First slide

Functions (XII)

Question

The range of the function $f (x) = \frac{1 - x^{2}}{x^{2} + 3}$ is

Easy

A

$(- 1, \frac{1}{3})$
B

$[- 1, \frac{1}{3}]$
C

$(- 1, \frac{1}{3}]$
D

$[- 1, \frac{1}{3})$

Solution

$\begin{array}{l} f (x) = y = \frac{1 - x^{2}}{x^{2} + 3} \\ \Rightarrow y x^{2} + 3 y = 1 - x^{2} \\ \Rightarrow x^{2} = \frac{1 - 3 y}{y + 1} \\ Now x^{2} \geq 0 \Rightarrow \frac{1 - 3 y}{y + 1} \geq 0 \Rightarrow \frac{3 y - 1}{y + 1} \leq 0 \\ \Rightarrow (3 y - 1) (y + 1) \leq 0 \\ \Rightarrow (y + 1) (y - \frac{1}{3}) \leq 0 \\ Hence, y \in (- 1, 1 / 3] \end{array}$

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