Question

In the expansion of ${(\frac{x+1}{x^{2/3}-x^{1/3}+1}-\frac{x-1}{x-x^{1/2}})}^{10}$, the term does not contain $x$ is -

• A. ${{}^{11}C}_4-{{}^{10}C}_3$
• B. ${{}^{10}C}_7$
• C. ${{}^{10}C}_4$
• D. ${{}^{11}C}_5-{{}^{10}C}_5$

Answer 1

Answer: (A)

${{}^{11}C}_4-{{}^{10}C}_3$

We have

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

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

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

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

$T_{r+1}{=}^{10}{C}_2{\left(-x^{\frac{-1}{2}}\right)}^r{\left(x^{\frac{1}{3}}\right)}^{10-r}$

${=}^{10}{C}_2{\left(-1\right)}^r{x}^{\frac{10}{3}-\frac{r}{3}-\frac{r}{2}}$

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

$20-2r-3r=0$

$$r=4$

${}^{10}{C}_4{\left(-1\right)}^4{=}^{10}{C}_4$

Hence, the option $\left(A\right)$ is the correct answer.