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Home  >>  CBSE XI  >>  Math  >>  Binomial Theorem
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If the $6^{th}$ term in the expansion of $\big(\large\frac{1}{x^{8/3}}+$$x^2\log _{10} x\big)^{8}$ is 5600 then the value of $x$ is

$\begin{array}{1 1}(A)\;8\\(B)\;9\\(C)\;10\\(D)\;11\end{array} $

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1 Answer

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  • $T_{r+1}=nC_r a^{n-r} b^r$
$T_6=T_{5+1}=8C_5 (\large\frac{1}{x^{8/3}})^{8-5}$$(x^2\log_{10}x)^5$
$\Rightarrow 56.\large\frac{1}{x^8}$$x^{10} (\log _{10} x)^5$
$\Rightarrow 56x^2(\log _{10} x)^5$
$\Rightarrow 56x^2(\log _{10} x)^5=5600$
$\Rightarrow x^2(\log _{10} x)^5=100$
$\Rightarrow 10^2 (\log _{10} 10)^5$
$\Rightarrow x=10$
Hence (C) is the correct answer.
answered Jun 25, 2014 by sreemathi.v

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