Relations and Functions — Maths STD 12 Science — Question
Gujarat BoardEnglish MediumSTD 12 ScienceMathsRelations and Functions3 Marks
Question
Prove that $f: R \rightarrow R , f(x)=x^3+x$ is one $-$ one onto function.
✓
Answer
One $-$ one : Suppose $x, y \in R$ such that $f(x)=f(y)$
$\because f(x)=f(y) $
$\Rightarrow x^3+x=y^3+y$
$ \Rightarrow x^3-y^3+x-y=0$
$ \Rightarrow(x-y)\left(x^2+x y+y^2\right)-(x-y)=0$
$ \Rightarrow(x-y)\left(x^2+x y+y^2+1\right)=0$
$ \Rightarrow x-y=0 $
$\quad\left[\begin{array}{l} \because x^2+x y+y^2 \geq 0 \forall x, y \in R \\ \therefore x^2+x y+y^2 \geq 1 \forall x, y \in R \end{array}\right] $
thus $f(x)=f(y) \Rightarrow x=y \forall x, y \in R$
$\therefore f$, is one $-$ one function.
Onto : Suppose $y$ is a arbitrary constant in $R$ then
$f(x)=y$
$ \Rightarrow x^3+x=y$
$\Rightarrow x^3+x-y=0 $
We know that is every equation of odd power has a real root then for every real value of $y$, equation $x^3+$
$x-y=0$ has a real root.Suppose this root is $\alpha$ then
$ \alpha^3+\alpha-y=0 $
$\Rightarrow \alpha^3+\alpha=y$
$ \Rightarrow f(x)=y $
thus for every $y \in R$, there exist $\alpha \in R$ then $f(x)= y$.
Hence $f$ is onto.
Thus $f: R \rightarrow R$ is one $-$ one onto function.
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