒ sin4 α = 1 = 4 cos4 β
sin α = 1 , cos β = ±

1

√2

⇒ sin β =

1

√2

and β ∊ [0 , π]

cos (α + β) - cos (α - β) = - 2 sin α sin β
= - √2