Find the range of the function f (x) = \(\frac{x^2 -2}{x^2 -3}\).

1.	Find the range of the function f (x) = \(\frac{x^2 -2}{x^2 -3}\).
Answer» Let y = \(\frac{x^2-2}{x^2-3}\) y = \(\frac{x^2-2}{x^2-3}\) is not defined for x² – 3 = 0, i.e., x = ± \(\sqrt{3}\) ∴ Domain of y = R - { \(-\sqrt{3}\) , +\(\sqrt{3}\) } Now y = \(\frac{x^2-2}{x^2-3}\) ⇒ yx² – 3y = x² – 2 ⇒ x² (y – 1) = 3y – 2 ⇒ \(x^2\) = \(\frac{3y-2}{y-1}\) ∵ LHS is a perfect square ⇒ \(\frac{3y-2}{y-1}\) ≥ 0 ⇒ (3y – 2) ≥ 0 and (y – 1) ≥ 0 or (3y – 2) ≤ 0, (y – 1) ≤ 0 and y ≠ 1 (Note) ⇒ y \(\geq \frac{2}{3}\) and y > 1 or y ≤ \(\frac{2}{3}\) , y < 1 ⇒ y ∈ (1, ∞) or y ∈ (- ∞ , \(\frac{2}{3}\)] ⇒ y ∈ (- ∞ , \(\frac{2}{3}\)] \(\cup\) (1, ∞) For x = ± \(\sqrt{3}\), y does not exist.

Answer»

Let y = \(\frac{x^2-2}{x^2-3}\)

y = \(\frac{x^2-2}{x^2-3}\) is not defined for x² – 3 = 0, i.e., x = ± \(\sqrt{3}\)

∴ Domain of y = R - { \(-\sqrt{3}\) , +\(\sqrt{3}\) }

Now y = \(\frac{x^2-2}{x^2-3}\) ⇒ yx² – 3y = x² – 2

⇒ x² (y – 1)

= 3y – 2

⇒ \(x^2\) = \(\frac{3y-2}{y-1}\)

∵ LHS is a perfect square ⇒ \(\frac{3y-2}{y-1}\) ≥ 0

⇒ (3y – 2) ≥ 0 and (y – 1) ≥ 0

or (3y – 2) ≤ 0, (y – 1) ≤ 0 and y ≠ 1 (Note)

⇒ y \(\geq \frac{2}{3}\) and y > 1 or y ≤ \(\frac{2}{3}\) , y < 1

⇒ y ∈ (1, ∞) or y ∈ (- ∞ , \(\frac{2}{3}\)]

⇒ y ∈ (- ∞ , \(\frac{2}{3}\)] \(\cup\) (1, ∞)

For x = ± \(\sqrt{3}\), y does not exist.

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