A) 198
B) 199
C) 200
D) 201
Correct Answer: B
Solution :
We have, \[{{x}^{2}}+10000\left[ x \right]=10000x\]\[\Rightarrow {{x}^{2}}+10000\left( x-\{x\} \right)=10000x\]\[\Rightarrow {{x}^{2}}+10000x-10000\{x\}=10000x\] |
\[\Rightarrow \,\{x\}=\frac{{{x}^{2}}}{10000}\Rightarrow 0\le \{x\}<1\] |
\[\therefore \,\,\,\,\,\,\,\,\,0\le \frac{{{x}^{2}}}{10000}<1\] |
\[-100<x<100\]Total number is 199 |
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