Binomial Theorem Ques 2
- The term independent of $x$ in expansion of $\left(\frac{x+1}{x^{2 / 3}-x^{1 / 3}+1}-\frac{x-1}{x-x^{1 / 2}}\right)^{10}$ is
(2013 Main)
(a) $4$
(b) $120$
(c) $210$
(d) $310$
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Answer:
Correct Answer: 2.(c)
Solution: (c)
$ \begin{aligned} & {\left[\frac{x+1}{x^{2 / 3}-x^{1 / 3}+1}-\frac{(x-1)}{x-x^{1 / 2}}\right]^{10}} \\ & =\left[\frac{\left(x^{1 / 3}\right)^3+1^3}{x^{33}-x^{1 / 3}+1}-\frac{\left\{(\sqrt{x})^2-1\right\}}{\sqrt{x}(\sqrt{x}-1)}\right]^{10} \\ & =\left[\frac{\left(x^{1 / 3}+1\right)\left(x^{23}+1-x^{13}\right)}{x^{33}-x^{1 / 3}+1}-\frac{\left\{(\sqrt{x})^2-1\right\}}{\sqrt{x}(\sqrt{x}-1)}\right]^{10} \\ & =\left[\left(x^{1 / 3}+1\right)-\frac{(\sqrt{x}+1)}{\sqrt{x}}\right]^{10}=\left(x^{1 / 3}-x^{-1 / 2}\right)^{10} \end{aligned} $
$\therefore \quad$ The general term is
$ T_{r+1}={ }^{10} C_r\left(x^{1 / a}\right)^{10-r}\left(-x^{-1 / 2}\right)^r={ }^{10} C_r(-1)^r x^{\frac{10-r}{3}-\frac{r}{2}} $
For independent of $x$, put
$ \begin{aligned} & & \frac{10-r}{3}-\frac{r}{2} & =0 \Rightarrow 20-2 r-3 r=0 \\ \Rightarrow & & 20 & =5 r \Rightarrow r=4 \\ \therefore & & T_5={ }^{10} C_4 & =\frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1}=210 \end{aligned} $
BETA
