JEE Main202431 Jan 2024Evening ShiftMathematicsLimitsActual
If lim x → 0 a x 2 e x − b log e 1 + x + c x e − x x 2 sin x = 1 , then 16 a 2 + b 2 + c 2 is equal to ______.
Correct answer
0
Step-by-step solution
Given: lim x → 0 a x 2 e x - b log e 1 + x + c x e - x x 2 sin x = 1 ⇒ lim x → 0 a x 2 e x - b log e 1 + x + c x e - x x 2 · x · sin x x = 1 ⇒ lim x → 0 a x 2 1 + x 1 ! + x 2 2 ! + . . . - b x - x 2 2 + x 3 3 + . . . + c x 1 - x 1 ! + x 2 2 ! + . . . x 2 × x = 1 Coefficient of x = 0 ⇒ - b + c = 0 . . . i i i Coefficient of x 2 = 0 ⇒ a + b 2 - c = 0 . . . i i ⇒ a - c 2 = 0 ⇒ a = c 2 Coefficient of x 3 = 1 ⇒ a - b 3 + c 2 = 1 . . . i i i ⇒ c 2 - c 3 + c 2 = 1 ⇒ 2 c 3 = 1 ⇒ c = 3 2 = b ⇒ a = 3 4 ⇒ 16 a 2 + b 2 + c 2 =