Let $$f:\left( { - {\pi \over 4},{\pi \over 4}} \right) \to R$$ be defined as $$f(x) = \left\{ {\matrix{ {{{(1 + |\sin x|)}^{{{3a} \over {|\sin x|}}}}} & , & { - {\pi \over 4} < x < 0} \cr b & , & {x = 0} \cr {{e^{\cot 4x/\cot 2x}}} & , & {0 < x < {\pi \over 4}} \cr } } \right.$$

If f is continuous at x = 0, then the value of 6a + b² is equal to :

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 :