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  • If <img alt="\mathrm{\lim _{x \rightarrow 0} \frac{\alpha x e^x-\beta \log _e(1+x)+\gamma x^2 e^{-x}}{x \sin ^2 x}=10, \alpha, \beta, \gamma \in R}" src="https://entrancecorner.oncodecogs.com/gif.l

If \mathrm{\lim _{x \rightarrow 0} \frac{\alpha x e^x-\beta \log _e(1+x)+\gamma x^2 e^{-x}}{x \sin ^2 x}=10, \alpha, \beta, \gamma \in R}, then the value of \mathrm{ \alpha+ \beta+ \gamma } is

Option: 1

3


Option: 2

0


Option: 3

1


Option: 4

2


Answers (1)

best_answer

Given, \mathrm{\lim _{x \rightarrow 0} \frac{\alpha x e^x-\beta \log _e(1+x)+\gamma x^2 e^{-x}}{x \sin ^2 x}=10}

\mathrm{\Rightarrow\frac{ \lim _{x \rightarrow 0} \alpha x\left(1+x+\frac{x^2}{\lfloor 2}+\frac{x^3}{\lfloor 3}+\ldots\right)-\beta\left(x-\frac{1}{2} x^2+\frac{1}{3} x^3\right.+\ldots .+\gamma x^2\left(1-x+\frac{1}{\lfloor 2} x^2-\frac{1}{\underline{\underline{3}}} x^3+\ldots\right)}{x \sin ^2 x}=10}

\mathrm{\Rightarrow \lim _{x \rightarrow 0} \frac{\alpha x\left(1+x+\frac{x^2}{2}+\ldots\right)-\beta\left(x-\frac{x^2}{2}+\frac{x^3}{3}-\ldots\right)+\gamma x^2(1-x+\ldots)}{x^3 x}=10 }

\mathrm{\Rightarrow \lim _{x \rightarrow 0} \frac{x(\alpha-\beta)+x^2\left(\alpha+\frac{\beta}{2}+\gamma\right)+x^3\left(\frac{\alpha}{2}-\frac{\beta}{3}-\gamma\right)}{x^3} }

                                                                                        \mathrm{+\text { Higher powers of } x=10 }

For existence of limit

\mathrm{ \alpha-\beta=0, \alpha+\frac{\beta}{2}+\gamma=0 \text { and } \frac{\alpha}{2}-\frac{\beta}{3}-\gamma=10 }       .......(i)

\mathrm{ \Rightarrow \beta=\alpha, \gamma=\frac{-3 \alpha}{2} . }
Put it in (i), we get

\mathrm{\frac{\alpha}{2}-\frac{\alpha}{3}+\frac{3 \alpha}{2}=10 }

\mathrm{ \Rightarrow \frac{5 \alpha}{3}=10 }

\mathrm{ \Rightarrow \alpha=6 }

\mathrm{ \therefore \alpha=6, \beta=6 \text { and } \gamma=-9 }

\mathrm{ \therefore \alpha+\beta+\gamma=6+6-9=3 . }






 



 

Posted by

himanshu.meshram

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