The derivative of
$${\tan ^{ - 1}}\left( {{{\sqrt {1 + {x^2}} - 1} \over x}} \right)$$ with
respect to $${\tan ^{ - 1}}\left( {{{2x\sqrt {1 - {x^2}} } \over {1 - 2{x^2}}}} \right)$$ at x = $${1 \over 2}$$ is :

If $$\alpha $$ and $$\beta $$ are the roots of the equation,
7x² – 3x – 2 = 0, then the value of
$${\alpha \over {1 - {\alpha ^2}}} + {\beta \over {1 - {\beta ^2}}}$$ is equal to :

If the length of the chord of the circle,
x² + y² = r² (r > 0) along the line, y – 2x = 3 is r,
then r² is equal to :

If for some $$\alpha $$ $$ \in $$ R, the lines

L₁ : $${{x + 1} \over 2} = {{y - 2} \over { - 1}} = {{z - 1} \over 1}$$ and

L₂ : $${{x + 2} \over \alpha } = {{y + 1} \over {5 - \alpha }} = {{z + 1} \over 1}$$ are coplanar,

then the line L₂ passes through the point :