Question

The greatest value of term independent of $x$ in the expansion of $(x\sin p+x^{-1}\cos p{)}^{10},p\in R$, is:

• A. $2^5$
• B. $\frac{10!}{(5!{)}^2}$
• C. $\frac{1}{2^5}\times \frac{10!}{(5!{)}^2}$
• D. None of the above.

Answer 1

The correct option is C $\frac{1}{2^5}\times \frac{10!}{(5!{)}^2}$

The genral term of the expansion can be written as $T_{r+1}=$ ${}^{10}{C}_r.{\left(x\sin p\right)}^{10-r}.{\left(\frac{1}{x}\cos p\right)}^r$$=$ ${}^{10}{C}_r.{\left(\sin p\right)}^{10-r}.{\left(\cos p\right)}^r.x^{10-2r}$

As the term is independent of $x$.

So, $10-2r=0$

$\Rightarrow$ $r=5$

Hence, $T_6=$ ${}^{10}{C}_5.{\left(\sin p\right)}^5.{\left(\cos p\right)}^5$

$=$ $\frac{10!}{{(5!)}^2}$$\frac{1}{2^5}.{(2\sin p.\cos p)}^5$

$=$ $\frac{10!}{{(5!)}^2}$.$\frac{1}{2^5}$.${\left(\sin 2p\right)}^5$ .....As $\sin x$ has a max value $1$.

So, the greatest value of term independent of $x$ is $\frac{10!}{{(5!)}^2}$.$\frac{1}{2^5}$.