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Relations and Functions — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsRelations and Functions1 Mark
MCQ
Assertion (A) : The function $f: R \rightarrow[0,1)$ defined by $f(x)=\frac{x^2}{x^2+1}$ is surjective.
Reason (R) : For surjection, Range of $f(x)=$ codomain of $f(x)$
  • ✓
    Both (A) and (R) are true and (R) is the correct explanation of (A).
  • B
    Both (A) and (R) are true but (R) is not the correct explanation of (A).
  • C
    (A) is true but (R) is false.
  • D
    (A) is false but (R) is true.
✓

Answer

Correct option: A.
Both (A) and (R) are true and (R) is the correct explanation of (A).
(a) : For onto function, codomain of $f=$ range of $f$. We have, $f(x)=\frac{x^2}{1+x^2}$. Then $y \geq 0$.
$
\begin{array}{l}
\Rightarrow y\left(1+x^2\right)=x^2 \Rightarrow x^2(y-1)=-y \\
\Rightarrow x^2=\frac{y}{1-y} \geq 0 \\
\Rightarrow \quad \frac{y-0}{y-1} \leq 0 \Rightarrow 0 \leq y<1 & (\because y \neq 1)
\end{array}
$
$\Rightarrow$ codomain of $f=$ Range of $f$, as $f: R \rightarrow[0,1)$.

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