Let f : R $$\to$$ R be a differentiable function such that $$f\left( {{\pi \over 4}} \right) = \sqrt 2 ,\,f\left( {{\pi \over 2}} \right) = 0$$ and $$f'\left( {{\pi \over 2}} \right) = 1$$ and

let $$g(x) = \int_x^{\pi /4} {(f'(t)\sec t + \tan t\sec t\,f(t))\,dt} $$ for $$x \in \left[ {{\pi \over 4},{\pi \over 2}} \right)$$. Then $$\mathop {\lim }\limits_{x \to {{\left( {{\pi \over 2}} \right)}^ - }} g(x)$$ is equal to :

Let f : R $$\to$$ R be a continuous function satisfying f(x) + f(x + k) = n, for all x $$\in$$ R where k > 0 and n is a positive integer. If $${I_1} = \int\limits_0^{4nk} {f(x)dx} $$ and $${I_2} = \int\limits_{ - k}^{3k} {f(x)dx} $$, then :

The area of the bounded region enclosed by the curve

$$y = 3 - \left| {x - {1 \over 2}} \right| - |x + 1|$$ and the x-axis is :

Let the slope of the tangent to a curve y = f(x) at (x, y) be given by 2 $$\tan x(\cos x - y)$$. If the curve passes through the point $$\left( {{\pi \over 4},0} \right)$$, then the value of $$\int\limits_0^{\pi /2} {y\,dx} $$ is equal to :