Question

Find the set of real values of x for which ${\log }_{12}$ ($x^2$ - x) < 1.

• A. x ∈ (-3, 0) ∪ (1, 4)
• B. x ∈ (-∞, 0) ∪ (1, ∞)
• C. (-∞, 0) ∪ (1, ∞)
• D. x ∈ (-3, 0)

Answer 1

The correct option is A

x ∈ (-3, 0) ∪ (1, 4)

${\log }_{12}$ ($x^2$ - x) < 1

For log to be define $x^2$ - x > 0

x ( x - 1) > 0

x ∈ $(-\infty ,0)$ U $(1,\infty )$ ______________(1)

Base of logarithm is greater than 1, then inequality is equivalent to

$x^2$ - x < $(12{)}^1$

$x^2$ - x < 12

$x^2$ - x - 12 < 0

$x^2$ - 4x + 3 $\times$ 12 < 0

(x - 4)(x + 3) < 0

x < 4 or x - 3

x ∈ (-3, 4) ____________(2)

From equation 1 and 2, common region is x ∈ (-3, 0) U(1, 4)