Let $\alpha $ and $\beta $ be the roots of $a\;{x^2} + b\;x + c = 0$, then $\mathop {\lim }\limits_{x \to a} \frac{{1 - \cos \left( {a\;{x^2} + bx + c} \right)}}{{{{\left( {x - \alpha } \right)}^2}}}$ is equal to

$\frac{1}{2}{\left( {\alpha - \beta } \right)^2}$

JEE MATHEMATICS MCQ: LIMITS

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Analyzing +4 | -1

Let

\alpha

and

\beta

be the roots of

a\;{x^2} + b\;x + c = 0

, then

\lim\limits_{x \to a} \frac{{1 - \cos \left( {a\;{x^2} + bx + c} \right)}}{{{{\left( {x - \alpha } \right)}^2}}}

is equal to

View answer and explanation

Correct option: C

\frac{{{\alpha ^2}}}{2}{\left( {\alpha - \beta } \right)^2}

Option C (

\frac{{{{\alpha ^2}}}}{2}{\left( {\alpha - \beta } \right)^2}

) is correct because after applying trigonometric identities and L'Hopital's rule, the limit evaluates to the given expression.