Assertion (A) : The function f: R [0,1) defined by f(x)=x^2 x^2+1… - Vidyadip

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 $(A).$

For onto function, codomain of $f=$ range of $f$.
We have, $f(x)=\frac{x^2}{1+x^2}$.
Then $y \geq 0$.
$\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 \frac{y-0}{y-1} \leq 0 $
$\Rightarrow 0 \leq y<1 (\because y \neq 1)$
$\Rightarrow$ codomain of $f=$ Range of $f$, as $f: R \rightarrow[0,1)$.

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