If y(x) is the solution of the differential equation $${{dy} \over {dx}} + \left( {{{2x + 1} \over x}} \right)y = {e^{ - 2x}},\,\,x > 0,\,$$ where $$y\left( 1 \right) = {1 \over 2}{e^{ - 2}},$$ then

The curve amongst the family of curves represented by the differential equation, (x² – y²)dx + 2xy dy = 0 which passes through (1, 1) is :

Let f be a differentiable function such that f '(x) = 7 - $${3 \over 4}{{f\left( x \right)} \over x},$$ (x > 0) and f(1) $$ \ne $$ 4. Then $$\mathop {\lim }\limits_{x \to 0'} \,$$ xf$$\left( {{1 \over x}} \right)$$ :

If  $${{dy} \over {dx}} + {3 \over {{{\cos }^2}x}}y = {1 \over {{{\cos }^2}x}},\,\,x \in \left( {{{ - \pi } \over 3},{\pi \over 3}} \right)$$  and  $$y\left( {{\pi \over 4}} \right) = {4 \over 3},$$  then  $$y\left( { - {\pi \over 4}} \right)$$   equals -