If y = |||x| 3| x|| , how many values are possible for y=1? 1) 0

Question

If y = |||x| – 3| – x| , how many values are possible for y=1?

A) 0
B) 1
C) 2
D) 4
E) 6

This Qs is from  VeritasKarishma  blog post. While I absolutely love her graphical approach in general, for this one particularly I find it way to complex and so looking for an alternate approach.

KarishmaB · Answer

The algebraic approach would be worse. This question is not GMAT type and is discussed for intellectual purposes only. Take some time to understand why the graph changes at each step as it does and then walk away from it. If what is done at each step is clear to you, that is all you need to take away from this question. You would not be required to solve such a question in the test.

lg800 · Answer

Could someone please explain how the answer is C?

Regor60 · Answer

Appears to be an absolute value sign missing in the question since there are only 7 appearing ?

Posted from my mobile device

Avneeshyadav01 · Answer

The algebraic approach is lengthier than the graphical solution. We need to solve the modulus from outside to inside. 
Given, Y = 1
Hence |||x|-3|-x| = 1

Case 1: ||x|-3|-x = 1 
||x|-3| = x+1

Subcase 1.1: 
|x|-3 = x+1
|x| = x+4

Subcase 1.1.1: 
X = x+4 (no solution)

Subcase 1.1.2:
X= -x-4
X= -2------------------------------(1)

Subcase 1.2:
|x|-3 = -x-1
|x| = 2-x

Subcase 1.2.1:
x=2-x
x=1---------------------------------(2)

Subcase 1.2.2:
X=x-2 (no solution)

Case 2: ||x|-3|-x = -1
||x|-3| = x-1

Subcase: 2.1
|x|-3 = x-1 
|x| = x+2

Subcase: 2.1.1:
X = x+2 (no solution)

Subcase: 2.1.2:
X= -x-2
X= -1 -----------------------------------------(3)

Subcase 2.2:
|x|-3 = -x+1
|x| = -x+4

Subcase 2.2.1:
X= -x+4
X= 2 --------------------------------------------(4)

Subcase 2.2.2:
X = x-4 (no solution)

Now, we have the solutions above. Since solving for modulus might give some extraneous solutions, we need to verify all possible solutions by putting values in the question stem
x y
1 1
-1 3
2 1
-2 3
 Hence, for y to be 1, x can have only two values i.e. x=1 & x=2. So option C is the right choice.

angysinghdhillon · Answer

KarishmaB  can you please post a link to the original question in this thread so that we can take a look at the graphical approach.

Oppenheimer1945 · Answer

modify the eq:

||x|-3|=x+1 , x-1

plot the graphs, we get 2 pts of intersection

bumpbot · Answer

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If y = |||x| 3| x|| , how many values are possible for y=1? 1) 0

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