Question

Find set of real values of x for which ${\log }_{(x+3)}$ $x^2$ - x - 11 < 0 if x > -2

• A. x ∈ (-3, 4)
• B. x ∈ (-2, 4)
• C. x ∈ (-2, 4)
• D. x ∈ (-3, 0)

Answer 1

The correct option is C

x ∈ (-2, 4)

${\log }_{(x+3)}$ ($x^2$ - x - 11 < 0

Since x > - 2

x + 3 will always be greater than 1

So, base of the logarithm is greater than 1, then inequality is equivalent to

Then $x^2$ - x - 11 < $(x+3{)}^{\circ }$

$x^2$ - x - 11 < 1

$x^2$ - x - 12 < 0

(x - 4) (x + 3) < 0 x ∈ (-3, 4)

Since x should be greater than -2

x lies between (-2 to 4)

x ∈ (-2, 4)