M42-06

Question

If  |p – |p|| = 3 , how many values of  p  satisfy this equation?

A. 0
B. 1
C. 2
D. 3
E. 4

Bunuel · Answer

Official Solution:

If  |p – |p|| = 3 , how many values of  p  satisfy this equation?

A. 0
B. 1
C. 2
D. 3
E. 4

When  p \geq 0 , then  |p| = p , so in this case we'd have  |p - p| = 3 , which gives  0 = 3 . This is obviously incorrect. Therefore,  p  cannot be 0 or positive.
 
Thus,  p < 0 . In this case,  |p| = -p , so we'd have  |p + p| = 3 , which gives  |2p| = 3 . This simplifies to  p = -\frac{3}{2}  (since  p = \frac{3}{2}  is not possible because we know that  p  is negative). Therefore, one value of  p  satisfies the given equation.

Answer: B

aaaunknown · Answer

In the second case, if |p|=-p then we should have |-p+p|=3 which will give us 0=3 "coz -p is already present in the equation"
Please let me know if I made a mistake?

Bunuel · Answer

p being negative does not mean that you should replace p with -p. For example, if p = -2, |p| equals -(-2), which is 2. The absolute value function always outputs a non-negative result, regardless of whether the input is positive or negative.

In the solution, when p is negative, |p| is correctly replaced with -p (because |p| = -p for p < 0), and the equation becomes |p + p| = 3, which simplifies to |2p| = 3. This leads to the valid solution p = -3/2.

It seems you misunderstood the behavior of the absolute value function. Please review the explanation above carefully. Let me know if you need further clarification!

Adit_ · Answer

There will be two cases:
When p = +ve:

Then we see that p-|p| = 3.
But since we took p as positive we get:
p-p = 0 which is not possible as RHS = 3.

So positive values are ruled out.

Taking negative values:

p-|p| = -3.
p-(-p) = -3.
2p = -3.
p = -3/2.

And as we see this is the only solution possible.

Answer: Option B

_____________________________________

IMO absolute value questions are a toss up between doing trial and error or solving it methodically.
In a time crunch situation for such a question you might miss the only case possible, in which case methodical work is the best. 
If there is only 1 mod then it is easier to visualise and thus better to do trial and error to get to the answer in time crunch situations.
But questions pertaining to number of solutions required would almost always need a proper approach IMO, so better to go with it.

M42-06

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