Let f : R $$\to$$ R be a function such that f(2) = 4 and f'(2) = 1. Then, the value of $$\mathop {\lim }\limits_{x \to 2} {{{x^2}f(2) - 4f(x)} \over {x - 2}}$$ is equal to :

Let f : R $$\to$$ R be defined as

$$f(x) = \left\{ {\matrix{ {{{\lambda \left| {{x^2} - 5x + 6} \right|} \over {\mu (5x - {x^2} - 6)}},} & {x < 2} \cr {{e^{{{\tan (x - 2)} \over {x - [x]}}}},} & {x > 2} \cr {\mu ,} & {x = 2} \cr } } \right.$$

where [x] is the greatest integer is than or equal to x. If f is continuous at x = 2, then $$\lambda$$ + $$\mu$$ is equal to :

Let f : R $$\to$$ R be defined as $$f(x) = \left\{ {\matrix{ {{{{x^3}} \over {{{(1 - \cos 2x)}^2}}}{{\log }_e}\left( {{{1 + 2x{e^{ - 2x}}} \over {{{(1 - x{e^{ - x}})}^2}}}} \right),} & {x \ne 0} \cr {\alpha ,} & {x = 0} \cr } } \right.$$

If f is continuous at x = 0, then $$\alpha$$ is equal to :

If $$f:R \to R$$ is given by $$f(x) = x + 1$$, then the value of $$\mathop {\lim }\limits_{n \to \infty } {1 \over n}\left[ {f(0) + f\left( {{5 \over n}} \right) + f\left( {{{10} \over n}} \right) + ...... + f\left( {{{5(n - 1)} \over n}} \right)} \right]$$ is :