The minimum value of the function $$f(x) = \int\limits_0^2 {{e^{|x - t|}}dt} $$ is :

The statement $$\left( {p \wedge \left( { \sim q} \right)} \right) \Rightarrow \left( {p \Rightarrow \left( { \sim q} \right)} \right)$$ is

Let $$x=2$$ be a local minima of the function $$f(x)=2x^4-18x^2+8x+12,x\in(-4,4)$$. If M is local maximum value of the function $$f$$ in ($$-4,4)$$, then M =

The mean and variance of the marks obtained by the students in a test are 10 and 4 respectively. Later, the marks of one of the students is increased from 8 to 12. If the new mean of the marks is 10.2, then their new variance is equal to :