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

limx→0α1+x1!+x22!+x33!+−−+β1−x1!+x22!−x33!+−−−+γx−x33!+x55!−−x3α+β=0,α−β+γ=0and limx→0x3α3!−β3!−γ3!+x4−α3!−β3!−γ3!x3=23α6−β6−γ6=23⇒α-β-γ=4⇒-2γ=4⇒γ=-2∴α+β=0,α-β=2⇒α=1,β=−1,γ=−2α2+β2+γ2=6,α2-β2+γ2=4,αβ+βγ+γα=-1+2-2=-1

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

limx→0α1+x1!+x22!+x33!+−−+β1−x1!+x22!−x33!+−−−+γx−x33!+x55!−−x3α+β=0,α−β+γ=0and limx→0x3α3!−β3!−γ3!+x4−α3!−β3!−γ3!x3=23α6−β6−γ6=23⇒α-β-γ=4⇒-2γ=4⇒γ=-2∴α+β=0,α-β=2⇒α=1,β=−1,γ=−2α2+β2+γ2=6,α2-β2+γ2=4,αβ+βγ+γα=-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 $\lim_{x \to 0} \frac{x^{3} (\frac{α}{3!} - \frac{β}{3!} - \frac{γ}{3!}) + x^{4} (- \frac{α}{3!} - \frac{β}{3!} - \frac{γ}{3!})}{x^{3}} = \frac{2}{3}$

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