The normal to the curve, $${x^2} + 2xy - 3{y^2} = 0$$, at $$(1,1)$$

If $$x=-1$$ and $$x=2$$ are extreme points of $$f\left( x \right) = \alpha \,\log \left| x \right|+\beta {x^2} + x$$ then

If $$f$$ and $$g$$ are differentiable functions in $$\left[ {0,1} \right]$$ satisfying
$$f\left( 0 \right) = 2 = g\left( 1 \right),g\left( 0 \right) = 0$$ and $$f\left( 1 \right) = 6,$$ then for some $$c \in \left] {0,1} \right[$$

The intercepts on $$x$$-axis made by tangents to the curve,
$$y = \int\limits_0^x {\left| t \right|dt,x \in R,} $$ which are parallel to the line $$y=2x$$, are equal to :