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$\left(\frac{\mathrm{x}}{\cos \theta}+\frac{1}{\mathrm{x} \sin \theta}\right)^{16}$ ના વિસ્તરણમાં જો $\frac{\pi}{8} \leq \theta \leq \frac{\pi}{4}$ હોય ત્યારે $\ell_{1}$ એ $x$ થી સ્વતંત્ર ન્યૂનતમ પદ છે અને જ્યારે $\frac{\pi}{16} \leq \theta \leq \frac{\pi}{8} $ હોય ત્યારે $\ell_{2}$ એ $x$ થી સ્વતંત્ર ન્યૂનતમ પદ છે તો $\ell_{2}: \ell_{1}$ ગુણોતર મેળવો.
$1 : 8$
$1 : 16$
$8 : 1$
$16 : 1$
Solution
$\mathrm{T}_{\mathrm{r}+1}=16 \mathrm{C}_{\mathrm{r}}\left(\frac{\mathrm{x}}{\cos \theta}\right)^{16-\mathrm{r}}\left(\frac{1}{\mathrm{x} \sin \theta}\right)^{\mathrm{r}}$
$=^{16} \mathrm{C}_{\mathrm{r}}(\mathrm{x})^{16-2 \mathrm{r}} \times \frac{1}{(\cos \theta)^{16-\mathrm{r}}(\sin \theta)^{\mathrm{r}}}$
For independent of $x ; 16-2 r=0 \Rightarrow r=8$
$\Rightarrow \mathrm{T}_{9}=^{16}\mathrm{C}_{8} \frac{1}{\cos ^{8} \theta \sin ^{8} \theta}$
$=^{16} \mathrm{C}_{8} \frac{2^{8}}{(\sin 2 \theta)^{8}}$
for $\theta \in\left[\frac{\pi}{8}, \frac{\pi}{4}\right] \ell_{1}$ is least for $\theta_{1}=\frac{\pi}{4}$
for $\theta \in\left[\frac{\pi}{16}, \frac{\pi}{8}\right] \ell_{2}$ is least for $\theta_{2}=\frac{\pi}{8}$
$\frac{\ell_{2}}{\ell_{1}}=\frac{\left(\sin 2 \theta_{1}\right)^{8}}{\left(\sin 2 \theta_{2}\right)^{8}}=(\sqrt{2})^{8}=\frac{16}{1}$
