Let A (3, 0, –1), B(2, 10, 6) and C(1, 2, 1) be the vertices of a triangle and M be the midpoint of AC. If G divides BM in the ratio, 2 : 1, then cos ($$\angle $$GOA) (O being the origin) is equal to :

Let f : R $$ \to $$ R be differentiable at c $$ \in $$ R and f(c) = 0. If g(x) = |f(x)| , then at x = c, g is :

The value of $$\int\limits_0^{2\pi } {\left[ {\sin 2x\left( {1 + \cos 3x} \right)} \right]} dx$$,
where [t] denotes the greatest integer function is :

If $$\int {{{dx} \over {{{\left( {{x^2} - 2x + 10} \right)}^2}}}} = A\left( {{{\tan }^{ - 1}}\left( {{{x - 1} \over 3}} \right) + {{f\left( x \right)} \over {{x^2} - 2x + 10}}} \right) + C$$

where C is a constant of integration then :