If $\lim _{x \rightarrow 0} \frac{\alpha e^{x}+\beta e^{-x}+\gamma \sin x}{x \sin ^{2} x}=\frac{2}{3}$, where $\alpha, \beta, \gamma \in \mathrm{R}$, then which of the following is NOT correct ?

(C) $\alpha \beta^{2}+\beta \gamma^{2}+\gamma \alpha^{2}+3=0$

(A) $\alpha^{2}+\beta^{2}+\gamma^{2}=6$

(B) $\alpha \beta+\beta \gamma+\gamma \alpha+1=0$

(D) $\alpha^{2}-\beta^{2}+\gamma^{2}=4$

JEE Main · 2022 · 29 Jul 2022 · Shift 1 · Limits

lim_{x \to 0} \frac{α e ^{x} + β e ^{- x} + γ s i n x}{x s i n ^{2} x} = \frac{2}{3}

,
where

α, β, γ \in R

, then which of the following is NOT correct ?

Answer: (C)

Expanding using Taylor series and comparing coefficients for the limit to exist:

For constant terms:

α + β = 0

For

x

terms:

α - β + γ = 0

For

x^{3}

coefficient:

\frac{α - β - γ}{6} = \frac{2}{3}

Solving:

α = 1, β = - 1, γ = - 2

Checking options:
(A)

1 + 1 + 4 = 6

✓
(B)

- 1 + 2 - 2 + 1 = 0

✓
(C)

- 1 - 4 - 2 + 3 = - 4 \neq = 0

✗
(D)

1 - 1 + 4 = 4

✓

Option (C) is NOT correct.

JEE Main · 2022 · 29 Jul 2022 · Shift 1 · Single correct option