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)

$T_{r+1}={}^{10}{C}_r(x\sin \alpha {)}^{10-r}{\left(\frac{a\cos \alpha }{x}\right)}^r$
$r=0,1,2,\dots ,10$
$T_{\gamma +1}$ will be independent of $x$
When $10-2r=0\Rightarrow r=5$
$T_6={}^{10}{C}_5(x\sin \alpha {)}^5\times {\left(\frac{a\cos \alpha }{x}\right)}^5$
$={}^{10}{C}_5\times a^5\times \frac{1}{2^5}(\sin 2\alpha {)}^5$
will be greatest when $\sin 2\alpha =1$
$\Rightarrow {}^{10}{C}_5\frac{a^5}{2^5}={}^{10}{C}_5$
$\Rightarrow a=2$