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Question

(A) The general term in the expansion of $(x \sin \alpha + a \frac{\cos \alpha}{x})^{10}$ is given by $T_{r+1} = {}^{10}C_r (x \sin \alpha)^{10-r} (a

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$2$

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(A) The general term in the expansion of $(x \sin \alpha + a \frac{\cos \alpha}{x})^{10}$ is given by $T_{r+1} = {}^{10}C_r (x \sin \alpha)^{10-r} (a \frac{\cos \alpha}{x})^r$.For the term to be independent of $x$,the power of $x$ must be zero,so $10-r-r = 0$,which implies $r = 5$.The term independent of $x$ is $T_6 = {}^{10}C_5 (\sin \alpha)^5 (a \cos \alpha)^5 = {}^{10}C_5 a^5 (\sin \alpha \cos \alpha)^5$.Using the identity $\sin \alpha \cos \alpha = \frac{\sin 2\alpha}{2}$,we get $T_6 = {}^{10}C_5 a^5 (\frac{\sin 2\alpha}{2})^5 = {}^{10}C_5 \frac{a^5}{2^5} (\sin 2\alpha)^5$.The greatest value occurs when $\sin 2\alpha = 1$,so the greatest value is ${}^{10}C_5 \frac{a^5}{32}$.Given that the greatest value is $\frac{10!}{(5!)^2} = {}^{10}C_5$,we have ${}^{10}C_5 \frac{a^5}{32} = {}^{10}C_5$.Thus,$\frac{a^5}{32} = 1$,which means $a^5 = 32$,so $a = 2$.

If the greatest value of the term independent of $x$ in the expansion of $(x \sin \alpha + a \frac{\cos \alpha}{x})^{10}$ is $\frac{10!}{(5!)^2}$,then the value of $a$ is equal to:

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