A) \[1:16\]
B) \[16:1\]
C) \[1:8\]
D) \[8:1\]
Correct Answer: B
Solution :
\[{{T}_{r+1}}={}^{16}{{C}_{r}}\cdot {{\left( \frac{x}{\cos \theta } \right)}^{16-r}}\cdot {{\left( \frac{1}{x\sin \theta } \right)}^{r}}\] \[={}^{16}{{C}_{r}}\cdot \frac{{{x}^{16-2r}}}{{{(\cos \theta )}^{16-r}}\cdot {{(\sin \theta )}^{r}}}\] \[\because \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,16-2r=0\Rightarrow r=8\] \[{{T}_{9}}=\frac{^{16}{{C}_{8}}\cdot {{2}^{8}}}{{{\sin }^{8}}2\theta }\] \[\because \,\,\,\,\,\,\,\,\,\,\,\sin 2\theta \] is increasing in \[\left[ \frac{\pi }{16},\frac{\pi }{4} \right]\] Hence, \[{{l}_{1}}=\frac{^{16}{{C}_{8}}\cdot {{2}^{8}}}{{{\sin }^{8}}\left( \frac{\pi }{2} \right)}\] and \[{{l}_{2}}=\frac{^{16}{{C}_{8}}\cdot {{2}^{8}}}{{{\sin }^{8}}\left( \frac{\pi }{4} \right)}\] \[\frac{{{l}_{2}}}{{{l}_{1}}}=\frac{{{\sin }^{8}}\left( \frac{\pi }{2} \right)}{{{\sin }^{8}}\left( \frac{\pi }{4} \right)}=\frac{16}{1}\]
You need to login to perform this action.
You will be redirected in
3 sec
