The greatest value of the term independent of $x$ in the expansion of $(x^{- 1} cos α + x sin α)^{10}, α ϵ R$ , is

\frac{10!}{32}

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\frac{10!}{(5!)^{2} \times 32}

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\frac{10!}{(5!)^{2}}

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\frac{10!}{25}

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Solution

The correct option is B $\frac{10!}{(5!)^{2} \times 32}$
The $6^{t h}$ term will independent of $x$ .
Hence, $T_{5 + 1}$
$=^{10} C_{5} (cos α sin α)^{5}$
$=^{10} C_{5} \frac{1}{32} (2 cos α sin α)^{5}$
$= \frac{1}{32}^{10} C_{5} {sin}^{5} (2 α)$
Substituting $α = \frac{π}{4}$ , we get the greatest value as $\frac{1}{32}^{10} C_{5}$ .
$= \frac{10!}{(5!) \times 32}$

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