Question

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:

• A. -1
• B. 1
• C. -2
• D. 2

Answer 1

Answer: (D)

$T_{r+1}={ }^{10} C_{r}(x \sin \alpha)^{10-r}\left(\frac{a \cos \alpha}{x}\right)^{r}$
$r=0,1,2, \ldots, 10$
$T_{\gamma+1}$ will be independent of $x$
When $10-2 r=0 \Rightarrow r=5$
$T_{6}={ }^{10} C_{5}(x \sin \alpha)^{5} \times\left(\frac{a \cos \alpha}{x}\right)^{5}$
$={ }^{10} C_{5} \times a^{5} \times \frac{1}{2^{5}}(\sin 2 \alpha)^{5}$
will be greatest when $\sin 2 \alpha=1$
$\Rightarrow{ }^{10} C_{5} \frac{a^{5}}{2^{5}}={ }^{10} C_{5}$
$\Rightarrow a=2$