MathematicsMedium179×since 2002Q4265Let α\alphaα and β\betaβ be the distinct roots of ax2+bx+c=0a{x^2} + bx + c = 0ax2+bx+c=0, then limx→α1−cos(ax2+bx+c)(x−α)2\mathop {\lim }\limits_{x \to \alpha } {{1 - \cos \left( {a{x^2} + bx + c} \right)} \over {{{\left( {x - \alpha } \right)}^2}}}x→αlim(x−α)21−cos(ax2+bx+c) is equal toAa2(α−β)22{{{a^2}{{\left( {\alpha - \beta } \right)}^2}} \over 2}2a2(α−β)2B0C−a2(α−β)22- {{{a^2}{{\left( {\alpha - \beta } \right)}^2}} \over 2}−2a2(α−β)2D(α−β)22{{{{\left( {\alpha - \beta } \right)}^2}} \over 2}2(α−β)2Check answerSkip