JEE Main 202120 Jul, Evening shiftMathematicsMediumIntegral Calculus369× since 2002Let g(t)=∫−π/2π/2cos(π4t+f(x))dxg(t) = \int_{ - \pi /2}^{\pi /2} {\cos \left( {{\pi \over 4}t + f(x)} \right)} dxg(t)=∫−π/2π/2cos(4πt+f(x))dx, where f(x)=loge(x+x2+1),x∈Rf(x) = {\log _e}\left( {x + \sqrt {{x^2} + 1} } \right),x \in Rf(x)=loge(x+x2+1),x∈R. Then which one of the following is correct?AOption A. g(1) = g(0)BOption B. 2g(1)=g(0)\sqrt 2 g(1) = g(0)2g(1)=g(0)COption C. g(1)=2g(0)g(1) = \sqrt 2 g(0)g(1)=2g(0)DOption D. g(1) + g(0) = 0CheckSkip · show solution
