If $\alpha ,\,\,\beta $ are the distinct roots of ${x^2} + bx + c = 0,$ then, $\mathop {\lim }\limits_{x \to \beta } \frac{{{e^{2({x^2} + bx + c)}} - 1 - 2({x^2} + bx + c)}}{{{{(x \to \beta )}^2}}}$ is equal to

JEE MATHEMATICS MCQ: LIMITS

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\alpha ,\,\,\beta

are the distinct roots of

{x^2} + bx + c = 0,

then,

\lim\limits_{x \to \beta } \frac{{{e^{2({x^2} + bx + c)}} - 1 - 2({x^2} + bx + c)}}{{{{(x \to \beta )}^2}}}

is equal to

View answer and explanation

Correct option: C

2\left( {{b^2} - 4c} \right)

Option C (

2\left( {{{b^2} - 4c}} \right)

) is correct because by factoring the quadratic and using L'Hopital's rule, the limit simplifies to

2(\beta - \alpha)^2

, which equals

2(b^2 - 4c)