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The fractional part of a real number $x$ is $x-[x]$, where $[\mathrm{x}]$ is the greatest integer less than or equal to $\mathrm{x}$. Let $\mathrm{F}_{1}$ and $\mathrm{F}_{2}$ be the fractional parts of $(44-\sqrt{2017})^{2017}$ and $(44+\sqrt{2017})^{2017}$ respectively. Then $\mathrm{F}_{1}+\mathrm{F}_{2}$ lies between the numbers
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The correct answer is:
$0.9$ and $1.35$
$\mathrm{I}+\mathrm{F}_{2}=(\sqrt{2017}+44)^{2017}$
$\quad \mathrm{~F}_{2}^{\prime}=(\sqrt{2017}-44)^{2017} ; 0 < \mathrm{F}_{2}^{\prime} < 1$
$\mathrm{I}+\mathrm{F}_{2}-\mathrm{F}_{2}^{\prime}=2\left[{ }^{2017} \mathrm{C}_{1}(\sqrt{2017})^{2016}(44)+\ldots .\right]$
$\mathrm{F}_{2}=\mathrm{F}_{2}{ }^{\prime}$
$\mathrm{F}_{2}=(0.911)^{2017}$
Now, $\mathrm{F}_{1}=(44-\sqrt{2017})^{2017}=-(0.911)^{2017}$
Fractional part can not $-\mathrm{ve}$.
So, $F_{1}=1-(0.911)^{2017}$
So, $F_{1}+F_{2}=1$
1 lie Between $0.9 \& 1.35$
$\quad \mathrm{~F}_{2}^{\prime}=(\sqrt{2017}-44)^{2017} ; 0 < \mathrm{F}_{2}^{\prime} < 1$
$\mathrm{I}+\mathrm{F}_{2}-\mathrm{F}_{2}^{\prime}=2\left[{ }^{2017} \mathrm{C}_{1}(\sqrt{2017})^{2016}(44)+\ldots .\right]$
$\mathrm{F}_{2}=\mathrm{F}_{2}{ }^{\prime}$
$\mathrm{F}_{2}=(0.911)^{2017}$
Now, $\mathrm{F}_{1}=(44-\sqrt{2017})^{2017}=-(0.911)^{2017}$
Fractional part can not $-\mathrm{ve}$.
So, $F_{1}=1-(0.911)^{2017}$
So, $F_{1}+F_{2}=1$
1 lie Between $0.9 \& 1.35$
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