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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
