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If \lim _{x \rightarrow 0} \frac{1}{x^2}\left(e^{\alpha x}-e^x-x\right)=\frac{3}{2} then the value of 1004|\alpha| must be 

 

Option: 1

2010


Option: 2

2008

 


Option: 3

1020

 


Option: 4

1012


Answers (1)

best_answer

 

\\\text { Let } P=\lim _{x \rightarrow 0}\left(\frac{e^{\alpha x}-e^x-x}{x^2}\right)\: \: \: \: \: \: \: \: \: \: \: ...(i)\\ \\\text{Replacing x by -x in Eq.(i) then}\\ \\P=\lim _{x \rightarrow 0}\left(\frac{e^{-\alpha x}-e^{-x}+x}{x^2}\right)........(ii)\\

Adding, Eqs. (i) and (ii), then 

\begin{aligned} 2 P & =\lim _{x \rightarrow 0}\left(\frac{e^{\alpha x}+e^{-\alpha x}-\left(e^x+e^{-x}\right)}{x^2}\right) \\ \\& =\lim _{x \rightarrow 0}\left(\frac{\left(e^{\alpha x}+e^{-\alpha x}-2\right)-\left(e^x+e^{-x}-2\right)}{x^2}\right) \\ \\& =\lim _{x \rightarrow 0}\left\{\frac{e^{-\alpha x}\left(e^{2 \alpha x}-2 e^{\alpha x}+1\right)-e^{-x}\left(e^{2 x}-2 e^x+1\right)}{x^2}\right. \end{aligned}

          \begin{aligned} & =\lim _{x \rightarrow 0}\left\{e^{-\alpha x}\left(\frac{e^{\alpha x}-1}{x}\right)^2-e^{-x}\left(\frac{e^x-1}{x}\right)^2\right\} \\ \\& =1 \cdot \alpha^2 \cdot 1^2-1 \cdot 1 \\ \\& =\alpha^2-1 \end{aligned}

\begin{aligned} & \therefore \quad P=\frac{\alpha^2}{2}-\frac{1}{2}=\frac{3}{2} \\ \\& \therefore \quad \alpha^2=4 \\ \\& \Rightarrow \quad \alpha= \pm 2 \\ & \end{aligned}

\begin{aligned} & \therefore \quad|\alpha|=2 \\ \\& \text { Then, } 1004|\alpha|=2008 \end{aligned}

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HARSH KANKARIA

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