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Functions — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsFunctions5 Marks
Question
Classify the following functions as injection, surjection or bijection: $f : R \rightarrow R,$ defined by $\text{f(x)}=\frac{\text{x}}{\text{x}^2+1}$
✓

Answer

$f : R \rightarrow R,$ defined by $\text{f(x)}=\frac{\text{x}}{\text{x}^2+1}$
Injection test: Let x and y be any two elements in the domain $(R),$
such that $f(x) = f(y).$
$f(x) = f(y)$
$\frac{\text{x}}{\text{x}^2+1}=\frac{\text{y}}{\text{y}^2+1}$
$xy^2 + x = x^2y + y$
$xy^2 - x^2y + x - y = 0$
$-xy(-y + x) + 1(x - y) = 0$
$(x - y)(1 - xy) = 0$
$x = y$ or $\text{x}=\frac{1}{\text{y}}$
So, $f$ is not an injection.
Surjection test: Let $y$ be any element in the co-domain $(R),$ such that $f(x) = y$ for some element $x$ in $R ($domain$).$
$f(x) = y$
$\frac{\text{x}}{\text{x}^2+1}=\text{y}$
$yx^2 - x + y = 0$
$\text{x}=\frac{-(-1)\pm\sqrt{1-4\text{y}^2}}{2\text{y}},$ if $\text{y}\neq0$
$=\frac{1\pm\sqrt{1-4\text{y}^2}}{2\text{y}},$ which may not be in $R$
For example, if $y = 1$, then
$\text{x}=\frac{1\pm\sqrt{1-4}}{2}=\frac{1\pm\text{i}\sqrt{3}}{2},$ which is not in $R$
So, $f$ is not surjection and $f$ is not bijection.

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