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Bunuel
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IMO D

When x<0=> |x|=-x => |x|+x=0
=>[(x+|x|)^2]/4 =0

When x>=0=> |x|=x=> |x|+x=2x
=>[(x+|x|)^2]/4 =(2x)^2/4=x^2
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Bunuel
Which of the following expressions equals 0 when x < 0 and equals x^2 when x ≥ 0?

A. \(x|x|\)

B. \(x + |x|\)

C. \(2x^2 - x|x|\)

D. \(\frac{1}{4}(x + |x|)^2\)

E. \((\frac{x - |x|}{2})^2\)

A) If x = -1 < 0, then -1|-1| = -1(1) = -1 ≠ 0.

B) If x = 1 ≥ 0, then 1 + |1| = 2 ≠ 1^2.

C) If x = -1 < 0, then 2(-1)^2 – (-1)|-1| = 2 – (-1) = 2 + 1 = 3 ≠ 0.

E) If x = 1 ≥ 0, then [(1 - |1|)/2]^2 = 0^2 = 0 ≠ 1^2

D) Since we eliminated four incorrect choices, the correct answer must be D.

Answer: D

Note:

Is D really the correct answer? Let’s check.

If x < 0, then (1/4)(x + |x|)^2 = (1/4)(x + (-x))^2 = (1/4)(0)^2 = 0.

If x ≥ 0, then (1/4)(x + |x|)^2 = (1/4)(2x)^2 = (1/4)(4x^2) = x^2.
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gmatophobia
Bunuel
Which of the following expressions equals 0 when x < 0 and equals x^2 when x ≥ 0?

A. \(x|x|\)

B. \(x + |x|\)

C. \(2x^2 - x|x|\)

D. \(\frac{1}{4}(x + |x|)^2\)

E. \((\frac{x - |x|}{2})^2\)

Let's take some values and test

For x ≥ 0 → we will use x = 3
For x < 0 → we will use x = -2

A. \(x|x|\)

If x < 0

\(x|x|\) = -2 * 2 = -4; The expected value = 0.

B. \(x + |x|\)

If x ≥ 0

\(x + |x|\) = 3 + 3 = 9; The expected value = 9.

C. \(2x^2 - x|x|\)

If x < 0

\(2x^2 - x|x|\) = 2(-2)^2 - (2*2) = 4; The expected value = 0.

D. \(\frac{1}{4}(x + |x|)^2\)

If x < 0

\(\frac{1}{4}(x + |x|)^2\) = \(\frac{1}{4}(-2 + 2)^2\); The expected value = 0.

If x ≥ 0

\(\frac{1}{4}(x + |x|)^2\) = \(\frac{1}{4}(3 + 3)^2\) = \(\frac{36}{4} = 9\); The expected value = 9.

Let's keep this.

E. \((\frac{x - |x|}{2})^2\)

If x ≥ 0

\((\frac{x - |x|}{2})^2\) = \((\frac{3 - 3}{2})^2 = 0\); The expected value = 9.

Option D

Hi, I believe there is a slight error:

Quote:
If x ≥ 0

\(x + |x|\) = 3 + 3 = 9; The expected value = 9.

You mean 3 + 3 = 6
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Hi, I believe there is a slight error:

Quote:
If x ≥ 0

\(x + |x|\) = 3 + 3 = 9; The expected value = 9.

You mean 3 + 3 = 6

Good catch ! Corrected the typo :)
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AryaSwagat your solution is really better and timesaving
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Bunuel
Which of the following expressions equals 0 when x < 0 and equals x^2 when x ≥ 0?

A. \(x|x|\)

B. \(x + |x|\)

C. \(2x^2 - x|x|\)

D. \(\frac{1}{4}(x + |x|)^2\)

E. \((\frac{x - |x|}{2})^2\)
­
Use the definition of absolute values here:
|x| = x if x >= 0
|x| = -x if x < 0

First we check for negative values because we should simply get 0. So we will be able to eliminate many options. 


A. \(x|x|\)
If x < 0, we get -x^2. This is not 0. Eliminate.

B. \(x + |x|\)
If x < 0, we get x - x = 0

C. \(2x^2 - x|x|\)
If x < 0, we get
\(2x^2 + x^2\). This is not 0. Eliminate.

D. \(\frac{1}{4}(x + |x|)^2\)
If x < 0, we get 

\(\frac{1}{4}(x - x)^2 = 0\)

E. \((\frac{x - |x|}{2})^2\)
If x < 0, we get x + x/2. This is not 0. Eliminate.

Now simply check for options (B) and (D).
B. \(x + |x|\)
If x >= 0, we get x + x = 2x. This is not x^2.
Eliminate.

Answer (D)

The definition of absolute values and its usage is discussed here:
https://anaprep.com/algebra-the-why-beh ... questions/
 
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AryaSwagat
\(|x| = x\) , if \(x ≥ 0\) and \((-x)\), if \(x <0\).

So
If \(x ≥ 0\)
Replace \(|x|\) with \(x \)in each expression.

A. \(x|x|= x^2\)

B. \(x+|x| = 2x\)

C. \(2x^2−x|x|= 2x^2-x^2= x^2\)

D. \(\frac{1}{4} (x+|x|)^2= \frac{1}{4} (2x)^2= x^2\)

E. \((\frac{(x−|x|)}{2})^2= 0 \)

If x<0
Replace \(|x|\) with \((-x)\) and calculate each expression.

A. \(x|x|= (-x^2)\)

B. \(x+|x| = 0\)

C. \(2x^2−x|x|= 2x^2- x*(-x)= 2x^2+x^2= 3x^2\)

D. \(\frac{1}{4} (x+|x|)^2= \frac{1}{4} (x-x)^2=0\)

E. \((\frac{(x−|x|)}{2})^2= (\frac{(x+x)}{2})^2= x^2 \);

So D
This is probably the most correct (and fastest) way to solve. But you can improve the time by calculating only A, C, D for the second case because all the others are not x^2­
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i tested -2 and 2 and i got B and moved on too fast. should you just never test numbers like 0 1 or 2?
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