A) not injective
B) surjective but not injective
C) injective but not surjective
D) neither injective nor surjective
Correct Answer: C
Solution :
\[f(x)=\frac{2x}{x-1}\] \[f(x)=2+\frac{2}{x-1}\] \[f'(x)=-\frac{2}{{{(x-1)}^{2}}}<0\,\forall \,x\in R\] Hence f(x) is strictly decreasing So, f(x) is one-one Range: \[let\text{ }y=\frac{2x}{x-1}\] \[xy-y=2x\] \[\Rightarrow \,\,\,\,\,x(y-2)=y\] \[\Rightarrow \,\,\,x=\frac{y}{y-2}\] given that \[x\text{ }\in \text{ }R\text{ }:\text{ }x\] is not a +ve integer \[\therefore \,\,\,\frac{y}{y-2}\,\,\ne \,\,N\,\,\,\,\,\,(N\to Natural\,\,number)\] \[\Rightarrow \,\,y\,\ne \,Ny\,-2N\] \[\Rightarrow \,\,y\,\ne \,\frac{2\,N}{N-1}\] So range \[\notin \] R (in to function)
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