JEE Advanced Mathematics Question (2023) — Solution

Step-by-step solution

Given, x 2 - 5 d y d x - 2 x y = - 2 x x 2 - 5 2 ⇒ d y d x + - 2 x x 2 - 5 y = - 2 x x 2 - 5 Which is a linear differential equation, So, integrating factor,  I.F. = e ∫ - 2 x x 2 - 5 d x = 1 x 2 − 5 Now solution of differential equation is given by, y ⋅ 1 x 2 − 5 = ∫ − 2 x ⋅ x 2 − 5 x 2 − 5 d x ⇒ y x 2 − 5 = x 2 − 5 x 2 − 5 − x 2 + C Now using the given value,  y ( 2 ) = 7 ⇒ C = 3 ⇒ y = - x 2 x 2 - 5 + 3 x 2 - 5 ⇒ y = f ( x )  is even function If 0 < x < 5 ,   y = - x 4 + 5 x 2 - 3 x 2 + 15 ⇒ y = - x 4 + 2 x 2 + 15 For increasing function d y d x > 0 ⇒ d y d x = - 4 x 3 + 4 x > 0 ⇒ x < 1 If x > 5 ,   y = - x 4 + 5 x 2 + 3 x 2 - 15 ⇒ y = - x 4 + 8 x 2 - 15 For increasing function d y d x > 0 ⇒ d y d x = - 4 x 3 + 16 x > 0 ⇒ - 4 x x 2 - 4 > 0 ⇒ 4 x x 2 - 4 < 0 ⇒ x ∈ - ∞ , - 2 ∪ 0 , 2  but  x > 5 ⇒ x = ϕ ⇒ y ( x ) is increasing over  ( 0 , 1 ) Now plotting the graph we get, Hence, maximum value of the function is given by, ⇒ f ( x ) max = 16