If x is satisfied the inequality #log_(x+3)(x^2-x) < 1#, the x may belongs to the set?

A) #x in (-3,-2)#
B) #x in (-1,3)#
C) #x in (1,3)#
D) #x in (-1,0)#
The answer is A), C) and D) for your reference

1 Answer
Noah G
Aug 19, 2017

I got #x in (-3, -2) uu (-1, 0) uu (0, 3)#, so #A, B and D#

Explanation:

By the definition of the logarithm, we have:

#x^2 - x < (x + 3)^1#

#x^2 - x < x + 3#

#x^2 - 2x - 3 < 0#

Solving like an equation:

#x^2 - 2x - 3 = 0#

#(x- 3)(x + 1) = 0#

#x= 3 or -1#

If we select a test point, say #x = 0#, we get:

#0^2 - 2(0) - 3 < 0 color(green)(√)#

However, if we use #x = -2#, we realize it doesn't work. So our solution set is #x in (-1, 3)#. This is only for the quadratic though. We haven't considered the the number in #x^2 - x# must be positive for the logarithm to be defined. Hence, we have:

#x^2 - x > 0#

#x^2 - x = 0#

#x(x - 1) = 0#

#x = 0 or 1#

If we repeat the process with test points, we realize that the solution is #(-oo, 0)# and #(1, oo)#. Therefore, we must eliminate #(0, 1)# from our solution set above.

Now, we must also guarantee that #x + 3 > 0#. Then #x > -3#. This means that we can include #x > -3# in our solution set. The set becomes #(-3, 3)#. However, if you try #x = -1.5#, (or any number in the interval #(-2, -1)#, you will realize that the result you get is not within the range of the problem, that' s to say it will be greater, instead of less than #1#.

The answer is therefore :

#x in (-3, -2) uu (-1, 0) uu (0, 3)#

A graphical verification yields the same results.

Hopefully this helps!

enter image source here

Related questions
Impact of this question
3808 views around the world
You can reuse this answer
Creative Commons License