Question

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

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

Answer 1

The correct option is A f is one-one and onto R

Let $x,y\in R$ be such that

$f\left(x\right)=f\left(y\right)$

$\Rightarrow$ $x^3+5x+1=y^3+5y+1$

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

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

$\Rightarrow$ $(x-y)(x^2+xy+y^2)+5(x-y)=0$

$\Rightarrow$ $(x-y)(x^2+xy+y^2+5)=0$

$\Rightarrow$ $(x-y)\{{(x+\frac{y}{2})}^2+\frac{3y^2}{4}+5\}=0$

$\Rightarrow$ $x=y$

Since, ${(x+\frac{y}{2})}^2+\frac{3y^2}{4}+5\ne 0$

$\therefore$ $f:R\to R$ is one-one.

Let $y$ be an arbitary element in $R$(co-domain ).

Then, $\left(x\right)=y$ i.e. $x^3+5x+1=y$ has at-least 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

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