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 $\frac{1}{2^{5}}{ }^{10} C_{5} \times(1)=\left(\frac{1}{2}\right)^{5} \times{ }^{10} C_{I}$