$$\mathop {\lim }\limits_{x \to {1 \over {\sqrt 2 }}} {{\sin ({{\cos }^{ - 1}}x) - x} \over {1 - \tan ({{\cos }^{ - 1}}x)}}$$ is equal to :

Let f, g : R $$\to$$ R be two real valued functions defined as $$f(x) = \left\{ {\matrix{ { - |x + 3|} & , & {x < 0} \cr {{e^x}} & , & {x \ge 0} \cr } } \right.$$ and $$g(x) = \left\{ {\matrix{ {{x^2} + {k_1}x} & , & {x < 0} \cr {4x + {k_2}} & , & {x \ge 0} \cr } } \right.$$, where k₁ and k₂ are real constants. If (gof) is differentiable at x = 0, then (gof) ($$-$$ 4) + (gof) (4) is equal to :

$$\mathop {\lim }\limits_{x \to {\pi \over 2}} \left( {{{\tan }^2}x\left( {{{(2{{\sin }^2}x + 3\sin x + 4)}^{{1 \over 2}}} - {{({{\sin }^2}x + 6\sin x + 2)}^{{1 \over 2}}}} \right)} \right)$$ is equal to

Let $$f(x) = \left\{ {\matrix{ {{{\sin (x - [x])} \over {x - [x]}}} & {,\,x \in ( - 2, - 1)} \cr {\max \{ 2x,3[|x|]\} } & {,\,|x| < 1} \cr 1 & {,\,otherwise} \cr } } \right.$$

where [t] denotes greatest integer $$\le$$ t. If m is the number of points where $$f$$ is not continuous and n is the number of points where $$f$$ is not differentiable, then the ordered pair (m, n) is :