A function $$y=f(x)$$ has a second order derivative $$f''\left( x \right) = 6\left( {x - 1} \right).$$ If its graph passes through the point $$(2, 1)$$ and at that point the tangent to the graph is $$y = 3x - 5$$, then the function is :

If $$2a+3b+6c=0$$, then at least one root of the equation
$$a{x^2} + bx + c = 0$$ lies in the interval

The normal to the curve x = a(1 + cos $$\theta $$), $$y = a\sin \theta $$ at $$'\theta '$$ always passes through the fixed point

Let $$A = \left( {\matrix{ 0 & 0 & { - 1} \cr 0 & { - 1} & 0 \cr { - 1} & 0 & 0 \cr } } \right)$$. The only correct

statement about the matrix $$A$$ is