The value of $$\int\limits_{ - \pi /2}^{\pi /2} {{{{{\sin }^2}x} \over {1 + {2^x}}}} dx$$ is

Let g(x) = cosx², f(x) = $$\sqrt x $$ and $$\alpha ,\beta \left( {\alpha < \beta } \right)$$ be the roots of the quadratic equation 18x² - 9$$\pi $$x + $${\pi ^2}$$ = 0. Then the area (in sq. units) bounded by the curve
y = (gof)(x) and the lines $$x = \alpha $$, $$x = \beta $$ and y = 0 is :

Let S = { t $$ \in R:f(x) = \left| {x - \pi } \right|.\left( {{e^{\left| x \right|}} - 1} \right)$$$$\sin \left| x \right|$$ is not differentiable at t}, then the set S is equal to

Let $$f\left( x \right) = {x^2} + {1 \over {{x^2}}}$$ and $$g\left( x \right) = x - {1 \over x}$$,
$$x \in R - \left\{ { - 1,0,1} \right\}$$.
If $$h\left( x \right) = {{f\left( x \right)} \over {g\left( x \right)}}$$, then the local minimum value of h(x) is