Matching Intervals with Properties of a Rational Function

Factorizing the rational function:
$f(x) = \frac{(x - 1)(x - 5)}{(x - 2)(x - 3)}$

Also express $f(x) - 1$:
$f(x) - 1 = \frac{(x^2 - 6x + 5) - (x^2 - 5x + 6)}{x^2 - 5x + 6} = \frac{-x - 1}{(x - 2)(x - 3)} = -\frac{x + 1}{(x - 2)(x - 3)}$

Evaluating intervals in Column I:

1. (A) $-1 < x < 1$:

• Numerator: $(x-1) < 0$ and $(x-5) < 0 \implies (x-1)(x-5) > 0$.
• Denominator: $(x-2) < 0$ and $(x-3) < 0 \implies (x-2)(x-3) > 0$.
• Hence $f(x) > 0$ ($r$).
• For $f(x) - 1$: $x + 1 > 0$, $(x-2)(x-3) > 0 \implies f(x) - 1 < 0 \implies f(x) < 1$ ($s$).
• Together, $0 < f(x) < 1$ ($p$).
• Matching: $(\text{A}) \rightarrow (\text{p}, \text{r}, \text{s})$.

1. (B) $1 < x < 2$:

• Numerator: $(x-1) > 0$ and $(x-5) < 0 \implies (x-1)(x-5) < 0$.
• Denominator: $(x-2) < 0$ and $(x-3) < 0 \implies (x-2)(x-3) > 0$.
• Thus $f(x) < 0$ ($q$), which also implies $f(x) < 1$ ($s$).
• Matching: $(\text{B}) \rightarrow (\text{q}, \text{s})$.

1. (C) $3 < x < 5$:

• Numerator: $(x-1) > 0$ and $(x-5) < 0 \implies (x-1)(x-5) < 0$.
• Denominator: $(x-2) > 0$ and $(x-3) > 0 \implies (x-2)(x-3) > 0$.
• Thus $f(x) < 0$ ($q$), which also implies $f(x) < 1$ ($s$).
• Matching: $(\text{C}) \rightarrow (\text{q}, \text{s})$.

1. (D) $x > 5$:

• All factors are positive $\implies f(x) > 0$ ($r$).
• For $f(x) - 1$: $x + 1 > 0$ and $(x-2)(x-3) > 0 \implies f(x) - 1 < 0 \implies f(x) < 1$ ($s$).
• Together, $0 < f(x) < 1$ ($p$).
• Matching: $(\text{D}) \rightarrow (\text{p}, \text{r}, \text{s})$.

Matching:
$(\text{A}) \rightarrow (\text{p}, \text{r}, \text{s}),\; (\text{B}) \rightarrow (\text{q}, \text{s}),\; (\text{C}) \rightarrow (\text{q}, \text{s}),\; (\text{D}) \rightarrow (\text{p}, \text{r}, \text{s})$

Correct Answer: A