A) one - one but not onto
B) one - one and onto
C) onto but not one - one
D) neither one - one nor onto
Correct Answer: C
Solution :
\[\because \]\[f\,(x)={{\log }_{0.5}}\,({{x}^{2}}-5x+6)\]\[\Rightarrow \]f(x) will be defined if \[{{x}^{2}}-5x+6>0\]\[\Rightarrow \]\[(x-2)\,\,(x-3)>0\] |
from graph of f(x) we can say that it will be a many-one and onto function. |
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