If the greatest value of the term independent of ' $x$ '
in the expansion of $\left(x \sin \alpha+a \frac{\cos \alpha}{x}\right)^{10}$ is $\frac{10 !}{(5 !)^{2}}$,
then the value of 'a' is equal to:
Correct Option: , 4
$\mathrm{T}_{\mathrm{r}+1}={ }^{10} \mathrm{C}_{\mathrm{r}}(\mathrm{x} \sin \alpha)^{10-\mathrm{r}}\left(\frac{\mathrm{a} \cos \alpha}{\mathrm{x}}\right)^{\mathrm{r}}$
$\mathrm{r}=0,1,2, \ldots, 10$
$\mathrm{T}_{r+1}$ will be independent of $\mathrm{x}$
when $10-2 r=0 \Rightarrow r=5$
$\mathrm{T}_{6}={ }^{10} \mathrm{C}_{5}(\mathrm{x} \sin \alpha)^{5} \times\left(\frac{\mathrm{a} \cos \alpha}{\mathrm{x}}\right)^{5}$
$={ }^{10} \mathrm{C}_{5} \times \mathrm{a}^{5} \times \frac{1}{2^{5}}(\sin 2 \alpha)^{5}$
will be greatest when $\sin 2 \alpha=1$
$\Rightarrow{ }^{10} \mathrm{C}_{5} \frac{\mathrm{a}^{5}}{2^{5}}={ }^{10} \mathrm{C}_{5} \Rightarrow \mathrm{a}=2$
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.