If limx→0αex+βe−x+γsin⁡xxsin2⁡x=23 where α,β,γ∈R, then which of the following is/are correct

α 2 + β 2 + γ 2 = 6 , α β 2 + β γ 2 + γ α 2 + 3 = 0 , α β + β γ + γ α + 1 = 0

If limx→0αex+βe−x+γsin⁡xxsin2⁡x=23 where α,β,γ∈R, then which of the following is/are correct

lim x → 0 α 1 + x 1 ! + x 2 2 ! + x 3 3 ! + − − + β 1 − x 1 ! + x 2 2 ! − x 3 3 ! + − − − + γ x − x 3 3 ! + x 5 5 ! − − x 3 α + β = 0 , α − β + γ = 0 and lim x → 0 x 3 α 3 ! − β 3 ! − γ 3 ! + x 4 − α 3 ! − β 3 ! − γ 3 ! x 3 = 2 3 α 6 − β 6 − γ 6 = 2 3 ⇒ α - β - γ = 4 ⇒ - 2 γ = 4 ⇒ γ = - 2 ∴ α + β = 0 , α - β = 2 ⇒ α = 1 , β = − 1 , γ = − 2 α 2 + β 2 + γ 2 = 6 , α 2 - β 2 + γ 2 = 4 , α β + β γ + γ α = - 1 + 2 - 2 = - 1

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If $\lim_{x \to 0} \frac{α e^{x} + β e^{- x} + γ \sin x}{x \sin^{2} x} = \frac{2}{3}$ where $α, β, γ \in R$ , then which of the following is/are correct

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$α^{2} + β^{2} + γ^{2} = 6$

$α β^{2} + β γ^{2} + γ α^{2} + 3 = 0$

$α^{2} - β^{2} + γ^{2} = 4$

$α β + β γ + γ α + 1 = 0$

answer is A, B, D.

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Detailed Solution

$\lim_{x \to 0} \frac{α (1 + \frac{x}{1!} + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + - -) + β (1 - \frac{x}{1!} + \frac{x^{2}}{2!} - \frac{x^{3}}{3!} + - - -) + γ (x - \frac{x^{3}}{3!} + \frac{x^{5}}{5!} - -)}{x^{3}}$

$α + β = 0, α - β + γ = 0$

and $\lim_{x \to 0} \frac{x^{3} (\frac{α}{3!} - \frac{β}{3!} - \frac{γ}{3!}) + x^{4} (- \frac{α}{3!} - \frac{β}{3!} - \frac{γ}{3!})}{x^{3}} = \frac{2}{3}$

$\begin{array}{l} \frac{α}{6} - \frac{β}{6} - \frac{γ}{6} = \frac{2}{3} \Rightarrow α - β - γ = 4 \\ \Rightarrow - 2 γ = 4 \Rightarrow γ = - 2 \\ ∴ α + β = 0, α - β = 2 \\ \Rightarrow α = 1, β = - 1, γ = - 2 \\ α^{2} + β^{2} + γ^{2} = 6, α^{2} - β^{2} + γ^{2} = 4, α β + β γ + γ α = - 1 + 2 - 2 = - 1 \end{array}$