JEE Main Mathematics Question (2026) — Solution

Step-by-step solution

Using the standard expansions for small x, 1 - ^2 2 and . The given limit can be written as: _ x 0 1 - (1 - ^2 x^2 2 ) (1 - ( +1)^2 x^2 2 ) (1 - ( +2)^2 x^2 2 ) (( +1)x)^2 Neglecting higher powers of x, the numerator simplifies to: 1 - (1 - x^2 2 ( ^2 + ( +1)^2 + ( +2)^2 ) ) = x^2 2 ( ^2 + ( +1)^2 + ( +2)^2 ) Substituting this back into the limit: _ x 0 x^2 2 ( ^2 + ( +1)^2 + ( +2)^2 ) ( +1)^2 x^2 = ^2 + ( +1)^2 + ( +2)^2 2( +1)^2 We are given that this limit is equal to 2. Therefore: ^2 + ( +1)^2 + ( +2)^2 2( +1)^2 = 2 ^2 + ( +1)^2 + ( +2)^2 = 4( +1)^2 Let y = +1. Then = y-1 and +2 = y+1. The equation becomes: (y-1)^2 + y^2 + (y+1)^2 = 4y^2 y^2 - 2y + 1 + y^2 + y^2 + 2y + 1 = 4y^2 3y^2 + 2 = 4y^2 y^2 = 2 Substituting y = +1 back into the equation: ( +1)^2 = 2 ^2 + 2 + 1 = 2 ^2 + 2 - 1 = 0 The roots of this quadratic equation represent all possible values of . The product of the roots is given by c a : Product = -1 1 = -1 Answer: -1