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Which of the following describes all values of x for which 1

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Which of the following describes all values of x for which 1 [#permalink] New post 20 Dec 2012, 08:15
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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
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Re: Which of the following describes all values of x for which 1 [#permalink] New post 20 Dec 2012, 08:16
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Walkabout 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


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

Answer: E.
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Re: Which of the following describes all values of x for which 1 [#permalink] New post 27 May 2014, 01:57
Bunuel wrote:
Walkabout 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


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

Answer: E.



Bunuel, Can you explain how we go from x^2 to x in last step
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Re: Which of the following describes all values of x for which 1 [#permalink] New post 27 May 2014, 02:17
Which of the following describes all values of x for which 1–x^2 >= 0?

(A) x >= 1

Plugged in 2. 1–(2)^2 >= 0 -3>=0? NO. Wrong

(B) x <= –1

Plugged in -2. 1–(-2)^2 >= 0 -3>=0? NO. Wrong

(C) 0 <= x <= 1
Plugged in 0, 1, and 1/2. All of them work. But E is better because it describes all the values of x

(D) x <= –1 or x >= 1
A and B answer this. Wrong.

(E) –1 <= x <= 1

X is a positive or negative fraction or a zero.
x = -1/2 x = 1/2 x = 0
Plug all of them. They work.
Answer is E.

Another solution:
1–x^2 >= 0
1>= x^2
Therefore, -1=<x=<1
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Re: Which of the following describes all values of x for which 1 [#permalink] New post 27 May 2014, 02:37
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jatinsachani wrote:
Bunuel wrote:
Walkabout 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


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

Answer: E.



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


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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Re: Which of the following describes all values of x for which 1 [#permalink] New post 26 Jun 2014, 15:16
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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
Re: Which of the following describes all values of x for which 1   [#permalink] 26 Jun 2014, 15:16
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