$$\mathop {\lim }\limits_{x \to {1 \over {\sqrt 2 }}} {{\sin ({{\cos }^{ - 1}}x) - x} \over {1 - \tan ({{\cos }^{ - 1}}x)}}$$ 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 :
Let S be the set of all the natural numbers, for which the line $${x \over a} + {y \over b} = 2$$ is a tangent to the curve $${\left( {{x \over a}} \right)^n} + {\left( {{y \over b}} \right)^n} = 2$$ at the point (a, b), ab $$\ne$$ 0. Then :