Range of function f(x) = frac{{{x^2}}}{{{x^2} + 1}} is

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1.Relation and Function

normal

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

A

$(-1, 0)$

B

$(-1, 1)$

C

$[0, 1)$

D

$(1, 1)$

Solution

(c) Let $y = \frac{{{x^2}}}{{{x^2} + 1}}$

==> $(y – 1){x^2} + 0x + y = 1,y \ne 1$ for real values of $x$,

we have $D \ge 0 \Rightarrow – 4y(y – 1) \ge 0 $

$\Rightarrow y(y – 1) \le 0 \Rightarrow y \in [0,\,1)$

$0 \le \frac{{{x^2}}}{{{x^2} + 1}} < 1$.

Standard 12

Mathematics

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