Question

For real $x,$ let $f\left(x\right)=x^3+5x+1$, then.

• A. f is one-one but not onto on R
• B. f is onto on R but not one-one
• C. f is one-one and onto on R
• D. f is neither one-one nor onto on R

Answer 1

Answer: (C)

f is one-one and onto on R

Let $x,y\in R$ be such that
$f\left(x\right)=f\left(y\right)$
$\Rightarrow x^3+5x+1=y^3+5y+1$

$\Rightarrow x^3-y^3+5x-5y=0$
$\Rightarrow (x^3-y^3)+5(x-y)=0$

$\Rightarrow (x-y)(x^2+xy+y^2)+5(x-y)=0$
$\Rightarrow (x-y)(x^2+xy+y^2+5)=0$
$\Rightarrow (x-y)\{{(x+\frac{y}{2})}^2+\frac{3y^2}{4}+5\}=0$
$\Rightarrow x=y$
$[\because {(x+\frac{y}{2})}^2+\frac{3y^2}{4}+5\ne 0]$
$\therefore f:R\to R$ is one one
Let $y$ be an arbitrary element in $R$(co-domain). Then, $f\left(x\right)=y$ i.e., $x^3+5x+1=y$ has atleast one real root, say $\alpha$, in R.
$\therefore {\alpha }^3+5\alpha +1=y$
$\Rightarrow f\left(\alpha \right)=y$.
Thus, for each $y\in R$ there exists $\alpha \in R$ such that
$f\left(\alpha \right)=y$.

So, $f:R\to R$ is onto.

Hence, $f:R\to R$ is one - one and onto.