JEE Main 2011Annual, Paper 1MathematicsMediumLimit, Continuity and Differentiability420× since 2002For x∈(0,5π2),x \in \left( {0,{{5\pi } \over 2}} \right),x∈(0,25π), define f(x)=∫0xtsint dt.f\left( x \right) = \int\limits_0^x {\sqrt t \sin t\,dt.}f(x)=0∫xtsintdt. Then fff hasAOption A. local minimum at π\piπ and 2π2\pi2πBOption B. local minimum at π\piπ and local maximum at 2π2\pi2πCOption C. local maximum at π\piπ and local minimum at 2π2\pi2πDOption D. local maximum at π\piπ and 2π2\pi2πCheckSkip · show solution
