Discussion Forum

1)

The term independent of x in expansion of 

$\left(\frac{x+1}{x^{\frac{2}{3}}-x^{\frac{1}{3}}+1}-\frac{x-1}{x-x^{\frac{1}{2}}}\right)^{10} is$

Answer:

Option C

Explanation:

$\therefore$    $\left[\frac{x+1}{x^{\frac{2}{3}}-x^{\frac{1}{3}}+1}-\frac{x-1}{x-x^{\frac{1}{2}}}\right]^{10}$

= $\left[\frac{(x^{\frac{1}{3}})^{3}+1^{3}}{x^{\frac{2}{3}}-x^{\frac{1}{3}}+1}-\frac{((\sqrt{x}))^{2}-1}{\sqrt{x}(\sqrt{x}-1)}\right]^{10}$

  = $\left[\frac{(x^{\frac{1}{3}}+1)(x^{2/3}+1-x^{1/3}}{x^{\frac{2}{3}}-x^{\frac{1}{3}}+1}-\frac{((\sqrt{x}))^{2}-1}{\sqrt{x}(\sqrt{x}-1)}\right]^{10}$

   =   $\left[(x^{1/3}+1)-\frac{(\sqrt{x}+1)}{\sqrt{x}}\right]^{10}$

       = $(x^{1/3}-x^{-1/2})^{10}$

  $\therefore$ The general term is

        $T_{r+1}=^{10}C_{r}(x^{1/3})^{10-r}(-x^{-1/2})^{r}$

        =  $^{10}C_{r}(-1)^{r}x^{\frac{10-r}{3}-\frac{r}{2}}$

 For independent of x, put

$\frac{10-r}{3}-\frac{r}{2}=0\Rightarrow 20-2r-3r=0$

 $\Rightarrow$    $20=5r\Rightarrow r=4$

  $\therefore$      $T_{5}=^{10}C_{4}=\frac{10\times9\times8\times7}{4\times3\times2\times1}=210$