JEE Main 2014Session, Shift 1MathematicsMediumLimit, Continuity and Differentiability420× since 2002If x=−1x=-1x=−1 and x=2x=2x=2 are extreme points of f(x)=α log∣x∣+βx2+xf\left( x \right) = \alpha \,\log \left| x \right|+\beta {x^2} + xf(x)=αlog∣x∣+βx2+x thenAOption A. α=2,β=−12\alpha = 2,\beta = - {1 \over 2}α=2,β=−21BOption B. α=2,β=12\alpha = 2,\beta = {1 \over 2}α=2,β=21COption C. α=−6,β=12\alpha = - 6,\beta = {1 \over 2}α=−6,β=21DOption D. α=−6,β=−12\alpha = - 6,\beta = -{1 \over 2}α=−6,β=−21CheckSkip · show solution
