Let \[\alpha \] and \[\beta \] be the roots of \[a{{x}^{2}}+bx+c=0.\]Then \[\underset{x\to \alpha }{\mathop{\lim }}\,\frac{1-\cos (a{{x}^{2}}+bx+c)}{{{(x-\alpha )}^{2}}}\] is equal to:

\[\frac{1}{2}{{(\alpha -\beta )}^{2}}\]

\[\frac{{{a}^{2}}}{2}{{(\alpha -\beta )}^{2}}\]

Let \[\alpha \] and \[\beta \] be the roots of \[a{{x}^{2}}+bx+c=0.\]T

1.	Let \[\alpha \] and \[\beta \] be the roots of \[a{{x}^{2}}+bx+c=0.\]Then \[\underset{x\to \alpha }{\mathop{\lim }}\,\frac{1-\cos (a{{x}^{2}}+bx+c)}{{{(x-\alpha )}^{2}}}\] is equal to:
A.	0
B.	\[\frac{1}{2}{{(\alpha -\beta )}^{2}}\]
C.	\[\frac{{{a}^{2}}}{2}{{(\alpha -\beta )}^{2}}\]
D.	None of these
Answer» D. None of these

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Let \[\alpha \] and \[\beta \] be the roots of \[a{{x}^{2}}+bx+c=0.\]Then \[\underset{x\to \alpha }{\mathop{\lim }}\,\frac{1-\cos (a{{x}^{2}}+bx+c)}{{{(x-\alpha )}^{2}}}\] is equal to:

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