(c) (– ∞, 2) ∪ [7, ∞)
\(\frac{x+3}{x-2}\) < 2 ⇒ \(\frac{x+3}{x-2}\) - 2 < 0 ⇒ \(\frac{x+3-2x+4}{x-2}\) < 0
⇒ \(\frac{7-x}{x-2}\) < 0 ⇒ \(\frac{x-7}{x-2}\) > 0 Here \(x\) ≠ 2
The critical points on putting (x – 7) and (x – 2) are obtained as equal to zero are 7 and 2.
The real number line is divided into 3 parts as shown below by these two critical points.
When x < 2, the expression \(\frac{x-7}{x-2}\) > 0.
When x lies between 2 and 7, the expression \(\frac{x-7}{x-2}\) < 0.
When x > 7, the expression \(\frac{x-7}{x-2}\) > 0.
∴ The inequality \(\frac{x+3}{x-2}\) < 2 holds when \(x\) < 2 or \(x\) > 7,
i.e., \(x\) ∈ (– ∞, 2) ∪ [7, ∞).
