A) \[f(0)=\frac{1}{e}\]
B) \[f(0)=0\]
C) \[f(0)=e\]
D) None of these
Correct Answer: C
Solution :
For continuity at \[x=0\], we must have \[f(0)=\underset{x\to 0}{\mathop{\text{lim}}}\,f(x)\]\[\underset{x\to \alpha }{\mathop{\lim }}\,\,\frac{1-\cos \,(a{{x}^{2}}+bx+c)}{{{(x-\alpha )}^{2}}}\]\[=\underset{x\to 0}{\mathop{\text{lim}}}\,{{\left\{ {{(1+x)}^{\frac{1}{x}}} \right\}}^{x\cot x}}\] \[=\underset{x\to 0}{\mathop{\text{lim}}}\,{{\left\{ {{(1+x)}^{\frac{1}{x}}} \right\}}^{\underset{x\to 0}{\mathop{\text{lim}}}\,\,\left( \frac{x}{\tan x} \right)}}\] \[={{e}^{1}}=e\].
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