$$\mathop {\lim }\limits_{x \to 0} {{x\left( {{e^{\left( {\sqrt {1 + {x^2} + {x^4}} - 1} \right)/x}} - 1} \right)} \over {\sqrt {1 + {x^2} + {x^4}} - 1}}$$

If the function
$$f\left( x \right) = \left\{ {\matrix{ {{k_1}{{\left( {x - \pi } \right)}^2} - 1,} & {x \le \pi } \cr {{k_2}\cos x,} & {x > \pi } \cr } } \right.$$ is
twice differentiable, then the ordered pair (k₁, k₂) is equal to :

If $$\alpha $$ is positive root of the equation, p(x) = x² - x - 2 = 0, then

$$\mathop {\lim }\limits_{x \to {\alpha ^ + }} {{\sqrt {1 - \cos \left( {p\left( x \right)} \right)} } \over {x + \alpha - 4}}$$ is equal to :

Let $$f:\left( {0,\infty } \right) \to \left( {0,\infty } \right)$$ be a differentiable function such that f(1) = e and
$$\mathop {\lim }\limits_{t \to x} {{{t^2}{f^2}(x) - {x^2}{f^2}(t)} \over {t - x}} = 0$$. If f(x) = 1, then x is equal to :