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(D) The general term in the expansion of ${\left( {x\sin 	heta  + \frac{{\cos 	heta }}{x}} ight)^{10}}$ is given by $T_{r + 1} = ^{10}C_r (x\sin \

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$\frac{^{10}C_5}{2^5}$

Vedclass · Answer

(D) The general term in the expansion of ${\left( {x\sin 	heta  + \frac{{\cos 	heta }}{x}} ight)^{10}}$ is given by $T_{r + 1} = ^{10}C_r (x\sin 	heta )^{10 - r} \cdot {\left( {\frac{{\cos 	heta }}{x}} ight)^r}$.Simplifying the expression,we get $T_{r + 1} = ^{10}C_r (\sin 	heta )^{10 - r} (\cos 	heta )^r \cdot x^{10 - 2r}$.For the term to be independent of $x$,the exponent of $x$ must be zero,so $10 - 2r = 0$,which gives $r = 5$.The independent term is $T_6 = ^{10}C_5 (\sin 	heta )^5 (\cos 	heta )^5 = ^{10}C_5 (\sin 	heta \cos 	heta )^5$.Using the identity $\sin 2	heta = 2\sin 	heta \cos 	heta$,we have $\sin 	heta \cos 	heta = \frac{\sin 2	heta}{2}$.Thus,$T_6 = ^{10}C_5 \left( \frac{\sin 2	heta}{2} ight)^5 = \frac{^{10}C_5}{2^5} (\sin 2	heta )^5$.The greatest value of this term occurs when $\sin 2	heta = 1$,which gives the value $\frac{^{10}C_5}{2^5}$.

The greatest value of the term independent of $x$ in the expansion of ${\left( {x\sin \theta + \frac{{\cos \theta }}{x}} \right)^{10}}$ is

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