JEE Advanced Mathematics Question (2022) — Solution

Step-by-step solution

β = lim x → 0 e x 3 - 1 - x 3 1 3 + 1 - x 2 1 2 - 1 sin x x · sin 2 x x 2 · x 2 β = lim x → 0 1 + x 3 + x 6 2 ! + ⋯ - 1 - 1 3 x 3 + - 1 9 x 6 + ⋯ + - 1 2 x 2 - 1 8 x 4 + ⋯ x - x 3 3 ! + ⋯ x 3 β = lim x → 0 x 3 1 + 1 3 - 1 2 x 3   (Neglecting higher powers of x ) β = 5 6   i.e.  6 β = 5