Is ( x^6 - x^4 ) ( x^5 - x^3 ) 1) x 1 2) x^4

Question

Is ( x^6  -  x^4 ) > ( x^5  -  x^3 )

1) x >1
2)  x^4  > x^3

scthakur · Answer

Answer should be D.

x^6 - x^4 > X^5 - x^3 can be simplified as
x^4(x^2-1) > x^3(x^2-1)

With (a), x>1 and hence x^2-1 will be positive. Hence we can divide both sides by x^2-1 and the result will be x^4 > x^3 or x > 1.

With (b) x^4 > x^3 only if x does not equal 1. Again divide both sides by x^2-1 and the result will be x^4 > x^3.

BTW. I do not know how to write square here and hence I used ^ sign.

bhatiagp · Answer

Answer is A ..

Because for (2) it doesnt hold when x <= -1/2

pmenon · Answer

got A.

stat 1 is pretty clear, for any value larger than 1 , the left side is always greater than the right side

for stat 2, the equality may or may not hold depending on whether x is larger than 1, or less than 1. we dont know this info, so this statement alone is insuff.

maximusdecimus · Answer

Answer is A.
For 2) 
consider the following example:

x=-1/2==> x^4 > x^3 and 
( x^6  -  x^4 ) < ( x^5  -  x^3 )

for x=-2 ==> x^4 > x^3 and 
( x^6  -  x^4 ) > ( x^5  -  x^3 )

jallenmorris · Answer

what do you mean it doesn't hold when x is less than or equal to negative 1?

If you simplify the inequaltiy out to  x^4(x^2 - 1) > x^3(x^2 - 1) , then put in -2 for x gives you

16(3) > -8(3) - how is this incorrect?

What about for -3

81(8) > -27(8) - again, this solves the equation.

Now, if you use  -\frac{1}{2}  or  \frac{1}{2}  it falls apart, and you see #2 is insufficient.

Don't you mean when  x <= 1  ?

brokerbevo · Answer

Yes, it doesn't hold if x is a fraction. Stick in -0.5 and you will see that the answer is  A

greenoak · Answer

Agree with those who post A. Another possible approach (basically involves the straightforward solving the inequalities): (x^6 - x^4) > (x^5 - x^3 ) = x^4(x^2-1) - x^3(x^2-1) = (x^4-x^3)(x-1)(x+1)=x^3(x-1)^2(x+1). Now, solving x^3(x-1)^2(x+1) > 0 with any sign-analysis method we have: **** + ****** - ****** + ****** + _____-1____0____1_____ => x < - 1 or 01. Statement 1: As was shown above, for x>1 the inequality is always true => sufficient. Statement 2: x^4 – x^3 = x^3(x-1). Again, let’s solve x^3(x-1)>0: ************ + ***** - ****** + ___________0____1_____ => x<0 or x>1. From here we cannot conclude that for such x the initial inequality will always be true. In particular, it does not hold for -1

bhatiagp · Answer

Apologize, for the error. I read that in the OE ... And just posted it.. without realising it doesnt work for fractions. Honestly speaking I was unable to answer it myself, and was looking at how to solve it, hence posted the question.

Irrespective, the answer remains A....

Is ( x^6 - x^4 ) > ( x^5 - x^3 ) 1) x >1 2) x^4 >

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