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
Let the shortest distance between the lines
$$L: \frac{x-5}{-2}=\frac{y-\lambda}{0}=\frac{z+\lambda}{1}, \lambda \geq 0$$ and
$$L_{1}: x+1=y-1=4-z$$ be $$2 \sqrt{6}$$. If $$(\alpha, \beta, \gamma)$$ lies on $$L$$,
then which of the following is NOT possible?
Let $$\vec{a}=2 \hat{i}+\hat{j}+\hat{k}$$, and $$\vec{b}$$ and $$\vec{c}$$ be two nonzero vectors such that $$|\vec{a}+\vec{b}+\vec{c}|=|\vec{a}+\vec{b}-\vec{c}|$$ and $$\vec{b} \cdot \vec{c}=0$$. Consider the following two statements:
(A) $$|\vec{a}+\lambda \vec{c}| \geq|\vec{a}|$$ for all $$\lambda \in \mathbb{R}$$.
(B) $$\vec{a}$$ and $$\vec{c}$$ are always parallel.
Then,