Question

The term independent of $x$ in $(\frac{1}{2}{x}^{\frac{1}{3}}+x^{\frac{-1}{5}}{)}^8$ is

• A. $\frac{35}{8}$
• B. $7$
• C. $\frac{7}{2}$
• D. $28$

Answer 1

Answer: (B)

$7$

We know that

$T_{r+1}{=}^n{C}_r{a}^{n-r}{b}^r$

So, $T_{r+1}=8C_r{\left(\frac{1}{2}{x}^{\frac{1}{3}}\right)}^{8-r}{\left(x^{\frac{-1}{5}}\right)}^r$

$=8C_r{\left(\frac{1}{2}\right)}^{8-r}{\left(x^{\frac{1}{3}}\right)}^{8-r}{\left(x^{\frac{-1}{5}}\right)}^r$

$=8C_r{\left(\frac{1}{2}\right)}^{8-r}{x}^{(\frac{8-r}{3}-\frac{r}{5})}$

The term will be independent of x if the index of x is zero,

i.e., $\frac{8-r}{3}-\frac{r}{5}=0$

$\Rightarrow 40-5r-3r=0$

$\Rightarrow r=5$

Hence $6^{th}$ term is independent of x and is given by $8C_5{\left(\frac{1}{2}\right)}^{8-5}{x}^{(\frac{8-5}{3}-\frac{5}{5})}$

$\Rightarrow 8C_5{\left(\frac{1}{2}\right)}^3{x}^0$

$\Rightarrow \frac{8!}{(8-5)!5!}\times \frac{1}{8}$

$\Rightarrow \frac{8!}{3!5!}\times \frac{1}{8}=7$