JEE Main 202127 Aug, Morning shiftMathematicsMediumLimit, Continuity and Differentiability420× since 2002If α\alphaα, β\betaβ are the distinct roots of x² + bx + c = 0, then limx→βe2(x2+bx+c)−1−2(x2+bx+c)(x−β)2\mathop {\lim }\limits_{x \to \beta } {{{e^{2({x^2} + bx + c)}} - 1 - 2({x^2} + bx + c)} \over {{{(x - \beta )}^2}}}x→βlim(x−β)2e2(x2+bx+c)−1−2(x2+bx+c) is equal to :AOption A. b² + 4cBOption B. 2(b² + 4c)COption C. 2(b² −-− 4c)DOption D. b² −-− 4cCheckSkip · show solution
