1 - x² ≥ 0 is equivalent to (1+x)(1-x) ≥ 0.
And that means either both (1+X) & (1-x) are negative or both (1+X) & (1-x) are negative...
That translates to x ≤ -1 or x ≥ 1 stated by option D

If x is less than or equal to -1 ... according to answer D ... then x could have a value of -2.

If x = -2, then is 1 - x² ≥ 0 ? NO 1 - 4 = -3 ... which is NOT ≥ 0.

IMO E

II wrote:Which of the following describes all values of x for which 1 - x² ≥ 0 ?

(A) x ≥ 1
(B) x ≤ -1
(C) 0 ≤ x ≤ 1
(D) x ≤ -1 or x ≥ 1
(E) -1 ≤ x ≤ 1

What is the best way to solve this question please ?
Thanks.
II

Let's rewrite the original as:

x^2 ≤ 1

and let's use a bit of logic.

For any number with magnitude greater than 1 (magnitude means we ignore the sign), the square of that number will also be greater than 1.

Both 1 and -1 squared = 1.

Fractions (positive and negative) squared are between 0 and 1.

So, for the statement to be true, |x| ≤ 1, which is just another way of saying -1 ≤ x ≤ 1: choose (e).

simplyjat wrote:1 - x² ≥ 0 is equivalent to (1+x)(1-x) ≥ 0.
And that means either both (1+X) & (1-x) are negative or both (1+X) & (1-x) are negative...
That translates to x ≤ -1 or x ≥ 1 stated by option D

Sorry messed up the last part of translation ... E is the correct answer.

As Stuart says:

If you are completely bowled by such a question, just use substitution which I believe will be fast enough to get this one right in less than 2 mins.

A. x =2 fails
B. x = -2 fails
C. x = 0 and x = 1 and x = 1/2 pass (lets keep it)
D. fails because A or B fails
E. x = -1 and C above pass so this is the right answer

Thanks guys ... as ever Stuart breaks it down so well !
Good to have different approaches to solve questions ... when in the midst of the pressure of the GMAT ... its good to have secondary plans which you can draw upon to help you out.
And yes ... plugging in numbers here would help a lot.

Which of the following describes all values of x for which 1 - x² ≥ 0 ?

(A) x ≥ 1
(B) x ≤ -1
(C) 0 ≤ x ≤ 1
(D) x ≤ -1 or x ≥ 1
(E) -1 ≤ x ≤ 1


We can just change the sign an say:
x²-1≤0.
(x+1)(x-1)≤0
or x lies between -1 and 1. Plot on the number line to verify.
My answer is E.

Cheers!
Aks

as we boil down to (x+1)(x-1)≤0 and x less than equal to 1 and -1. Why isnt b correct where it satisifies both the 1 and -1? I beleive I am getting the question wrong again. If we graph out x less than or equal to 1 and -1 we see that -1 is the statement where both are satisfied. I understand the math but mixed up on last step!

So we have 2 good approaches here (thanks to Stuart and Simplyjat)

1) Rewrite the original as: x^2 ≤ 1 and use a bit of logic.

For any number with magnitude greater than 1 (magnitude means we ignore the sign), the square of that number will also be greater than 1.
Both 1 and -1 squared = 1.
Fractions (positive and negative) squared are between 0 and 1.

So, for the statement to be true, |x| ≤ 1, which is just another way of saying -1 ≤ x ≤ 1: choose (e).

2) 1-x^2 ≥ 0

Rewrite 1-x^2 as (1+x)(1-x).
In order for (1+x)(1-x) to be ≥ 0, then x has an upper limit of 1, and a lower limit of -1 ... in other words it is between 1 and -1; -1 ≤ x ≤ 1 : choose E

I can eliminate a, b, and d easily. But when it comes down to choosing between C and E, I'm a bit torn. C isn't actually wrong, is it? It just neglects to include -1 as answer choice E does... would that be fair to say.

Is the reason that E is correct just because it's a better answer (and incorporates -1), not necessarily because C is wrong?


Basically, plugging in gets me through AB and D. How do i decide between C and E?

pbrmoney wrote:I can eliminate a, b, and d easily. But when it comes down to choosing between C and E, I'm a bit torn. C isn't actually wrong, is it? It just neglects to include -1 as answer choice E does... would that be fair to say.

Is the reason that E is correct just because it's a better answer (and incorporates -1), not necessarily because C is wrong?


Basically, plugging in gets me through AB and D. How do i decide between C and E?

Let's consider y=x^2 and y<=1

If you put a negative or a positive value of x of same magnitude the value of y remains the same. Such a function is also known as even function.

Hence, a situation describing x^2<=1 is analogous to |x|<=1.

This means we must have symmetry wrt values of x on both sides of '0'. Also the absolute value of x should not exceed 1.

Hence, -1<=x<=1

The problem with choice C is that it fails to take into account all the symmetrically placed negative values of x that I discussed above.

Note: Any even power of x makes the places the values of x symmetric to y axis.

II wrote:Which of the following describes all values of x for which 1 - x^2 ≥ 0 ?

(A) x ≥ 1
(B) x ≤ -1
(C) 0 ≤ x ≤ 1
(D) x ≤ -1 or x ≥ 1
(E) -1 ≤ x ≤ 1

We are given that 1-x^2 >= 0 and need to determine an answer that describes all values of x. Let's isolate x in our inequality.

1-x^2 ≥ 0

1 ≥ x^2

Taking the square root of both sides of the inequality gives us:

1 ≥ x

x ≤ 1

OR

1 ≥ - x

-1 ≤ x

Thus, -1 ≤ x ≤ 1.

Answer: E