$${x^m}.{y^n} = {\left( {x + y} \right)^{m + n}}$$

$$ \Rightarrow m\ln x + n\ln y = \left( {m + n} \right)\ln \left( {x + y} \right)$$

Differentiating both sides.

$$\therefore$$ $${m \over x} + {n \over y}{{dy} \over {dx}} = {{m + n} \over {x + y}}\left( {1 + {{dy} \over {dx}}} \right)$$

$$ \Rightarrow \left( {{m \over x} - {{m + n} \over {x + y}}} \right) = \left( {{{m + n} \over {x + y}} - {n \over y}} \right){{dy} \over {dx}}$$

$$ \Rightarrow {{my - nx} \over {x\left( {x + y} \right)}} = \left( {{{my - nx} \over {y\left( {x + y} \right)}}} \right){{dy} \over {dx}}$$

$$ \Rightarrow {{dy} \over {dx}} = {y \over x}$$

$$f\left( x \right) = \left\{ {\matrix{ {{x \over {1 - x}},} & {x < 0} \cr {{x \over {1 + x}},} & {x \ge 0} \cr } } \right.$$

$$ \Rightarrow f'\left( x \right) = \left\{ {\matrix{ {{x \over {{{\left( {1 - x} \right)}^2}}},} & {x < 0} \cr {{x \over {{{\left( {1 + x} \right)}^2}}}} & {x \ge 0} \cr } } \right.$$

$$\therefore$$ $$f'\left( x \right)$$ exist at everywhere.

Let $$\alpha ,\,\,\alpha + 1\,\,$$ be roots

Then $$\alpha + \alpha + 1 = b = $$ sum of -

roots $$\alpha \left( {\alpha + 1} \right) = c$$

$$=$$ product of roots

$$\therefore$$ $${b^2} - 4c $$

$$ = {\left( {2\alpha + 1} \right)^2} - 4\alpha \left( {\alpha + 1} \right)$$

$$ = 1.$$

$${x^2} - \left( {a - 2} \right)x - a - 1 = 0$$

$$ \Rightarrow \alpha + \beta = a - 2;\,\,\alpha \beta = - \left( {a + 1} \right)$$

$${\alpha ^2} + {\beta ^2} = {\left( {\alpha + \beta } \right)^2} - 2\alpha \beta $$

$$ = {a^2} - 2a + 6 = {\left( {a - 1} \right)^2} + 5$$

For min. value of $${\alpha ^2} + {\beta ^2}$$ where $$\alpha $$ is an integer

$$ \Rightarrow \,\,a = 1.$$