(C) The general term $T_{r+1}$ in the expansion of $\left(\sqrt{x}-\frac{k}{x^2}\right)^{10}$ is given by:$T_{r+1} = \binom{10}{r} (x^{1/2})^{10-r} (-

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(C) The general term $T_{r+1}$ in the expansion of $\left(\sqrt{x}-\frac{k}{x^2}\right)^{10}$ is given by:$T_{r+1} = \binom{10}{r} (x^{1/2})^{10-r} (-k x^{-2})^r$$T_{r+1} = \binom{10}{r} (-k)^r x^{\frac{10-r}{2} - 2r}$For the term to be independent of $x$,the exponent of $x$ must be $0$:$\frac{10-r}{2} - 2r = 0$$10 - r = 4r$$5r = 10 \Rightarrow r = 2$Now,substitute $r=2$ into the expression for the term:$T_3 = \binom{10}{2} (-k)^2 = 405$$45 k^2 = 405$$k^2 = 9$$k = \pm 3$

Class 11 Mathematics Binomial Theorem General term, Coefficient of any power of x, Independent term, Middle term and Greatest term and Greatest coefficient

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

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$3$ only
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$-3$ only
C
$\pm 3$
D
$0$

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

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