JEE Advanced Mathematics Question (2022) — Solution

Step-by-step solution

Given, d y d x + 12 y = cos π x 12 I . F = e 12 x Now solution of the differential equation is given by, ⇒ y · e 12 x = ∫ e 12 x · cos π x 12 d x + C ⇒ y · e 12 x = e 12 x 12 2 + π 12 2 12 cos π x 12 + π 12 sin π x 12 + C Now given curve is passing through  0 , 0 So,  y 0 = 0 ⇒ C = - 12 12 2 + π 12 2 So y = 1 λ 12 cos π x 12 + π 12 sin π x 12 ⏟ f 1 x - 12 e - 12 x Where  λ = 1 12 2 + π 12 2 Now on differentiating we get, d y d x = 1 λ ' - π sin π x 12 + π 2 12 2 cos π x 12 ⏟ f 2 x + 12 e - 12 x where   λ ' = 12 12 2 + π 12 2 Now when x is large then 12 e - 12 x tends to zero. But f 2 x varies in - π 2 + π 12 4   ,   π 2 + π 12 4   by using  - a 2 + b 2   &   a 2 + b 2 Hence d y d x is changing its sign. So y x is non monotonic for all real number. Also when x is very large then again - 12 e - 12 x is almost zero but f 1 x is periodic, so there exist some β for which y = β intersect y = y x at infinitely many points.