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

The sum of the absolute minimum and the absolute maximum values of the

function f(x) = |3x $$-$$ x² + 2| $$-$$ x in the interval [$$-$$1, 2] is :

The area bounded by the curve y = |x² $$-$$ 9| and the line y = 3 is :