Question

The greatest value of the term independent of $x$ , as $\alpha$ varies over $R$ , in the expansion of ${\left(x\cos \alpha +\frac{\sin \alpha }{x}\right)}^{10}$ is

• A. ${}^{10}{C}_5$
• B. ${\left(\frac{1}{2}\right)}^5$ ${}^{10}{C}_5$
• C. ${\left(\frac{1}{2}\right)}^4$ ${}^{10}{C}_5$
• D. ${\left(\frac{1}{2}\right)}^3$ ${}^{10}{C}_5$

Answer 1

Answer: (B)

Given, ${\left(x\cos \alpha +\frac{\sin \alpha }{x}\right)}^{10}$
$=\frac{1}{x^{10}}{\left(x^2\cos \alpha +\sin \alpha \right)}^{10}$
General term of the expansion
$T_{t+1}=\frac{1}{x^{10}}\times {}^{10}{C}_t{\left(x^2\cos \alpha \right)}^{10-t}(\sin \alpha {)}^t$
$=\frac{1}{x^{10}}\times {}^{10}{C}_t{\left(x^2\right)}^{10-t}(\cos \alpha {)}^{10-r}(\sin \alpha {)}^t$
For this term to be independent of $x$,
$10-r=5$
$\Rightarrow r=5$
$\therefore T_{5+1}=\frac{1}{x^{10}}\times {}^{10}{C}_r{x}^{10}\cdot {\cos }^5\alpha \cdot {\sin }^5\alpha$
$={}^{10}{C}_r(\cos \alpha \cdot \sin \alpha {)}^5$
$={}^{10}{C}_r{\left(\frac{2\sin \alpha \cdot \cos \alpha }{2}\right)}^5$
$=\frac{1}{2^5}{}^{10}{C}_r(\sin 2\alpha {)}^5$
$\therefore$ Greatest value of $T_6$ (term independent of $x$ ) will be $\frac{1}{2^5}{}^{10}{C}_5\times (1)={\left(\frac{1}{2}\right)}^5\times {}^{10}{C}_I$