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Home  >>  CBSE XI  >>  Math  >>  Binomial Theorem
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If the middle term of $(\large\frac{1}{x}$$+x\sin x)^{10}$ is equal to $7\large\frac{7}{8}$ then value of $x$ is

$\begin{array}{1 1}(A)\;2n\pi+\large\frac{\pi}{6}\\(B)\;n\pi+\large\frac{\pi}{6}\\(C)\;n\pi+(-1)^n\large\frac{\pi}{6}\\(D)\;n\pi+(-1)^n\large\frac{\pi}{3}\end{array} $

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1 Answer

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  • $T_{r+1}=nC_r a^{n-r}b^r$
  • Middle term :-If $n$ is even then the total number of term is n+1.
  • Hence there is only one middle term
  • (i.e) $(\large\frac{n}{2}$$+1)^{th}$
$(\large\frac{10}{2}$$+1)^{th}=6^{th}$ term
$T_6=10C_5.\large\frac{1}{x^5}.$$x^5 \sin ^5x$
$10C_5=\large\frac{10!}{5!5!}=\frac{10\times 9\times 8\times 7\times 6\times 5!}{5\times 4\times 3\times 2\times 5!}$
$\Rightarrow 252$
$252 \sin ^5 x=\large\frac{63}{8}$
$\Rightarrow \sin ^5 x=\large\frac{1}{8}$
$\Rightarrow \sin ^5 x=\large\frac{1}{2^5}$
$\Rightarrow \sin x=\large\frac{1}{2}$
$\Rightarrow x=n\pi +(-1)^n \large\frac{\pi}{6}$
Hence (C) is the correct answer.
answered Jun 25, 2014 by sreemathi.v

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