The function \(f(x) = \displaystyle\frac{x-2}{x+1};\ x\ne-1\) is increasing when (Where R is a set of real numbers)
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(x\in \mathrm{R}-\{-1\}\)
By the quotient rule:
\(\displaystyle f'(x) = \frac{(x + 1)(1) - (x - 2)(1)}{(x + 1)^2}\)
\(\displaystyle = \frac{3}{(x + 1)^2}\)
\(f'(x) \gt 0\) for every \(x \ne -1\)
So f is increasing on each of the intervals \((-\infty, -1)\) and \((-1, \infty)\), that is, at every point of its domain \(\mathrm{R} - \{-1\}\).
Across the break at x = −1 the values do not keep rising (f(−2) = 4, but f(0) = −2), so “increasing” here means increasing on each interval of the domain.
Why the other options are incorrect
Option (1) \(x\in \mathrm{R}\) — Incorrect. It includes x = −1, where f is not defined.
Option (3) \(x\in \mathrm{R}-\{1,-1\}\) — Incorrect. It leaves out x = 1, where \(f'(1) = \frac{3}{4} \gt 0\); f is increasing there too, so this is not the full set.
Option (4) \(x\in \mathrm{R}-\{0\}\) — Incorrect. It includes x = −1, where f is not defined, and leaves out x = 0, where \(f'(0) = 3 \gt 0\).