Value of x^2-x when x is between 0 and 1.

lenagmat wrote:Which of the folloing describes value of x^2-x when x is between 0 and 1?

A). 1>x^2-x>0
B). 1>x^2-x>=-1/4
C). 0>x^2-x>=-1/4
D). 1>x^2-x>-1
E). 0>x^2->=-1

x^2-x = x * (x-1)

0<x<1 subtract 1 from all the terms 0-1<x-1<1-1 => -1<x-1<0
0<x<1 ; -1<x-1<0

So, 0>x*(x-1)>-1/4

Why the answer is C. (After preveuos explanation I am completly confused).

Let y=x^2-x

y=x(x-1)

y=0 at x=0 and x=1

The graph of y=x^2-x is a parabola that opens up. The vertex of the parabola is where it reaches its minimum. From the symmetry of parabolas we can argue that the x-coordinate of the vertex must be halfway between the x-intercepts, or at x=1/2. Plug in x=1/2 to y=x(x-1) to get y=-1/4. This must be the minimum value, so 0>x^2-x>=-1/4

Ans: C

lenagmat wrote:Which of the folloing describes value of x^2-x when x is between 0 and 1?

A). 1>x^2-x>0
B). 1>x^2-x>=-1/4
C). 0>x^2-x>=-1/4
D). 1>x^2-x>-1
E). 0>x^2->=-1

Since 0<x<1, x²-x = (smaller positive fraction) - (greater positive fraction).

Thus, x²-x<0.
Eliminate answer choice whose range includes positive values.
Eliminate A, B and D.

Since x² and x are both fractions are between 0 and 1, the distance between them must be less than 1: |x²-x| < 1.
Thus, it is not possible that x²-x = -1.
Eliminate E.

The correct answer is C.

A less GMAT-friendly approach:
The graph of y= x²-x opens upward, since the coefficient of x² is positive.
The minimum value of y= x²-x occurs when its first derivative = 0.
The first derivative of x²-x is 2x-1.
Setting 2x-1 equal to 0, we get:
2x-1 = 0.
x = 1/2.
Thus, the minimum value of x²-x occurs when x = 1/2:
y = (1/2)² - 1/2 = -1/4.
As shown above, when 0<x<1, x²-x<0.
Thus, the range is -1/4≤x<0.

How we from 0<x<1 ; -1<x-1<0 got So, 0>x*(x-1)>-1/4

neelgandham wrote:

lenagmat wrote:
0<x<1 ; -1<x-1<0

So, 0>x*(x-1)>-1/4

We can do all the guess work .. but for a systematic simple approach (without using calculus and cordinate geometry fundas )

let x^2-x = y
=> x^2-x -y =0

now for a quadratic equation ax2+bx+c if Roots are real
then b^2-4ac =>0

=> 1+4y =>0 -> y>= -1/4 ----- (A)

here roots (x) lie between 0 and 1
so product of roots must be positive and lie between 0 and 1 too

so 0<c/a<1
0<-y<1
-1<y<0 ---- (B)

from A and B
-1/4 <=y < 0 ( -1/4 > - 1 so we can narrow down the lower range here to -1/4)

or
-1/4 <= x^2-x < 0