In the expansion of left(frac{x}{cos theta}+f | vedclass

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(D) The general term is given by $T_{r+1} = ^{16}C_{r} \left(\frac{x}{\cos 	heta}ight)^{16-r} \left(\frac{1}{x \sin 	heta}ight)^{r}$.Simplifying

Vedclass · Accepted Answer

$16 : 1$

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(D) The general term is given by $T_{r+1} = ^{16}C_{r} \left(\frac{x}{\cos 	heta}ight)^{16-r} \left(\frac{1}{x \sin 	heta}ight)^{r}$.Simplifying,$T_{r+1} = ^{16}C_{r} x^{16-2r} \frac{1}{(\cos 	heta)^{16-r} (\sin 	heta)^{r}}$.For the term to be independent of $x$,we set $16-2r = 0$,which gives $r = 8$.Thus,the independent term is $T_{9} = ^{16}C_{8} \frac{1}{\cos^{8} 	heta \sin^{8} 	heta} = ^{16}C_{8} \frac{2^{8}}{(\sin 2	heta)^{8}}$.Let $f(	heta) = \frac{^{16}C_{8} \cdot 2^{8}}{(\sin 2	heta)^{8}}$.For $	heta \in [\frac{\pi}{8}, \frac{\pi}{4}]$,$\sin 2	heta$ is increasing,so $f(	heta)$ is least when $\sin 2	heta$ is maximum,i.e.,at $	heta = \frac{\pi}{4}$. Thus,$\ell_{1} = f(\frac{\pi}{4}) = ^{16}C_{8} \cdot 2^{8}$.For $	heta \in [\frac{\pi}{16}, \frac{\pi}{8}]$,$f(	heta)$ is least when $\sin 2	heta$ is maximum,i.e.,at $	heta = \frac{\pi}{8}$. Thus,$\ell_{2} = f(\frac{\pi}{8}) = ^{16}C_{8} \frac{2^{8}}{(\sin \frac{\pi}{4})^{8}} = ^{16}C_{8} \frac{2^{8}}{(1/\sqrt{2})^{8}} = ^{16}C_{8} \cdot 2^{8} \cdot 2^{4} = ^{16}C_{8} \cdot 2^{12}$.The ratio $\frac{\ell_{2}}{\ell_{1}} = \frac{^{16}C_{8} \cdot 2^{12}}{^{16}C_{8} \cdot 2^{8}} = 2^{4} = 16$.

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