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If the third term in the expansion of $\left[\mathrm{x}+\mathrm{x}^{\log _{10} \mathrm{x}}\right]^{5}$
is $10^{6},$ then $\mathrm{x}$ may be
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is $10^{6},$ then $\mathrm{x}$ may be
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The correct answer is:
10
Put $\log _{10} x=y,$ the given expression becomes $\left(x+x^{y}\right)^{5}$
$\mathrm{T}_{3}={ }^{5} \mathrm{C}_{2} \cdot \mathrm{x}^{3}\left(\mathrm{x}^{\mathrm{y}}\right)^{2}=10 \mathrm{x}^{3+2 \mathrm{y}}=10^{6}$ (given)
$\Rightarrow(3+2 \mathrm{y}) \log _{10} \mathrm{x}=5 \log _{10} 10=5$
$\Rightarrow(3+2 \mathrm{y}) \mathrm{y}=5$
$\Rightarrow \mathrm{y}=1,-\frac{5}{2}$
$\Rightarrow \log _{10} \mathrm{x}=1$ or $\log _{10} \mathrm{x}=-\frac{5}{2}$
$\therefore \mathrm{x}=10$ or $\mathrm{x}=(10)^{-5 / 2}$
$\mathrm{T}_{3}={ }^{5} \mathrm{C}_{2} \cdot \mathrm{x}^{3}\left(\mathrm{x}^{\mathrm{y}}\right)^{2}=10 \mathrm{x}^{3+2 \mathrm{y}}=10^{6}$ (given)
$\Rightarrow(3+2 \mathrm{y}) \log _{10} \mathrm{x}=5 \log _{10} 10=5$
$\Rightarrow(3+2 \mathrm{y}) \mathrm{y}=5$
$\Rightarrow \mathrm{y}=1,-\frac{5}{2}$
$\Rightarrow \log _{10} \mathrm{x}=1$ or $\log _{10} \mathrm{x}=-\frac{5}{2}$
$\therefore \mathrm{x}=10$ or $\mathrm{x}=(10)^{-5 / 2}$
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