If −1 < x < 0 and 0 < y < 1

Hi..

never cross multiply negative values..

x is NEGATIVE, y is positive......
So x/y is -/+ or a negative quantity..
Also x^2 will be positive and y^3 is also POSITIVE..
x^2/y^3 means -*-/+*+*+ , this will be positive..
So A is NEGATIVE and E is positive..
So A<E..

Hope it helps

Notice that x is negative and y is positive and that all of the answer choices involve powers or division. Thus, to determine the expression that has the least value, we want the expression to be negative. We see that choices B, C, and E are all positive, so we can eliminate these choices. Choices A and D are both negative. To determine the one with the least value, let’s use numbers for x and y. Let’s say x = -½ and y = ½:

A. x/y = (-½)/(½) = -1

D. x^3/y = (-½)^3/(½) = (-⅛)/(½) = (-⅛) x 2 = -¼

We see that choice A has the least value.

Answer: A

Hi..

It will be true whenever the value is between 0 and -1...
Say the value is -1/2... (-1/2)^3=-1/8..
And -1/8 >-1/2

As per the inequality, x is negative and y is positive. So to get the least value we will have to choose the operation where x is negative.
Options - B, C & E have x^2, which is positive hence they are out.
Option A & D are the contenders. Easiest and shortest way to solve could be substitution -
let x = -0.5 & y =0.5. Now evaluate option A & D
Option A: x/y = -0.5/0.5 = -1
Option D: x^3/y = (-0.5)^3/0.5 = -0.125/0.5 = - 0.25

Clearly Option A is least as -1<-0.25

Just FYI -
as x is negative and has to be a decimal number greater than -1, so higher powers will of x will move it towards 0. therefore higher powers of x will be greater than x itself, for example (-0.5)^11 will be -0.00049 which is greater than -0.5

Thanks for your response. I guess the same rule applies here that applies for numbers between 0 and 1 i.e. if x is 0<x<1 then x^2<x. But conversely for negative number it'll go like x^3<x<x^2.

Kindly correct me if I am wrong.

Could you elaborate more on the above? Thanks!

This is related to number properties.

When x > 1
\(x = 3\)
\(x^3 = 27\) (greater than x)

When 0 < x < 1
\(x = \frac{1}{3}\)
\(x^3 = \frac{1}{27}\) (smaller than x)

When -1 < x < 0
\(x = \frac{-1}{3}\)
\(x^3 = \frac{-1}{27}\) (absolute value of x^3 is smaller than absolute value of x. Since both are negative, x^3 > x)

When x < -1
\(x = -3\)
\(x^3 = -27\) (absolute value of x^3 is greater than absolute value of x. Since both are negative, x^3 < x)

Coming back to our question, when -1 < x < 0
\(x^3 > x\)
Dividing both sides by y (which is positive so the inequality sign stays the same),
\(\frac{x^3}{y} > \frac{x}{y}\)

hi,
karishma , although your explanation is quite clear , but how i usually decide which is grater or less, such as

say x=1/3 and x^3= 1/27 then x>x^3

now for the -ve one x=-1/3 then x^3= -1/27

so signs change -1/3 < -1/27

is my way of understanding is correct ?

please comment .

many thanks.

The most efficient way to solve this problem is possibly through both plugging in and testing values and process of elimination (made possible through knowing properties such as (-x)^2 is positive)

B, C and E all result in positive numbers because the exponent is even so the only choices are A & D

If we use (-.5) and (.5)

-.5/.5= -1
(-.5)^3/ (.5)
(-.0125)/(.5) = -.025

Hence "A" is correct

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