A) \[0\]
B) \[\frac{1}{2}{{(\alpha -\beta )}^{2}}\]
C) \[\frac{{{a}^{2}}}{2}{{(\alpha -\beta )}^{2}}\]
D) \[(\alpha -\beta )\]
E) \[1\]
Correct Answer: C
Solution :
Since, \[\alpha \]and\[\beta \]are the roots of\[a{{x}^{2}}+bx+c=0\]. \[\therefore \] \[a(x-\alpha )(x-\beta )=a{{x}^{2}}+bx+c\] Now, \[\underset{x\to \alpha }{\mathop{\lim }}\,=\frac{1-\cos (a{{x}^{2}}+bx+c)}{{{(x-\alpha )}^{2}}}\] \[=\underset{x\to \alpha }{\mathop{\lim }}\,=\frac{2{{\sin }^{2}}\left( \frac{a{{x}^{2}}+bx+c}{2} \right)}{{{(x-\alpha )}^{2}}}\] \[=\underset{x\to \alpha }{\mathop{\lim }}\,=\frac{2{{\sin }^{2}}\frac{a(x-\alpha )(x-\beta )}{2}}{{{\left( \frac{a(x-\alpha )(x-\beta )}{2} \right)}^{2}}}\times \frac{{{a}^{2}}{{(x-\beta )}^{2}}}{4}\] \[=\frac{{{a}^{2}}{{(\alpha -\beta )}^{2}}}{2}\]
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