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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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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?

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?

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

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

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

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

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

Re: Which of the following describes all values of x for which [#permalink]

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16 May 2015, 10:58

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Re: Which of the following describes all values of x for which [#permalink]

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28 Jun 2016, 17:52

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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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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