If $${\sin ^{ - 1}}\left( {x - {{{x^2}} \over 2} + {{{x^3}} \over 4} - ....} \right)$$ $$$ + {\cos ^{ - 1}}\left( {{x^2} - {{{x^4}} \over 2} + {{{x^6}} \over 4} - ....} \right) = {\pi \over 2}$$$
for $$0 < \left| x \right| < \sqrt 2 ,$$ then $$x$$ equals

If $$f\left( x \right) = x{e^{x\left( {1 - x} \right)}},$$ then $$f(x)$$ is

The triangle formed by the tangent to the curve $$f\left( x \right) = {x^2} + bx - b$$ at the point $$(1, 1)$$ and the coordinate axex, lies in the first quadrant. If its area is $$2$$, then the value of $$b$$ is

Let $$f\left( x \right) = \left( {1 + {b^2}} \right){x^2} + 2bx + 1$$ and let $$m(b)$$ be the minimum value of $$f(x)$$. As $$b$$ varies, the range of $$m(b)$$ is