JEE Main Mathematics Question (2026) — Solution

Step-by-step solution

Let x - 2 = t. As x 2, t 0. The given limit can be written as: _ t 0 ((t+2)^3 - 5(t+2)^2 + a(t+2) + b) ( t+1 - 1) _e(t+1) = m The denominator can be approximated for small t as: _ t 0 t+1 - 1 t _e(t+1) t t^2 = 1 2 1 t^2 = t^2 2 For the limit to be finite, the argument of the sine function in the numerator must have t^2 as its lowest degree term in its expansion. Let P(t) = (t+2)^3 - 5(t+2)^2 + a(t+2) + b P(t) = t^3 + 6t^2 + 12t + 8 - 5(t^2 + 4t + 4) + at + 2a + b P(t) = t^3 + t^2 + (a - 8)t + (2a + b - 12) For the limit to exist, the constant and linear terms must be zero: a - 8 = 0 a = 8 2a + b - 12 = 0 16 + b - 12 = 0 b = -4 Substituting a and b back into P(t): P(t) = t^3 + t^2 The limit becomes: m = _ t 0 (t^3 + t^2) t^2 / 2 m = _ t 0 (t^2(t+1)) t^2(t+1) t^2(t+1) t^2 / 2 = 1 2 = 2 Therefore, a + b + m = 8 - 4 + 2 = 6. Answer: 6