A) Not one-one
B) Not onto
C) Many-one and onto
D) One-one and onto
Correct Answer: D
Solution :
[d] Let \[{{x}_{1}},{{x}_{2}}\in A\] and let \[f({{x}_{1}})=f({{x}_{2}})\] Or \[\frac{{{x}_{1}}-2}{{{x}_{1}}-3}=\frac{{{x}_{2}}-2}{{{x}_{2}}-3}\] Or \[{{x}_{1}}={{x}_{2}}\]so, f is one-one. To find whether f is onto or not, first let us find the range of f. Let \[y=f(x)=\frac{x-2}{x-3}orx=\frac{3y-2}{y-1}\] X is defined if \[y\ne 1,\] i.e, the range of f is \[R-\{1\}\] Which is also the co-domain of f. Also, for no value of y, x can be 3 i.e., if we put 3 \[=x=\frac{3y-2}{y-1}then3y-3=3y-2or-3=-2\] Which is not possible. Hence, f is onto.
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