Question

If the greatest value of the term independent of $x$ in the expansion of ${(x\sin \alpha +a\frac{\cos \alpha }{x})}^{10}$ is $\frac{10!}{(5!{)}^2},$ then the value of $a$ is equal to

• A. $-1$
• B. $-2$
• C. $2$
• D. $1$

Answer 1

The correct option is C $2$

The general term of ${(x\sin \alpha +a\frac{\cos \alpha }{x})}^{10}$ is
$T_{r+1}={}^{10}{C}_r\cdot (x\sin \alpha {)}^{10-r}\cdot {\left(\frac{a\cos \alpha }{x}\right)}^r$
$\Rightarrow T_{r+1}={}^{10}{C}_r\cdot (\sin \alpha {)}^{10-r}\cdot {\left(a\cos \alpha \right)}^r\cdot (x{)}^{10-2r}$

For term independent of $x,$ $r=5$
$\therefore T_{5+1}={}^{10}{C}_5\cdot (\sin \alpha {)}^5\cdot a^5\cdot (\cos \alpha {)}^2$
$\Rightarrow T_{5+1}={}^{10}{C}_5\cdot \frac{a^5}{2^5}\cdot (\sin 2\alpha {)}^2$
$\Rightarrow {\left(T_{5+1}\right)}_{max}={}^{10}{C}_5\cdot \frac{a^5}{2^5}=\frac{10!}{(5!{)}^2}$
$\Rightarrow a^5=2^5$
$\therefore a=2$