question_answer

Let \[\alpha \] and \[\beta \] be the distinct 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     AIEEE  Solved  Paper-2005

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

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

C) 0                             

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

Correct Answer: D

Solution :

              Note Ifandare the roots of the equation then this equation can be rewritten as .