Question:
If $\mathrm{n}$ is the number of irrational terms in the expansion of $\left(3^{1 / 4}+5^{1 / 8}\right)^{60}$, then $(\mathrm{n}-1)$ is divisible by:
Correct Option: 1
Solution:
Here $\mathrm{AO}+\mathrm{OD}-1$ or $(\sqrt{2}+1) \mathrm{r}=1$
$\Rightarrow \quad \mathbf{r}=\sqrt{2-1}$
equation of circle $(x-r)^{2}+(y-r)^{2}=r^{2}$
Equation of CE
$y-1-m(x-1)$
$m x-y+1-M=0$
It is tangent to circle
$\therefore \quad\left|\frac{m r-r+1-m}{\sqrt{m^{2}+1}}\right|=r$
$\left|\frac{(m-1) r+1-m}{\sqrt{m^{2}+1}}\right|=r$
$\frac{(m-1)^{2}(r-1)^{2}}{m^{2}+1}=r^{2}$
Put $\mathrm{r}=\sqrt{2}-1$
On solving $\mathrm{m}-2-\sqrt{3}, 2+\sqrt{3}$