A) \[\sqrt{A+21}\]
B) \[\sqrt{A}\]
C) \[\sqrt{A+1}\]
D) \[\sqrt{A+5}\]
Correct Answer: C
Solution :
\[\underset{x\to 0}{\mathop{Lim}}\,\,\,x\left[ \frac{4}{x} \right]=A\] \[\Rightarrow \,\,\,\,\,\underset{x\to 0}{\mathop{Lim}}\,\,\,\,x\left( \frac{4}{x}-\left\{ \frac{4}{x} \right\} \right)=A\] \[\Rightarrow \,\,\,\,\,4-\underset{x\to 0}{\mathop{Lim}}\,\,\,\,x\,\left\{ \frac{4}{x} \right\}=A\,\,\,\,\,\,\,\Rightarrow A=4\] Now, at \[x=\sqrt{A+1}\] i.e. \[x=\sqrt{5},\] \[f(x)=[{{x}^{2}}]\,\sin \pi x\] is discontinuous Whereas at \[x=5,\] 2 and 3, \[f(x)\] is continuous.You need to login to perform this action.
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