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505-555 Level|   Inequalities|                  
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droopy57
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

Please expand on answers

E.

1-x^2 >= 0 ---> x^2-1<=0
--> (x+1)(x-1)<=0
Above equation true for
i) x+1<=0 and x-1>=0 ---> x<= -1 and x>=1 ---> this is not possible ---Strike out this solution
ii) x+1>=0 and x-1<=0 ---> x>=-1 and x<=1 --> -1<=x<=1
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droopy57
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

Please expand on answers

E.

1-x^2 >= 0 ---> x^2-1<=0
--> (x+1)(x-1)<=0
Above equation true for
i) x+1<=0 and x-1>=0 ---> x<= -1 and x>=1 ---> this is not possible ---Strike out this solution
ii) x+1>=0 and x-1<=0 ---> x>=-1 and x<=1 --> -1<=x<=1

Can someone please explain the signs in red above? this is not absolute value, why do we need to test these?
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Walkabout
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

    \(1-x^2\geq{0}\);

    \(x^2\leq{1}\);

    \(-1\leq{x}\leq{1}\).

Answer: E.
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Bumping for review and further discussion*. Get a kudos point for an alternative solution!

*New project from GMAT Club!!! Check HERE

Theory on Inequalities:
x2-4x-94661.html#p731476
inequalities-trick-91482.html
data-suff-inequalities-109078.html
range-for-variable-x-in-a-given-inequality-109468.html
everything-is-less-than-zero-108884.html
graphic-approach-to-problems-with-inequalities-68037.html

All DS Inequalities Problems to practice: search.php?search_id=tag&tag_id=184
All PS Inequalities Problems to practice: search.php?search_id=tag&tag_id=189

700+ Inequalities problems: inequality-and-absolute-value-questions-from-my-collection-86939.html
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(1-x^2) >= 0 can be expressed as (1-x) (1+x) >=0

So 1-x>=0 (OR) 1+x>=0.

Therefore, 1 >= x (OR) x >= -1
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Hi All,

With inequality-based questions, sometimes the easiest approach is just to come up with a few examples that 'fit' the given prompt and then use those examples to eliminate answer choices.

Here, we're told that 1 - X^2 >= 0. We're asked for ALL of the possible values that fit this inequality.

The 'easiest' value that most Test Takers would immediately 'see' is 1 (since 1 - 1^2 = 0), so X COULD be 1.

Next, since we're dealing with a squared term, -1 would also be a solution (since 1 - [-1]^2 = 0).

So we immediately have at least two solutions: 1 and -1. We can eliminate Answers A, B and C.

For the last step, we have to determine what OTHER solutions are possible. You can either prove that fractions fit (try using X = 1/2) or proving that larger integers do NOT fit (try using X = 2). Either way, you can eliminate the final incorrect answer.

Final Answer:
GMAT assassins aren't born, they're made,
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droopy57
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

To solve, we first isolate the x^2 in the inequality 1 – x^2 ≥ 0. So we have:

1 ≥ x^2

Next, we take the square root of both sides, to isolate x.

√1 ≥ √x^2

This gives us:

1 ≥ |x|

Because the variable x is inside the absolute value sign, we must consider that x can be either positive or negative. Therefore, we’ll need to solve the inequality twice.

When x is positive:

1 ≥ |x| means

1 ≥ x

This can be re-expressed as x ≤ 1.

When x is negative:

1 ≥ |x| means

1 ≥ -x (Divide both sides by -1 and switch the inequality sign)

-1 ≤ x

We combine the two resulting inequalities to get:

-1 ≤ x ≤ 1

Answer is E.
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Walkabout
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


In such inequalities, as long as one can factorize the expression into linear factors, the most methodical way to approach such questions is to use the wavy line approach.

You can refer to the following posts for a comprehensive treatment of the Wavy Line Approach:

https://gmatclub.com/forum/inequalities-trick-91482-80.html?sid=fde22066899cf98d261c2d987caa4509#p1465609

https://gmatclub.com/forum/wavy-line-method-application-complex-algebraic-inequalities-224319.html


Let’s apply this approach in the given question.

Given:

\(1–x^2 \geq{0}\)
\(x^2 – 1 \leq{0}\)

\((x + 1)*(x – 1) \leq{0}\) ………… (1)

Approach:

Apply the wavy line approach.


    Mark the zero points on the number and draw the wavy line.
    Identify the \(+ve\) and \(-ve\) regions of the curve.
    Since we need the range of values of \(x\), for which the expression in (1) is less than or equal to zero, we consider the \(-ve\) region(s) along with the zero points.

Working Out:



The range of values of \(x\) for which the given inequality is satisfied is \(-1 \leq x \leq 1\)

Correct Answer: Option E
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Hi All,

With inequality-based questions, sometimes the easiest approach is just to come up with a few examples that 'fit' the given prompt and then use those examples to eliminate answer choices.

Here, we're told that 1 - X^2 >= 0. We're asked for ALL of the possible values that fit this inequality.

The 'easiest' value that most Test Takers would immediately 'see' is 1 (since 1 - 1^2 = 0), so X COULD be 1.

Next, since we're dealing with a squared term, -1 would also be a solution (since 1 - [-1]^2 = 0).

So we immediately have at least two solutions: 1 and -1. We can eliminate Answers A, B and C.

For the last step, we have to determine what OTHER solutions are possible. You can either prove that fractions fit (try using X = 1/2) or proving that larger integers do NOT fit (try using X = 2). Either way, you can eliminate the final incorrect answer.

Final Answer:

GMAT assassins aren't born, they're made,
Rich
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Which of the following describes all values of x for which 1−x2≥0?

1-x^2 ≥ 0
or,x^2-1≤0
or, (x+1)(x-1)≤0
so,-1≤x≤1

correct answer E
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mgoblue123
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


By plugging in values, I see that option c also suits well
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Hi Jaya6,

With inequality-based questions, sometimes the easiest approach is just to come up with a few examples that 'fit' the given prompt and then use those examples to eliminate answer choices.

Here, we're told that 1 - X^2 >= 0. We're asked for ALL of the possible values that fit this inequality.

The 'easiest' value that most Test Takers would immediately 'see' is 1 (since 1 - 1^2 = 0), so X COULD be 1.

Next, since we're dealing with a squared term, -1 would also be a solution (since 1 - [-1]^2 = 0).

So we immediately have at least two solutions: 1 and -1. We can eliminate Answers A, B and C.

For the last step, we have to determine what OTHER solutions are possible. You can either prove that fractions fit (try using X = 1/2) or proving that larger integers do NOT fit (try using X = 2). Either way, you can eliminate the final incorrect answer.

GMAT assassins aren't born, they're made,
Rich
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EMPOWERgmatRichC
Hi Jaya6,

With inequality-based questions, sometimes the easiest approach is just to come up with a few examples that 'fit' the given prompt and then use those examples to eliminate answer choices.

Here, we're told that 1 - X^2 >= 0. We're asked for ALL of the possible values that fit this inequality.

The 'easiest' value that most Test Takers would immediately 'see' is 1 (since 1 - 1^2 = 0), so X COULD be 1.

Next, since we're dealing with a squared term, -1 would also be a solution (since 1 - [-1]^2 = 0).

So we immediately have at least two solutions: 1 and -1. We can eliminate Answers A, B and C.

For the last step, we have to determine what OTHER solutions are possible. You can either prove that fractions fit (try using X = 1/2) or proving that larger integers do NOT fit (try using X = 2). Either way, you can eliminate the final incorrect answer.

GMAT assassins aren't born, they're made,
Rich


Thanks Rich!

That was useful. I had not taken -1 into consideration and hence felt option C was right.
But now my confusion is cleared
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To determine the values of x for which 1−x^2≥0, we need to find the values of x that make the expression greater than or equal to zero.

To do that, we can solve the inequality:

1 - x^2 ≥ 0

Rearranging the inequality:

x^2 - 1 ≤ 0

Now, factor the expression:

(x - 1)(x + 1) ≤ 0

To solve this inequality, we consider the signs of each factor:

When x < -1, both (x - 1) and (x + 1) are negative, so their product is positive.

When -1 ≤ x ≤ 1, (x - 1) is negative, and (x + 1) is positive, so their product is negative.

When x > 1, both (x - 1) and (x + 1) are positive, so their product is positive.

We are looking for the values where the product is less than or equal to zero, which occurs when:

-1 ≤ x ≤ 1

Therefore, the correct answer is E. -1 ≤ x ≤ 1, which describes all values of x for which 1−x^2≥0.
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Quote:
Actually you can transform it to an absolute value problem: \(1-x^2\geq{0}\) --> \(x^2\leq{1}\), since both parts of the inequality are non-negative then we can take square root: \(|x|\leq{1}\) --> \(-1\leq{x}\leq{1}\).

Now, other approach would be: \(1-x^2\geq{0}\) --> \(x^2-1\leq{0}\) --> \((x+1)(x-1)\leq{0}\) --> the roots are -1 and 1 --> "<" sign indicates that the solution lies between the roots, so \(-1\leq{x}\leq{1}\).


Solving inequalities:
https://gmatclub.com/forum/x2-4x-94661.html#p731476 (check this one first)
https://gmatclub.com/forum/inequalities ... 91482.html
https://gmatclub.com/forum/data-suff-in ... 09078.html
https://gmatclub.com/forum/range-for-va ... me#p873535
https://gmatclub.com/forum/everything-i ... me#p868863

Now, about x2suresh's approach: we have \((x+1)(x-1)\leq{0}\), so the product of two multiples is less than (or equal to) zero, which means that the multiples must have opposite signs. Then x2suresh checks the case A. when the first multiple (x+1) is negative and the second (x-1) is positive and the case B. when the first multiple (x+1) is positive and the second (x-1) is negative to get the range for which \((x+1)(x-1)\leq{0}\) holds true. Notice that, for this particular problem, we don't realy need to test case A, since it's not possible (x+1), the larger number, to be negative and (x-1), the smaller number to be positive. As for case B, it gives: \(x+1\geq{0}\) and \(x-1\leq{0}\) --> \(x1\geq{-1}\) and \(x\leq{1}\) --> \(-1\leq{x}\leq{1}\).

Hope it helps.

Hi Bunuel -
I can get to the answer when I look at in the sense of x^2-1<=0, hence x will have a + and - value.

But when I look at it, this way:
(x+1)(x−1)≤0
Simplifying each gives me
(x+1)≤0 = x≤1
(x−1)≤0 = x≤-1 (But I know this is not right, so I wonder, on a fundamental level, why is the inequality sign flipped here, even though we do not, divide or multiply by a negative)?

Thanks for your help as always.
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Quote:
1. Actually you can transform it to an absolute value problem: \(1-x^2\geq{0}\) --> \(x^2\leq{1}\), since both parts of the inequality are non-negative then we can take square root: \(|x|\leq{1}\) --> \(-1\leq{x}\leq{1}\).

2. Now, other approach would be: \(1-x^2\geq{0}\) --> \(x^2-1\leq{0}\) --> \((x+1)(x-1)\leq{0}\) --> the roots are -1 and 1 --> "<" sign indicates that the solution lies between the roots, so \(-1\leq{x}\leq{1}\).


Solving inequalities:
https://gmatclub.com/forum/x2-4x-94661.html#p731476 (check this one first)
https://gmatclub.com/forum/inequalities ... 91482.html
https://gmatclub.com/forum/data-suff-in ... 09078.html
https://gmatclub.com/forum/range-for-va ... me#p873535
https://gmatclub.com/forum/everything-i ... me#p868863

3. Now, about x2suresh's approach: we have \((x+1)(x-1)\leq{0}\), so the product of two multiples is less than (or equal to) zero, which means that the multiples must have opposite signs. Then x2suresh checks the case A. when the first multiple (x+1) is negative and the second (x-1) is positive and the case B. when the first multiple (x+1) is positive and the second (x-1) is negative to get the range for which \((x+1)(x-1)\leq{0}\) holds true. Notice that, for this particular problem, we don't realy need to test case A, since it's not possible (x+1), the larger number, to be negative and (x-1), the smaller number to be positive. As for case B, it gives: \(x+1\geq{0}\) and \(x-1\leq{0}\) --> \(x\geq{-1}\) and \(x\leq{1}\) --> \(-1\leq{x}\leq{1}\).

Hope it helps.

Hi Bunuel -
I can get to the answer when I look at in the sense of x^2-1<=0, hence x will have a + and - value.

But when I look at it, this way:
(x+1)(x−1)≤0
Simplifying each gives me
(x+1)≤0 = x≤1
(x−1)≤0 = x≤-1
(But I know this is not right, so I wonder, on a fundamental level, why is the inequality sign flipped here, even though we do not, divide or multiply by a negative)?

Thanks for your help as always.

Yes, the red parts in your solution are not correct at all. That is not how this inequality can be solved. I have highlighted above three different approaches you can use to solve it, the first one being the easiest. As for your question, I'm not sure I follow – where is the sign flipped, in which solution?
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Yes, the red parts in your solution are not correct at all. That is not how this inequality can be solved. I have highlighted above three different approaches you can use to solve it, the first one being the easiest. As for your question, I'm not sure I follow – where is the sign flipped, in which solution?

Thanks Bunuel. I am actually able to solve using the 3 approaches above and get to the answer, but let me clarify my question. I am trying to clarify this, as I feel, there is a flaw in my understanding of fundamentals somewhere, which is why the below is not making sense to me.

if we have x+1<0, we can move 1 to the other side, and get x<-1. We dont flip signs, as we are not multiplying or divide.
Now coming to this question (Sorry I wrote the signs upside down in my last post by mistake), we have:
(x+1)≤0 = if I move 1 to the other side I should get x≤-1
(x−1)≤0 = similarly, here I get x≤1

Now, if I take those two values of X, I get x≤-1 or x≤1, but I do not get a compound equality, i.e. -1≤x≤1, which is the correct answer.
This means x≤-1 is obv. not correct and the sign is being flipped in the final answer. So why is this being flipped considering there is no multiplication or division by a negative involved?
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