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

A) $2^{5}$
B) $\ ^{10}C_{5}$
C) $\frac{1}{2^{5}}{\ (}^{10}C_{5})$
D) None of these
Explanation

(c)
The general term in the expansion of ${(x\sin\alpha + x^{- 1}\cos\alpha)}^{10}$ is
$T_{r + 1} = \ ^{10}C_{r}{(x\sin\alpha)}^{10 - r}{(x^{- 1}\cos\alpha)}^{r}$
$= \ ^{10}C_{r}{(\sin\alpha)}^{10 - r}{(\cos\alpha)}^{r}x^{10 - 2r}$
For the term independent of $x$, put
$10 - 2r = 0 \Rightarrow r = 5$
$\therefore$ Coefficient of term independent of $x$, is
$\ ^{10}C_{5}{(\sin\alpha)}^{5}{(\cos\alpha)}^{5} = \ ^{10}C_{5}\left( \frac{1}{2^{5}} \right){(\sin{2\alpha})}^{5}$
$\leq \frac{1}{2^{5}}\left( \ ^{10}C_{5} \right)\ \ \lbrack\because\ \sin{(2\alpha) \leq 1}\rbrack$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.