Manhattan GMAT Challenge Problem of the Week – 14 Feb 2011
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Which of the following represents the complete range of x over which [pmath]x^3[/pmath] 4[pmath]x^5[/pmath] < 0?
(A) 0 < |x| <
(B) |x| >
(C) < x < 0 or < x
(D) x < or 0 < x <
(E) x < or x > 0
One way to attack this problem is to factor the given expression:
[pmath]x^3[/pmath] 4[pmath]x^5[/pmath] = [pmath]x^3[/pmath](1 4[pmath]x^2[/pmath])
Notice that 1 4[pmath]x^2[/pmath] is a difference of squares. This part of the expression factors into
(1 2x)(1 + 2x)
So the whole expression becomes
[pmath]x^3[/pmath](1 2x)(1 + 2x).
To trace the sign changes of the whole expression, track what happens to each part of the product.
- [pmath]x^3[/pmath] is negative when x is less than zero, but its positive when x is greater than zero.
- (1 2x) is positive when x is less than , but its negative when x is greater than . (Careful about the sign change.)
- (1 + 2x) is negative when x is less than , but its positive when x > .
Now you have three break points where signs change: , 0, . This means that you have four regions to examine. You might set up a quick table to take care of the cases, or you can just talk your way through them.
1) x is less than : first term is negative, second is positive, third is negative, so the product is positive.
2) x is between and 0: first term is still negative, second is still positive, but third is now positive. So the product is negative.
3) x is between 0 and : first term is now positive. Second is still positive, third is positive, so the product is positive.
4) x is greater than : first term is positive, but now second term is negative. Third is still positive, so the product is negative.
Cases 2 and 4 give us a negative product. You can also test numbers, of course, but given the high powers, you might not want to raise fractions to these powers.
The correct answer is C.
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