Mathematics - Binomial Theorem Question with Solution | TestHub

We have, $r^{\mathrm{th}}$ term in the expansion of ${\left(x\sin \alpha +a\frac{\cos \alpha }{x}\right)}^{10}$ is

$T_{r+1}={}^{10}C_r{\left(x\sin \alpha \right)}^{10-r}{\left(\frac{a\cos \alpha }{x}\right)}^r$

$r=0,1,2,\dots ,10$

$T_{r+1}$ will be independent of $x,$ when $10-2r=0$

$\Rightarrow r=5$

Now, $T_6={}^{10}C_5{\left(x\sin \alpha \right)}^5\times {\left(\frac{a\cos \alpha }{x}\right)}^5$

$={}^{10}C_5\times a^5\times \frac{1}{2^5}{\left(\sin 2\alpha \right)}^5$

$T_6$ will be greatest, when $\sin 2\alpha =1$

Therefore, ${}^{10}C_5\frac{a^5}{2^5}={}^{10}C_5$

$\Rightarrow a=2$