(i) \(\frac{x-6}{x-11}>0\)
Equating (x – 6) and (x – 11) to zero, we obtain x = 6, 11 as the critical points.
Now we plot these points on the real number line as shown:
The real number line is divided into three regions. Now check the sign of the expression \(\frac{x-6}{x-11}\) in all the three regions.
When x < 6, both numerator and denominator are negative, so \(\frac{x-6}{x-11}\) is +ve.
When x > 11, both numerator and denominator are positive, so \(\frac{x-6}{x-11}\) is +ve.
When 6 < x < 11, the expression \(\frac{x-6}{x-11}\) becomes –ve and hence < 0.
So, the solution set of the given inequations is the union of regions containing positive signs.
∴ \(\frac{x-6}{x-11}>0\) ⇒ \(x\) ∈ (– ∞, 6) ∪ (11, ∞)
(ii) \(\frac{x-3}{x+5}>2\) ⇒ \(\frac{x-3}{x+5}-2>0\)
⇒ \(\frac{x-3-2x-10}{x+5}>0\) ⇒ \(\frac{-x-13}{x+5}>0\)
\(\frac{x+13}{x+5}<0\) (Multiplying by – 1 to make co-efficient of x positive in the expression in the numerator)
Now, putting, (x + 13) and (x + 5) equal to zero, we get the critical points as x = –13, –5. Now plot these points on the real number line as shown and divide the line into three parts.
When x lies between –∞ and –13, the expression \(\frac{x+13}{x+5}\) becomes +ve.
Similarly, when x lies between –5 and ∞, the expression becomes +ve.
The expression is negative or (<0) when x lies between –13 and –5.
Hence \(\frac{x+13}{x+5}\) < 0 ⇒∈ + (–13, – 5)
(iii) \(\frac{x-1}{x-3}≥2\) ⇒ \(\frac{x-1}{x-3}-2≥0\)
⇒ \(\frac{x-1-2x-6}{x+3}≥0\) ⇒ \(\frac{-x-7}{x+3}≥0\) ⇒ \(\frac{x+7}{x+3}≤0\)
The critical points are –7 and –3, (on equating x + 7 = 0 and x + 3 = 0)
When x < –7, \(\frac{x+7}{x+3}\) becomes +ve
When x > –3, \(\frac{x+7}{x+3}\) becomes + ve
The expression \(\frac{x+7}{x+3}\) < 0 when x lies between –7 and –3
Also \(\frac{x+7}{x+3}\) < 0 when –7 is also included
∴ \(\frac{x+7}{x+3}\) < 0 \(x\)∈ [–7, –3)
(iv) \(\frac{5x-6}{x+6}≤1\) ⇒ \(\frac{5x-6}{x+6}-1 ≤0\)
⇒ \(\frac{5x-6-x-6}{x+6}≤0\) ⇒ \(\frac{4x-12}{x+6}\) < 0 ⇒ \(\frac{4(x-3)}{x+6}\) < 0 ⇒ \(\frac{x-3}{x+6}\) < 0
The critical points are x = 3, – 6. On the real number line they can be shown as:
When x < – 6 or x > 3, then the expresion \(\frac{x-3}{x+6}\) becomes positive.
When x lies between –6 and 3 (included), i.e., – 6 < x < 3, \(\frac{x-3}{x+6}\) < 0.
⇒ \(\frac{x-3}{x+6}\) < 0 \(x\)∈ (–6, 3].
