A) f is one-one but not onto R
B) f is onto R but not one-one
C) f is one-one and onto R
D) f is neither one-one nor onto R
Correct Answer: C
Solution :
Let, \[x,y\in R\]be such that \[f(x)=f(y)\] \[\Rightarrow \] \[{{x}^{3}}+5x+1={{y}^{3}}+5y+1\] \[\Rightarrow \] \[({{x}^{3}}-{{y}^{3}})+5(x-y)=0\] \[\Rightarrow \] \[(x-y)({{x}^{2}}+xy+{{y}^{2}}+5)=0\] \[\Rightarrow \] \[(x-y)\left\{ {{\left( x+\frac{y}{2} \right)}^{2}}+\frac{3{{y}^{2}}}{4}+5 \right\}=0\] \[\Rightarrow \] \[x=y\] \[\left[ \because {{\left( x+\frac{y}{2} \right)}^{2}}+\frac{3{{y}^{2}}}{4}+5\ne 0 \right]\] \[\therefore \] \[f:R\to R\]is one Let y be an arbitrary element in R (codomain) .Then, \[f(x)=y,i.e.{{x}^{3}}+5x+1=y\]has atleast one real root, say \[\alpha \]in R. \[\therefore \] \[{{\alpha }^{3}}+5\alpha +1=y\] \[\Rightarrow f(\alpha )=y\] Thus, for each \[y\in R\]there exists \[\alpha \in R\]such that \[f(\alpha )=y.\]So, \[f:R\to R\]is onto. Hence, \[f:R\to R\]is one-one and onto.
You need to login to perform this action.
You will be redirected in
3 sec
