A) \[f(0)=\frac{5}{2}\]
B) \[[f(0)]=-2\]
C) \[\{f(0)\}=-0.5\]
D) \[\{f(0)].\{f(0)\}=-1.5\]
Correct Answer: D
Solution :
[d]:\[\underset{x\to 0}{\mathop{Lim}}\,\frac{x-{{e}^{x}}+1-(1-cos2x)}{{{x}^{2}}}=-\frac{1}{2}-2\] \[=-\frac{5}{2}\] hence for continuity \[f(0)=-\frac{5}{2}\] \[\therefore \]\[[f(x)]=\left[ -\frac{5}{2} \right]=-3;\{f(0)\}=\left\{ -\frac{5}{2} \right\}=\frac{1}{2}\] Hence \[[f(0)].\{f(0)\}=-\frac{3}{2}=-1.5\]
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