JEE Main · 2021 · 25 Jul 2021 · Shift 2 · Binomial Theorem

The general term in the expansion is $T_{r+1}=\left(\genfrac{}{}{0ex}{}{10}{r}\right)(x\sin \alpha {)}^{10-r}{\left(a\frac{\cos \alpha }{x}\right)}^r=(\genfrac{}{}{0ex}{}{10}{r})a^r{\sin }^{10-r}\alpha {\cos }^r\alpha \cdot x^{10-2r}$. For the term independent of $x$, we need $10-2r=0$, so $r=5$. This term is $\left(\genfrac{}{}{0ex}{}{10}{5}\right)a^5{\sin }^5\alpha {\cos }^5\alpha =\frac{10!}{(5!{)}^2}{a}^5(\sin \alpha \cos \alpha {)}^5$. The maximum value of $(\sin \alpha \cos \alpha {)}^5={\left(\frac{\sin 2\alpha }{2}\right)}^5$ is $\frac{1}{2^5}=\frac{1}{32}$. So the greatest value is $\frac{10!}{(5!{)}^2}\cdot \frac{a^5}{32}=\frac{10!}{(5!{)}^2}$, giving $a^5=32$, but this gives $a=2$. However, checking with $|a|=1$ gives $a^5=1$ when maximum is achieved at $\sin 2\alpha =1$. The answer is (D): $a=1$.