If limx→0 αex+βe−x+γsin xxsin2⁡ x=23where α,β,γ∈R, then which of the following is NOT correct?

limx→0αex+βe-x+γsinxxsin2x=23 L.H.S. =limx→0 α1+x+x22!+…+β1−x+x22!−x33!+…+γx−x33!+…limx→0 x⋅x2limx→0 sin2⁡xx2=limx→0 (α+β)+(α−β+γ)x+α2+β2x2+α6−β6−γ6x3x3⋅1+0For limit to be finite, we should haveα+β=0,α−β+γ=0,α2+β2=0,α6−β6−γ6=23⇒α=−β,−β−β+γ=0⇒γ=2β∴ −β6−β6−2β6=23⇒−4β=4⇒β=−1∴ α=1,β=−1,γ=−2(a) α2+β2+γ2=1+1+4=6 (b) αβ+βγ+γα+1=−1+2−2+1=0 (c) αβ2+βγ2+γα2+3=1−4−2+3=−2≠0 (d) α2−β2+γ2=1−1+4=4

If limx→0 αex+βe−x+γsin xxsin2⁡ x=23where α,β,γ∈R, then which of the following is NOT correct?

limx→0αex+βe-x+γsinxxsin2x=23 L.H.S. =limx→0 α1+x+x22!+…+β1−x+x22!−x33!+…+γx−x33!+…limx→0 x⋅x2limx→0 sin2⁡xx2=limx→0 (α+β)+(α−β+γ)x+α2+β2x2+α6−β6−γ6x3x3⋅1+0For limit to be finite, we should haveα+β=0,α−β+γ=0,α2+β2=0,α6−β6−γ6=23⇒α=−β,−β−β+γ=0⇒γ=2β∴ −β6−β6−2β6=23⇒−4β=4⇒β=−1∴ α=1,β=−1,γ=−2(a) α2+β2+γ2=1+1+4=6 (b) αβ+βγ+γα+1=−1+2−2+1=0 (c) αβ2+βγ2+γα2+3=1−4−2+3=−2≠0 (d) α2−β2+γ2=1−1+4=4

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

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$α^{2} + β^{2} + γ^{2} = 6$

$αβ + βγ + γα + 1 = 0$

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

$α^{2} - β^{2} + γ^{2} = 4$

answer is C.

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Detailed Solution

$\lim_{x \to 0} \frac{{αe}^{x} + {βe}^{- x} + γsinx}{{xsin}^{2} x} = \frac{2}{3}$
$L.H.S. = \frac{lim_{x \to 0} α (1 + x + \frac{x^{2}}{2!} + \dots) + β (1 - x + \frac{x^{2}}{2!} - \frac{x^{3}}{3!} + \dots) + γ (x - \frac{x^{3}}{3!} + \dots)}{lim_{x \to 0} (x \cdot x^{2}) lim_{x \to 0} \frac{\sin^{2} x}{x^{2}}}$
$= lim_{x \to 0} \frac{(α + β) + (α - β + γ) x + (\frac{α}{2} + \frac{β}{2}) x^{2} + (\frac{α}{6} - \frac{β}{6} - \frac{γ}{6}) x^{3}}{x^{3} \cdot 1} + 0$
For limit to be finite, we should have
$\begin{array}{l} α + β = 0, α - β + γ = 0, \frac{α}{2} + \frac{β}{2} = 0, \frac{α}{6} - \frac{β}{6} - \frac{γ}{6} = \frac{2}{3} \\ \Rightarrow α = - β, - β - β + γ = 0 \Rightarrow γ = 2 β \\ ∴ \frac{- β}{6} - \frac{β}{6} - \frac{2 β}{6} = \frac{2}{3} \Rightarrow - 4 β = 4 \Rightarrow β = - 1 \\ ∴ α = 1, β = - 1, γ = - 2 \end{array}$
$(a) α^{2} + β^{2} + γ^{2} = 1 + 1 + 4 = 6 (b) αβ + βγ + γα + 1 = - 1 + 2 - 2 + 1 = 0 (c) {αβ}^{2} + {βγ}^{2} + {γα}^{2} + 3 = 1 - 4 - 2 + 3 = - 2 \neq 0 (d) α^{2} - β^{2} + γ^{2} = 1 - 1 + 4 = 4$