Suppose $$f(x)$$ is differentiable at x = 1 and

$$\mathop {\lim }\limits_{h \to 0} {1 \over h}f\left( {1 + h} \right) = 5$$, then $$f'\left( 1 \right)$$ equals

Let $$\alpha$$ and $$\beta$$ be the distinct roots of $$a{x^2} + bx + c = 0$$, then

$$\mathop {\lim }\limits_{x \to \alpha } {{1 - \cos \left( {a{x^2} + bx + c} \right)} \over {{{\left( {x - \alpha } \right)}^2}}}$$ is equal to

Let f be differentiable for all x. If f(1) = -2 and f'(x) $$ \ge $$ 2 for
x $$ \in \left[ {1,6} \right]$$, then

If $$f$$ is a real valued differentiable function satisfying

$$\left| {f\left( x \right) - f\left( y \right)} \right|$$ $$ \le {\left( {x - y} \right)^2}$$, $$x, y$$ $$ \in R$$
and $$f(0)$$ = 0, then $$f(1)$$ equals