Question

# If the term independent of $$x$$ in the expansion of $${ \left( \sqrt { x } -\cfrac { k }{ { x }^{ 2 } } \right) }^{ 10 }$$ is $$405$$, then $$k$$ equals:

A
2,2
B
3,3
C
4,4
D
1,1

Solution

## The correct option is B $$3,-3$$ $${(r+1)}^{th}$$ term of $${\left( \sqrt { x } -\dfrac { k }{ x^2 } \right) }^{10}$$ is $$^{ 10}{C }_{r}{({\sqrt { x }) }^{10-r }}{({-\dfrac { k }{ x^2 })}^{r}}.$$$${T}_{r+1} = ^{ 10}{C }_{r}{({\sqrt { x }) }^{10-r }}{({-\dfrac { k }{ x^2 })}^{r}}$$If the (r+1)th term is independent of x, then$$\dfrac { 10-r }{ 2 } -2r = 0$$$$\Rightarrow r=2$$Hence,  $$^{ 10}{C }_{2}{(-k)}^{2}=405$$$$\Rightarrow 45{(-k)}^{2}=405$$$$\Rightarrow k=\pm 3$$Mathematics

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