If $$z=\frac{1}{2}-2 i$$ is such that $$|z+1|=\alpha z+\beta(1+i), i=\sqrt{-1}$$ and $$\alpha, \beta \in \mathbb{R}$$, then $$\alpha+\beta$$ is equal to

Consider the function $$f:\left[\frac{1}{2}, 1\right] \rightarrow \mathbb{R}$$ defined by $$f(x)=4 \sqrt{2} x^3-3 \sqrt{2} x-1$$. Consider the statements

(I) The curve $$y=f(x)$$ intersects the $$x$$-axis exactly at one point.

(II) The curve $$y=f(x)$$ intersects the $$x$$-axis at $$x=\cos \frac{\pi}{12}$$.

Then

Let $$P Q R$$ be a triangle with $$R(-1,4,2)$$. Suppose $$M(2,1,2)$$ is the mid point of $$\mathrm{PQ}$$. The distance of the centroid of $$\triangle \mathrm{PQR}$$ from the point of intersection of the lines $$\frac{x-2}{0}=\frac{y}{2}=\frac{z+3}{-1}$$ and $$\frac{x-1}{1}=\frac{y+3}{-3}=\frac{z+1}{1}$$ is

$$\mathop {\lim }\limits_{x \to {\pi \over 2}} \left( {{1 \over {{{\left( {x - {\pi \over 2}} \right)}^2}}}\int\limits_{{x^3}}^{{{\left( {{\pi \over 2}} \right)}^3}} {\cos \left( {{t^{{1 \over 3}}}} \right)dt} } \right)$$ is equal to