The minimum value of f(x) = |10 – x| + |x – 2| – |4 – x| is attained

No since the function is of the form |x| + |y| - |z|

Method 1

Substitute values of x from options

Answer: Option D


Method 2

Draw graph and check the value of x for which graph has min. value

for given function value to get min value ; minimise the max possible in this case its 10
so \(f(x) = |10 – x| + |x – 2| – |4 – x|\) at x=10 gives minimum value
IMO D

A - 2 => 8-2 = 6
B - 4 => 6+2 = 8
C - 8 => 2+6-4 = 4
D - 10 => 8-6 = 2
E - 12 => 2+10-8 = 4

Plug in the options and check

(A) 8 + 0 - 2 = 6
(B) 6 + 2 - 0 = 8
(C) 2 + 6 - 4 = 4
(D) 0 + 8 - 6 = 2
(E) 2 + 10 - 8 = 4

Thus, Answer must be (D)

Is there any logic for similar questions?
|x| + |y| - |z|

I also solved using the substitution method.

for this to be minimum -> \(f(x) = |10 – x| + |x – 2| – |4 – x|\) we need |10-x|+|x-2| to be minimum and |4-x| to be maximum
now, if we see the distance method approach:
|10-x|+|x-2| this is same as |x-10|+|x-2| ;
i.e. the sum of distance of x from ten and distance of x from 2, and, this remains at a constant value of 8 for all values of x from 2 to 10 & increase if x >10 or x<2
thus for |10-x|+|x-2| to be minimum we have to select an x value ranging from 2 to 10.
Now, |4-x| = |x-4| , i.e. this is the distance of x from 4,
So, for this value to be maximum we must select a value for x which is as farthest away from 4 as possible
given both these value conditions, we have x = 10 which is the farthest from 4 and gives minimum value for sum of distances from 2 and 10.
This would be the general approach,
Here however, substituting the values is a good and fast option, which may not be the case every time.
D

How to solve this question with the distance method approach like done in the previous question?

Hi GMATinsight,

Could you please advise on how to make these graphs quickly and efficiently on the test day?

The question has a linear equation and we are asked to minimum value attainable. Just plug-in options, compare the results then we get the answer

Thanks prathyushaR

Also, I've seen the graphical approach posted here and in another thread for an equation with three moduli. The other thread indicated to use 'plug and chug' to find the intersection of a two mod line with a one mod line. Here we magically combine three moduli into a single line to find the minimum. Bunuel chetan2u Can a step-by-step breakdown please be given for how to make 2 and 3 mod graphs in under 2 minutes?

Additionally, how is the critical value method used here?
Bunuel has used it to find a minimum, but it was for two mods not three mods. The x's nicely cancel out in two cases in that example
https://gmatclub.com/forum/what-is-the- ... l#p2514397
It seems to me (a complete non-expert) that case two in that post means that the question in this thread is limited to the critical values and not lower

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