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Mathematics · 2026 · 5 marks
CBSE 2026 · Region 3 · Set 1 · Q32
Show that f: R → R defined as f ( x )=x/(√(1+ x^2)) is one-one but not onto.
Show that $\displaystyle \mathrm{f}: \mathrm{R} \rightarrow \mathrm{R}$ defined as $\displaystyle \mathrm{f}(\mathrm{x})=\frac{\mathrm{x}}{\sqrt{1+\mathrm{x}^{2}}}$ is one-one but not onto.
Marking-scheme solution
One one: For $\displaystyle x_{1}, x_{2} \in R$; let $\displaystyle f\left(x_{1}\right)=f\left(x_{2}\right) \Rightarrow \dfrac{x_{1}}{\sqrt{1+x_{1}^{2}}}=\dfrac{x_{2}}{\sqrt{1+x_{2}^{2}}}$
$\displaystyle \left[\Rightarrow x_{1}\right.$ and $\displaystyle x_{2}$ must be of same sign.$\displaystyle ]$
$\displaystyle \Rightarrow x_{1}^{2}\left(1+x_{2}^{2}\right)=x_{2}^{2}\left(1+x_{1}^{2}\right)$
$\displaystyle \Rightarrow x_{1}^{2}=x_{2}^{2} \Rightarrow x_{1}=x_{2}$ (Rejecting $\displaystyle x_{1}=-x_{2}$)
Hence, f is one $\displaystyle -$ one.
Onto: $\displaystyle y=\dfrac{x}{\sqrt{1+x^{2}}} \Rightarrow x= \pm \sqrt{\dfrac{y^{2}}{1-y^{2}}}$
x is defined $\displaystyle \forall y \in(-1,1) \Rightarrow$ Range $\displaystyle =(-1,1)$
$\displaystyle \because$ Range $\displaystyle \neq$ codomain $\displaystyle \Rightarrow$ f is not onto.
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CBSE Class 12 Mathematics past-paper question from the 2026 board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.