MathematicsMedium175×since 2002Q2729Let g(x)=f(x)+f(1−x)\mathrm{g}(x)=f(x)+f(1-x)g(x)=f(x)+f(1−x) and f′′(x)>0,x∈(0,1)f^{\prime \prime}(x) > 0, x \in(0,1)f′′(x)>0,x∈(0,1). If g\mathrm{g}g is decreasing in the interval (0,a)(0, a)(0,a) and increasing in the interval (α,1)(\alpha, 1)(α,1), then tan−1(2α)+tan−1(1α)+tan−1(α+1α)\tan ^{-1}(2 \alpha)+\tan ^{-1}\left(\frac{1}{\alpha}\right)+\tan ^{-1}\left(\frac{\alpha+1}{\alpha}\right)tan−1(2α)+tan−1(α1)+tan−1(αα+1) is equal to :A3π4\frac{3 \pi}{4}43πBπ\piπC5π4\frac{5 \pi}{4}45πD3π2\frac{3 \pi}{2}23πCheck answerSkip