JEE Main Mathematics Question (2026) — Solution

Step-by-step solution

For the logarithms to be defined, we must satisfy the following conditions: 1. Base of the first logarithm: x+1 > 0 x > -1 and x+1 1 x 0. 2. Base of the second logarithm: 2x+3 > 0 x > - 3 2 and 2x+3 1 x -1. 3. Arguments must be positive: 2x^2+5x+3 > 0 and x^2+2x+1 > 0. Taking the intersection of all these conditions, the domain of the equation is x (-1, 0) (0, ). Now, factorizing the arguments of the logarithms: 2x^2+5x+3 = (2x+3)(x+1) x^2+2x+1 = (x+1)^2 Substitute these into the given equation: _ (x+1) ((2x+3)(x+1)) = 4 - _ (2x+3) ((x+1)^2) Using the properties of logarithms: _ (x+1) (2x+3) + _ (x+1) (x+1) = 4 - 2 _ (2x+3) (x+1) _ (x+1) (2x+3) + 1 = 4 - 2 _ (x+1) (2x+3) Let t = _ (x+1) (2x+3). The equation becomes: t + 1 = 4 - 2 t t - 3 + 2 t = 0 t^2 - 3t + 2 = 0 (t-1)(t-2) = 0 t = 1 or t = 2 Case 1: t = 1 _ (x+1) (2x+3) = 1 2x+3 = x+1 x = -2 This value is rejected because x = -2 does not fall in the domain x > -1. Case 2: t = 2 _ (x+1) (2x+3) = 2 2x+3 = (x+1)^2 2x+3 = x^2+2x+1 x^2 = 2 x = 2 Since x > -1, x = - 2 is rejected. The only valid solution is x = 2 . The sum of squares of all the real solutions is ( 2 )^2 = 2. Answer: 2