(B) Let $y = \frac{x^2}{x^2+1}$.$y(x^2+1) = x^2$$yx^2 + y = x^2$$x^2(y-1) = -y$$x^2 = \frac{y}{1-y}$.For $x$ to be a real number,$x^2 \geq 0$.Therefor

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(B) Let $y = \frac{x^2}{x^2+1}$.$y(x^2+1) = x^2$$yx^2 + y = x^2$$x^2(y-1) = -y$$x^2 = \frac{y}{1-y}$.For $x$ to be a real number,$x^2 \geq 0$.Therefore,$\frac{y}{1-y} \geq 0$.This implies $y(1-y) \geq 0$ and $y \neq 1$.Solving the inequality $y(y-1) \leq 0$,we get $0 \leq y Thus,the range of the function is $[0, 1)$.

Class 12 Mathematics Relation and Function Domain and Range

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The range of the function $f(x) = \frac{x^2}{x^2+1}$ is

MHT CET 2023 All MHT CET

A
$(0, 1)$
B
$[0, 1)$
C
$(0, 1]$
D
$[0, 1]$

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