Quote:
If –1 < x < 1 and x ≠ 0, which of the following inequalities must be true?
I. x^3 < x
II. x^2 < |x|
III. x^4 – x^5 > x^3 – x^2
(A) I only
(B) II only
(C) III only
(D) II and III only
(E) I, II, and III
Can someone try doing this algebraically?
The algebraic approach to this question will be as follows:
I. x^3 < xWe will first find the range of values of x for which this inequality holds true.
So, let's solve this inequality: x^3 - x < 0
=> x(x^2 - 1) < 0
=> x(x+1)(x-1) < 0
Let's draw the Wavy Line for this inequality.
The expression x(x+1)(x-1) will be less than zero for those values of x where the Wavy Line goes below the number line.
So, we can say that the inequality x(x+1)(x-1) < 0 holds true for x < -1 or for 0 < x < 1
But, we know that x lies between -1 and 1.
This means, that for some possible values of x (values that lie between 0 and 1), Expression 1 will hold true and for other possible values of x (values that lie between -1 and 0), Expression 1 will NOT hold true. So, it's not a MUST BE TRUE expression.
II. x^2 < |x|Case 1: x is positive.This means, |x| = x
So, Expression 2 becomes: x^2 < x
That is, x^2 - x < 0
Or x(x-1) < 0
Again, by drawing the Wavy Line, we can see that the values of x that satisfy Case 1 are 0 < x < 1.
Case 2: x is negative.This means, |x| = -x
So, Expression 2 becomes: x^2 < -x
That is, x^2 + x < 0
Or x(x+1) < 0
From the Wavy Line, it's clear that the values of x that satisfy Case 2 are -1 < x < 0
Now, we are given that the range of possible values of x are between -1 and 1, excluding 0.
This means, that x will either lie between -1 and 0, exclusive, in which case, Case 2 above applies and Expression 2 holds true.
OR x will lie between 0 and 1, exclusive, in which case, Case 1 above applies and Expression 2 holds true.
This means, Expression 2 holds true for all possible values of x. So, it is a MUST BE TRUE expression.
III. x^4 – x^5 > x^3 – x^2We'll first find the range of values for which this expression holds true.
x^4(1-x) > x^2(x-1)
Since x is not equal to zero, it's safe to divide both sides by x^2, which being a positive number, will not impact the sign of inequality.
We get: x^2(1-x) > x-1
Or, x^2(1-x) - (x-1) > 0
That is, x^2(1-x) + (1-x) > 0
(1-x)(x^2+1) > 0
Since x^2 + 1 will always be positive, the above inequality will hold true when 1 - x > 0. That is, 1 > x
Thus we see that Expression 3 will hold true for all values of x that are less than 1.
Now, we are given that the only possible values of x are -1 < x < 1. Since all these possible values fall within the range in which Expression 3 holds true, we can conclude that Expression 3 will hold true for all possible values of x. So, it is a MUST BE TRUE expression.
Hope this solution was useful!
Best Regards
Japinder
1. why have we not flipped the inequality sign for stmt 2 when changing the sign to negative?
2. for stmt 3, we have x values less than 1 till negative infinity. how can this be sufficient as the range we want it to be is from -1 to +1