Question

The term independent of $x$ in expansion of $\bigg(\frac{x+1}{x^{2/3}-x^{1/3}+1}-\frac{x-1}{x-x^{1/2}}\bigg)^{10}$ is

• A. 4
• B. 120
• C. 210
• D. 310

Answer 1

The Correct Option is C

$\bigg[ \frac{x+1}{x^{2/3}-x^{1/3}+1}-\frac{(x-1)}{x-x^{1/2}}\bigg]^{10}$
=$\bigg[ \frac{(x^{1/3})^3+1^3}{x^{2/3}-x^{1/3}+1}-\frac{ \{ (\sqrt x)^2 \}}{\sqrt x (\sqrt x-1)}\bigg]^{10}$
=$\bigg[ \frac{(x^{1/3}+1)(x^{2/3}+1-x^{1/3})}{x^{2/3-x^{1/3}+1}}-\frac{ \{(\sqrt x)^2-1 \}}{\sqrt x(\sqrt x-1)}\bigg]^{10}$
$=\bigg[ (x^{1/3}+1)-\frac{(\sqrt x+1)}{\sqrt x}\bigg]^{10} \, \, \, = (x^{1/3}-x^{-1/2})^{10}$
$\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
$\frac{10-r}{3}-\frac{r}{2}=0 \, \, \, \Rightarrow \, \, \, 20-2r-3r=0$
$\Rightarrow \, \, \, \, \, \, \, \, \, \, \, \, 20=5r \, \, \Rightarrow \, \, 20-2r-3r=0$
$\therefore \, \, T_5= \, ^{10}C_4=\frac{10\times \, 9 \times \, 8 \times 7}{4 \times 3 \times 2 \times 1}=210$