If `1lt(x-1)/(x+2)lt7` then find the range of
`(i) x` `(ii) x^(2)` `(iii) (1)/(x)`

Answer

To solve the inequality \( \frac{x-1}{x+2} < 7 \) and \( \frac{x-1}{x+2} > 1 \), we will break it down step by step.

### Step 1: Solve the inequality \( \frac{x-1}{x+2} > 1 \)

1. Rearranging the inequality:
\[ \frac{x-1}{x+2} - 1 > 0 \]
This simplifies to:
\[ \frac{x-1 - (x+2)}{x+2} > 0 \]
Which further simplifies to:
\[ \frac{-3}{x+2} > 0 \]

2. The inequality \( \frac{-3}{x+2} > 0 \) holds when the denominator is negative:
\[ x + 2 < 0 \implies x < -2 \]

### Step 2: Solve the inequality \( \frac{x-1}{x+2} < 7 \)

1. Rearranging the inequality:
\[ \frac{x-1}{x+2} - 7 < 0 \]
This simplifies to:
\[ \frac{x-1 - 7(x+2)}{x+2} < 0 \]
Which further simplifies to:
\[ \frac{x - 1 - 7x - 14}{x+2} < 0 \]
This simplifies to:
\[ \frac{-6x - 15}{x+2} < 0 \]
Or:
\[ \frac{-3(2x + 5)}{x+2} < 0 \]

2. The inequality \( \frac{-3(2x + 5)}{x+2} < 0 \) holds when:
- The numerator \( 2x + 5 > 0 \) and the denominator \( x + 2 > 0 \)
- Or the numerator \( 2x + 5 < 0 \) and the denominator \( x + 2 < 0 \)

Solving \( 2x + 5 > 0 \):
\[ 2x > -5 \implies x > -\frac{5}{2} \]

Solving \( x + 2 > 0 \):
\[ x > -2 \]

Now, solving \( 2x + 5 < 0 \):
\[ 2x < -5 \implies x < -\frac{5}{2} \]

Solving \( x + 2 < 0 \):
\[ x < -2 \]

### Step 3: Combine the results

From Step 1, we have \( x < -2 \) and from Step 2, we have two cases:
1. \( x > -\frac{5}{2} \) and \( x > -2 \) (which is not possible)
2. \( x < -\frac{5}{2} \) and \( x < -2 \) (which is valid)

Thus, the valid solution is:
\[ x < -\frac{5}{2} \]

### Step 4: Find the ranges for \( x^2 \) and \( \frac{1}{x} \)

#### (i) Range of \( x \):
\[ x \in (-\infty, -\frac{5}{2}) \]

#### (ii) Range of \( x^2 \):
Since \( x \) is negative, squaring will yield positive values:
\[ x^2 \in \left(\left(-\frac{5}{2}\right)^2, \infty\right) = \left(\frac{25}{4}, \infty\right) \]

#### (iii) Range of \( \frac{1}{x} \):
Since \( x < -\frac{5}{2} \), \( \frac{1}{x} \) will be negative:
\[ \frac{1}{x} \in \left(-\frac{2}{5}, 0\right) \]

### Final Answers:
(i) \( x \in (-\infty, -\frac{5}{2}) \)
(ii) \( x^2 \in \left(\frac{25}{4}, \infty\right) \)
(iii) \( \frac{1}{x} \in \left(-\frac{2}{5}, 0\right) \)