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 $f : R\to [0, 1)$