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Consider  \mathrm{P(x)=a x^2+b x+c}  where  \mathrm{a, b, c \in \mathbb{R} \text { and } P(2)=\mathbf{9}}. Let \mathrm{\alpha } and \mathrm{\beta } be the

roots of equation \mathrm{P(x)=0}.

If \mathrm{\alpha \rightarrow \infty} and \mathrm{P^{\prime}(3)=5} , then  \mathrm{\lim _{x \rightarrow \infty}\left(\frac{P(x)}{5(x-1)}\right)^x}  is ?

Option: 1

\mathrm{e^{\frac{4}{5}}}


Option: 2

\mathrm{e^{\frac{2}{5}}}


Option: 3

\mathrm{e^{\frac{3}{5}}}


Option: 4

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Answers (1)

best_answer

                        \mathrm{\lim _{x \rightarrow \infty}\left(\frac{P(x)}{5(x-1)}\right)^x=\lim _{x \rightarrow \infty}\left(\frac{a x^2+b x+c}{5 x-5}\right)^x=\lim _{x \rightarrow \infty}\left(\frac{a}{5} x+\frac{o(x)}{5(x-1)}\right)^x}

if \mathrm{a> 0} the limit is \mathrm{+\infty} and if \mathrm{a< 0} the limit doesn't exist therefore \mathrm{a= 0} and we have

                      \mathrm{\lim _{x \rightarrow \infty}\left(\frac{P(x)}{5(x-1)}\right)^x=\lim _{x \rightarrow \infty}\left(\frac{b x+c}{5 x-5}\right)^x=\lim _{x \rightarrow \infty}\left(\frac{b}{5}+\frac{c+b}{5 x-5}\right)^x}

if \mathrm{b> 5}  the limit is \mathrm{\infty}, for \mathrm{-5< b< 5} the limit is 0 and doesn't exist for \mathrm{b< -5}  therefore \mathrm{b= 5}.

substituting this in the equations you derived yeilds \mathrm{c=-1} and we have

                                    \mathrm{\lim _{x \rightarrow \infty}\left(\frac{P(x)}{5(x-1)}\right)^x=\lim _{x \rightarrow \infty}\left(1+\frac{4}{5 x-5}\right)^x=e^{\frac{4}{5}}}

Posted by

Gautam harsolia

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