Let two points be A(1, –1) and B(0, 2). If a point P(x', y') be such that the area of $$\Delta $$PAB = 5 sq. units and it lies on the line, 3x + y – 4$$\lambda $$ = 0, then a value of $$\lambda $$ is :

Let y = y(x) be a solution of the differential equation,

$$\sqrt {1 - {x^2}} {{dy} \over {dx}} + \sqrt {1 - {y^2}} = 0$$, |x| < 1.

If $$y\left( {{1 \over 2}} \right) = {{\sqrt 3 } \over 2}$$, then $$y\left( { - {1 \over {\sqrt 2 }}} \right)$$ is equal to :

$$\mathop {\lim }\limits_{x \to 0} {\left( {{{3{x^2} + 2} \over {7{x^2} + 2}}} \right)^{{1 \over {{x^2}}}}}$$ is equal to

Let ƒ(x) = (sin(tan^–1x) + sin(cot^–1x))² – 1, |x| > 1.
If $${{dy} \over {dx}} = {1 \over 2}{d \over {dx}}\left( {{{\sin }^{ - 1}}\left( {f\left( x \right)} \right)} \right)$$ and $$y\left( {\sqrt 3 } \right) = {\pi \over 6}$$, then y($${ - \sqrt 3 }$$) is equal to :