Which of the following represents the complete set of real values of x that satisfy the in… — JAMB UTME

Which of the following represents the complete set of real values of $x$ that satisfy the inequality $\frac{x^2 - x - 6}{x - 1} \le 0$ ?

The complete solution set is

x \le -2

1 < x \le 3

The expression

\frac{(x-3)(x+2)}{x-1}

evaluates to a non-positive value (

\le 0

) when the numerator and denominator have opposite signs or when the numerator is zero. Evaluating across the critical boundaries

x = -2, 1, 3

while excluding

x = 1

yields the solution set

x \le -2

1 < x \le 3

Factor the quadratic numerator and state the domain restriction.

The numerator factors as

x^2 - x - 6 = (x - 3)(x + 2)

, giving the rational inequality

\frac{(x - 3)(x + 2)}{x - 1} \le 0

with

x \neq 1

Factoring isolates the critical boundary points where the expression can change sign.

Identify all critical numbers.

The critical values are

x = -2

x = 1

, and

x = 3

These points partition the real number line into sub-intervals.

Test points in each interval to determine the sign of the rational function

f(x) = \frac{(x - 3)(x + 2)}{x - 1}

For

x < -2

f(x) \le 0

; for

-2 < x < 1

f(x) > 0

; for

1 < x < 3

f(x) \le 0

; for

x > 3

f(x) > 0

Determining where the function is negative or zero identifies the regions satisfying

\le 0

Combine intervals and include non-undefined endpoints.

Endpoints

x = -2

and

x = 3

make the numerator zero (included), while

x = 1

makes the denominator zero (excluded). Thus,

x \le -2

1 < x \le 3

Division by zero must be excluded from the solution set.

Solving rational inequalities by finding critical values, using test intervals, and respecting domain restrictions.

Question