A wire of length $$20 \mathrm{~m}$$ is to be cut into two pieces. A piece of length $$l_{1}$$ is bent to make a square of area $$A_{1}$$ and the other piece of length $$l_{2}$$ is made into a circle of area $$A_{2}$$. If $$2 A_{1}+3 A_{2}$$ is minimum then $$\left(\pi l_{1}\right): l_{2}$$ is equal to :

Let a circle $$C_{1}$$ be obtained on rolling the circle $$x^{2}+y^{2}-4 x-6 y+11=0$$ upwards 4 units on the tangent $$\mathrm{T}$$ to it at the point $$(3,2)$$. Let $$C_{2}$$ be the image of $$C_{1}$$ in $$\mathrm{T}$$. Let $$A$$ and $$B$$ be the centers of circles $$C_{1}$$ and $$C_{2}$$ respectively, and $$M$$ and $$N$$ be respectively the feet of perpendiculars drawn from $$A$$ and $$B$$ on the $$x$$-axis. Then the area of the trapezium AMNB is :

Let $$\alpha \in (0,1)$$ and $$\beta = {\log _e}(1 - \alpha )$$. Let $${P_n}(x) = x + {{{x^2}} \over 2} + {{{x^3}} \over 3}\, + \,...\, + \,{{{x^n}} \over n},x \in (0,1)$$. Then the integral $$\int\limits_0^\alpha {{{{t^{50}}} \over {1 - t}}dt} $$ is equal to