Q68 / 170 · Manipal (MET) 2016 · Mathematics
The coefficient of the term independent of $x$ in the expansion of $\left(\frac{x+1}{x^{2 / 3}-x^{1 / 3}+1}-\frac{x-1}{x-x^{1 / 2}}\right)^{10}$ is
- Correct answerA$210$
- B$105$
- C$70$
- D$112$
Solution
$\frac{x+1}{x^{2 / 3}-x^{1 / 3}+1}-\frac{x-1}{x-x^{1 / 2}}$
$\begin{aligned} & =\frac{\left(x^{1 / 3}\right)^3+1^3}{x^{2 / 3}-x^{1 / 3}+1}-\frac{x-1}{x^{1 / 2}\left(x^{1 / 2}-1\right)} \\ & =\frac{\left(x^{1 / 3}+1\right)\left(x^{2 / 3}-x^{1 / 3}+1\right)}{x^{2 / 3}-x^{1 / 3}+1}-\frac{x^{1 / 2}+1}{x^{1 / 2}}\end{aligned}$
$\begin{aligned} & =x^{1 / 3}+1-1-x^{-1 / 2} \\ & =x^{1 / 3}-x^{-1 / 2}\end{aligned}$
$\therefore\left(\frac{x+1}{x^{2 / 3}-x^{1 / 3}+1}-\frac{x-1}{x-x^{1 / 2}}\right)^{10}=\left(x^{1 / 3}-x^{-1 / 2}\right)^{10}$
$\begin{aligned} & T_{r+1} \text { for }\left(x^{1 / 3}-x^{-1 / 2}\right) \text { is } \\ & { }^{10} C_r\left(x^{1 / 3}\right)^{10-r}(-1)^r\left(x^{-1 / 2}\right)^r .\end{aligned}$
For term independent of $x$,
$\frac{10-r}{3}-\frac{r}{2}=0$
$\begin{array}{rlrl}\Rightarrow & 20-2 r-3 r & =0 \\ \Rightarrow & & r & =4\end{array}$
Hence, required coefficient $={ }^{10} C_4(-1)^4=210$.
Hence, the correct answer is option: 'A'.