If limx-0 x -x 1 - 0 1-x - 1 0 1-x-5- where - are

The correct option is D αϵ R , β=1 , γ=5 α #160; x #160; amp; 1 amp; γ #160; 0 amp; 1 x #160; amp; β #160; ...

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Byju's Answer

Standard XII

Mathematics

Factorization Method Form to Remove Indeterminate Form

If limx→0 x...

Question

If $lim x \to 0 x ∣ ∣ ∣ ∣ ∣ \begin{matrix} α / x & 1 & γ 0 & 1 / x & β 1 & 0 & 1 / x \end{matrix} ∣ ∣ ∣ ∣ ∣ = - 5$ , where $α, β, γ$ are

α = 2

β = 1

γ ϵ R

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α = 2

β = 2

γ = - 5

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α ϵ R

β = 1

γ ϵ R

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α ϵ R

β = 1

γ = 5

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Solution

The correct option is D $α ϵ R$ , $β = 1$ , $γ = 5$
$∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} \frac{α}{x} & 1 & γ 0 & \frac{1}{x} & β 1 & 0 & \frac{1}{x} \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ = \frac{α}{x} (\frac{1}{x^{2}} = 0) - 1 (- β) + γ (- \frac{1}{x})$
$= \frac{α}{x^{3}} + β - \frac{γ}{x}$
${lim}_{x \to 0} x (\frac{α}{x^{3}} + β - \frac{γ}{x})$
$= {lim}_{x \to 0} x (\frac{α}{x^{3}} + β - \frac{γ}{x}) = {lim}_{x \to 0} x (\frac{α + β x^{3} - γ x^{2}}{x^{2}})$
Givin limit exists
$\Rightarrow α + β (0) - γ (0) = 0$
${lim}_{x \to 0} \frac{3 β x^{2} - 2 γ x}{2 x} = {lim}_{x \to 0} \frac{3 β x - 2 γ}{2} = - 5$
$\frac{0 - 2 γ}{2} = - 5 \Rightarrow γ = 5$

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If limx→0 x∣∣ ∣ ∣∣α/x1γ01/xβ101/x∣∣ ∣ ∣∣=−5, where α,β,γ are

If $lim x \to 0 x ∣ ∣ ∣ ∣ ∣ \begin{matrix} α / x & 1 & γ 0 & 1 / x & β 1 & 0 & 1 / x \end{matrix} ∣ ∣ ∣ ∣ ∣ = - 5$ , where $α, β, γ$ are