Question

Assertion (A) : The function $f:R\to [0,1)$ defined by $f(x)=\frac{x^2}{x^2+1}$ is surjective.
Reason (R) : For surjection. Range of $f(x)=$ co-domain of $f(x)$

• A. Both (A) & (R) are individually true & (R) is correct 3. explanation of (A).
• B. Both (A) & (R) are individually true but (R) is not the correct (proper) explanation of (A).
• C. (A) is true but (R) is false.
• D. (A) is false but (R) is true.

Answer 1

Answer: (A)

For onto function, codomain of $f=$ range of $f$
Let $f(x)=y$ then $y\le 0$ and $y=f(x)=\frac{x^2}{1+x^2}$
$\Rightarrow y(1+x^2)=x^2$
$\Rightarrow x^2(y-1)=-y$
$\Rightarrow x^2=\frac{y}{1-y}\ge 0$
$\Rightarrow \frac{y-0}{y-1}\le 0(y\ne 1)$
$\Rightarrow 0\le y<1$
$\Rightarrow$ codomain of $f$ = Range of $f$ as $f:R\to [0,1)$