A) \[f(x)\] is a many-one and into function
B) \[f(x)=0\] for exactly five number of values of x
C) \[f(x)=0\]for only two real values
D) none of these
Correct Answer: A
Solution :
| \[f(x)={{[x]}^{2}}+[x+1]-3=\{[x]+2\}\{[x]-1\}\]So, \[x=1,1.1,1.2,......\]\[\Rightarrow f(x)=0\] |
| \[\therefore f(x)\] is many one only integral values will be attained |
| \[\therefore f(x)\] is into |
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