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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
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B
3,3
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C
4,4
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D
1,1
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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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