Let $$\alpha $$ and $$\beta $$ be the roots of x² - 3x + p=0 and $$\gamma $$ and $$\delta $$ be the roots of x² - 6x + q = 0. If $$\alpha, \beta, \gamma, \delta $$ form a geometric progression.Then ratio (2q + p) : (2q - p) is:

Let f be a twice differentiable function on (1, 6). If f(2) = 8, f’(2) = 5, f’(x) $$ \ge $$ 1 and f''(x) $$ \ge $$ 4, for all x $$ \in $$ (1, 6), then :

Let $$f\left( x \right) = \int {{{\sqrt x } \over {{{\left( {1 + x} \right)}^2}}}dx\left( {x \ge 0} \right)} $$. Then f(3) – f(1) is eqaul to :

Let $$f(x) = \left| {x - 2} \right|$$ and g(x) = f(f(x)), $$x \in \left[ {0,4} \right]$$. Then
$$\int\limits_0^3 {\left( {g(x) - f(x)} \right)} dx$$ is equal to: