Q168 / 270 · NDA 2015 (Phase 2) · Mathematics

In the expansion of $\left(\sqrt{\mathrm{x}}+\frac{1}{3 \mathrm{x}^2}\right)^{10}$ the value of constant term (independent of x ) is

  • Correct answer
    A
    5
  • B
    8
  • C
    45
  • D
    90

Solution

Let $\mathrm{r}^{\text {th }}$ term is independent of x . $\begin{aligned} & T_r={ }^n C_r x^r y^{n-r} \\ & ={ }^{10} C_r(\sqrt{x})^r\left(\frac{1}{3 x^2}\right)^{10-r} \\ & ={ }^{10} C_r\left(\frac{1}{3}\right)^{10-r} \cdot(\sqrt{x})^r\left(\frac{1}{x^2}\right)^{10-r} \end{aligned}$ Equating the coefficient of $x$ to zero. $\begin{aligned} & \Rightarrow \quad x^{\mathrm{r} / 2} \cdot \mathrm{x}^{-2(10-\mathrm{r})}=\mathrm{x}^0 \\ & \Rightarrow \quad \frac{\mathrm{r}}{2}-20+2 \mathrm{r}=0 \\ & \Rightarrow \quad \frac{5}{2} \mathrm{r}=20 \Rightarrow \mathrm{r}=8 \end{aligned}$ $\begin{aligned} \text { Coefficient } & ={ }^{10} \mathrm{C}_{\mathrm{r}}\left(\frac{1}{3}\right)^{10-\mathrm{r}} \\ & ={ }^{10} \mathrm{C}_8\left(\frac{1}{3}\right)^{10-8} \end{aligned}$ $=\frac{10 \times 9}{2} \times \frac{1}{9}=5$

Hence, the correct answer is option: 'A'.

Relevant Questions