If limx→0αex+βe−x+γsinxxsin2x=23, where α, β, γ∈R, then which of the following is NOT correct?

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If $\lim_{x \to 0} \frac{{αe}^{x} + {βe}^{- x} + γsinx}{{xsin}^{2} x} = \frac{2}{3},$ where $α, β, γ \in R,$ then which of the following is NOT correct?

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$αβ + βγ + γα + 1 = 0$

$α^{2} + β^{2} + γ^{2} = 6$

$α^{2} - β^{2} + γ^{2} = 4$

${αβ}^{2} + {βγ}^{2} + {γα}^{2} + 3 = 0$

answer is C.

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Detailed Solution

$\begin{array}{l} lim_{x \to 0} \frac{{αe}^{x} + {βe}^{- x} + γsin x}{x^{3} {(\frac{\sin x}{x})}^{2}} = \frac{2}{3} \\ \Rightarrow lim_{x \to 0} \frac{{αe}^{x} + {βe}^{- x} + γsin x}{x^{3}} = \frac{2}{3} \\ ∴ α + β = 0 \dots \dots (1) (\frac{0}{0} f 0 rm) \\ lim_{x \to 0} \frac{{αe}^{x} - {βe}^{- x} + γcos x}{3 x^{2}} = \frac{2}{3} \\ ∴ α - β + γ = 0 \dots (2) (\frac{0}{0} form) \\ lim_{x \to 0} \frac{{αe}^{x} + {βe}^{- x} - γsin x}{6 x} = \frac{2}{3} \\ ∴ α + β = 0 (\frac{0}{0} form) \\ lim_{x \to 0} \frac{{αe}^{x} - {βe}^{- x} - γcos x}{6} = \frac{2}{3} \\ \Rightarrow \frac{α - β - γ}{6} = \frac{2}{3} \Rightarrow α - β - γ = 4 . \\ Solving (1), (2),(3) \\ γ = - 2, α = 1, β = - 1 . \end{array}$