if the constant term in the binomial expansion

Question:

if the constant term in the binomial expansion

of $\left(\sqrt{x}-\frac{k}{x^{2}}\right)^{10}$ is 405, then $|k|$ equals :

 

  1. 2

  2. 1

  3. 3

  4. 9


Correct Option: , 3

Solution:

$\left(\sqrt{\mathrm{x}}-\frac{\mathrm{k}}{\mathrm{x}^{2}}\right)^{10}$

$\mathrm{T}_{\mathrm{r}+1}={ }^{10} \mathrm{C}_{\mathrm{r}}(\sqrt{\mathrm{x}})^{10-\mathrm{r}}\left(\frac{-\mathrm{k}}{\mathrm{x}^{2}}\right)^{\mathrm{r}}$

$\mathrm{T}_{\mathrm{r}+1}={ }^{10} \mathrm{C}_{\mathrm{r}} \cdot \mathrm{x}^{\frac{10-\mathrm{r}}{2}} \cdot(-\mathrm{k})^{\mathrm{r}} \cdot \mathrm{x}^{-2 \mathrm{r}}$

$\mathrm{T}_{\mathrm{r}+1}={ }^{10} \mathrm{C}_{\mathrm{r}} \mathrm{x}^{\frac{10-5 \mathrm{r}}{2}}(-\mathrm{k})^{\mathrm{r}}$

Constant term : $\frac{10-5 \mathrm{r}}{2}=0 \Rightarrow \mathrm{r}=2$

$\mathrm{T}_{3}={ }^{10} \mathrm{C}_{2} \cdot(-\mathrm{k})^{2}=405$

$\mathrm{k}^{2}=\frac{405}{45}=9$

$\mathrm{k}=\pm 3 \Rightarrow|\mathrm{k}|=3$

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