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:
x2-4x-94661.html#p731476 (check this one first)
inequalities-trick-91482.html
data-suff-inequalities-109078.html
range-for-variable-x-in-a-given-inequality-109468.html?hilit=extreme#p873535
everything-is-less-than-zero-108884.html?hilit=extreme#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.
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Hello,

You have \(1-x^2\geq{0}\).
Since LHS and RHS are non-negative,we can take square root on both sides and get

\(1\geq{\sqrt{x^2}}\)

Also, \(\sqrt{x^2}\)=|x| so we have \(|x|\leq{1}\)
So x is between \(-1\leq{x}\leq{1}\)

Also, you can do it as \(1-x^2\geq{0}\) or \((1-x)(1+x)\geq{0}\) (using \(a^2-b^2=(a-b)(a+b)\) )

We need to find in which region does the equation hold true...try values of x <-1, -1<x<1 and x>1 to see where the relationship holds true

You need to brush your basics on mod values. Check out below links

graphic-approach-to-problems-with-inequalities-68037.html
math-number-theory-88376.html
if-x-is-an-integer-what-is-the-value-of-x-1-x-2-4x-94661.html#p731476
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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?

\(1-x^2\geq{0}\) --> \(x^2\leq{1}\) --> \(-1\leq{x}\leq{1}\).

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


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Bunuel, Can you explain how we go from \(x^2\) to x in last step

(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

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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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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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:

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

http://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
_________________

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