Question

Find the set of real values of x for which ${\log }_{(x+3)}$ ($x^2$ - x) < 1________.

• A. (-1, 3) U (-3, -2)
• B. (-1, 0) U (-3, -2)
• C. (-3, 3)
• D. (-1, 0) U (1, 3) U (-3, -2)

Answer 1

The correct option is D

(-1, 0) U (1, 3) U (-3, -2)

${\log }_{(x+3)}$ ($x^2$ - x) < 1 ------------(1)

For log to be defined

$x^2$ - x > 0

x(x-1)> 0

x < 0 or x > 1 --------------------------(2)

Let's case 1 when x+3 > 1

x > -2 ---------------------------(3)

${\log }_{(x+3)}$ ($x^2$ - x) < 1

We have

$x^2$ - x < $(x+3{)}^1$

$x^2$ - x < x+3

$x^2$ - 2x - 3 < 0

Hence x > 3 and x < -1

-1 < x < 3 --------------(4)

Common region in equation 2,3 and 4 we set,

x ∈ (-1,0) U (1,3)

Case 2 : When x + 3 lies between 0 and 1

0 < x + 3 < 1

-3 < x < -2 ------------------------(5)

Then ${\log }_{(x+3)}$ ($x^2$ - x) < 1
$lo{g}_{x+3}(x^2-x)$ < 1 $\Rightarrow x^2-x$ > x +3
$\Rightarrow x^2-2x-3$ >0
$\Rightarrow (x-3)(x+2)$ > 0
$\Rightarrow$ x < -1 or x > 3 ___(6)
From (5) & (6)
$xϵ(-3,-2)$
$\therefore$ From cases (1) & (2) $xϵ(-3,-2)\cup (-1,0)\cup (1,3)$