MathematicsMedium210×since 2002Q3428Let f : R →\to→ R be a differentiable function such that f(π4)=2, f(π2)=0f\left( {{\pi \over 4}} \right) = \sqrt 2 ,\,f\left( {{\pi \over 2}} \right) = 0f(4π)=2,f(2π)=0 and f′(π2)=1f'\left( {{\pi \over 2}} \right) = 1f′(2π)=1 and let g(x)=∫xπ/4(f′(t)sect+tantsect f(t)) dtg(x) = \int_x^{\pi /4} {(f'(t)\sec t + \tan t\sec t\,f(t))\,dt}g(x)=∫xπ/4(f′(t)sect+tantsectf(t))dt for x∈[π4,π2)x \in \left[ {{\pi \over 4},{\pi \over 2}} \right)x∈[4π,2π). Then limx→(π2)−g(x)\mathop {\lim }\limits_{x \to {{\left( {{\pi \over 2}} \right)}^ - }} g(x)x→(2π)−limg(x) is equal to :A2B3C4D−-−3Check answerSkip