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

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Joined: 02 Sep 2009
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The minimum value of f(x) = |10 – x| + |x – 2| – |4 – x| is attained  [#permalink]

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10 Mar 2020, 01:47
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45% (medium)

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61% (01:42) correct 39% (01:31) wrong based on 201 sessions

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

A. 2
B. 4
C. 8
D. 10
E. 12

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Re: The minimum value of f(x) = |10 – x| + |x – 2| – |4 – x| is attained  [#permalink]

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10 Mar 2020, 03:17
2
Bunuel wrote:
The minimum value of $$f(x) = |10 – x| + |x – 2| – |4 – x|$$ is attained at x = ?

A. 2
B. 4
C. 8
D. 10
E. 12

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Method 1

Substitute values of x from options

Method 2

Draw graph and check the value of x for which graph has min. value
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Screenshot 2020-03-10 at 4.45.54 PM.png [ 252.07 KiB | Viewed 1525 times ]

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Re: The minimum value of f(x) = |10 – x| + |x – 2| – |4 – x| is attained  [#permalink]

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10 Mar 2020, 01:55
1
Bunuel wrote:
The minimum value of $$f(x) = |10 – x| + |x – 2| – |4 – x|$$ is attained at x = ?

A. 2
B. 4
C. 8
D. 10
E. 12

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Asked: The minimum value of $$f(x) = |10 – x| + |x – 2| – |4 – x|$$ is attained at x = ?

At x=10
$$f(x) = |10 – x| + |x – 2| – |4 – x| = 0+8-6 = 2$$ which is the minimum value of the function.

IMO D
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Re: The minimum value of f(x) = |10 – x| + |x – 2| – |4 – x| is attained  [#permalink]

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10 Jun 2020, 08:47
1
Bunuel wrote:
The minimum value of $$f(x) = |10 – x| + |x – 2| – |4 – x|$$ is attained at x = ?

A. 2
B. 4
C. 8
D. 10
E. 12

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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
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Re: The minimum value of f(x) = |10 – x| + |x – 2| – |4 – x| is attained  [#permalink]

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10 Mar 2020, 02:25
Shouldn't be 0+8+6 instead of 0+8-6?

Thank you
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Re: The minimum value of f(x) = |10 – x| + |x – 2| – |4 – x| is attained  [#permalink]

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10 Mar 2020, 02:38
moneysniper365 wrote:
Shouldn't be 0+8+6 instead of 0+8-6?

Thank you

No since the function is of the form |x| + |y| - |z|
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Re: The minimum value of f(x) = |10 – x| + |x – 2| – |4 – x| is attained  [#permalink]

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10 Mar 2020, 06:45
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

Bunuel wrote:
The minimum value of $$f(x) = |10 – x| + |x – 2| – |4 – x|$$ is attained at x = ?

A. 2
B. 4
C. 8
D. 10
E. 12

Project PS Butler

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Re: The minimum value of f(x) = |10 – x| + |x – 2| – |4 – x| is attained  [#permalink]

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30 May 2020, 05:35
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
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Re: The minimum value of f(x) = |10 – x| + |x – 2| – |4 – x| is attained  [#permalink]

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30 May 2020, 07:09
Bunuel wrote:
The minimum value of $$f(x) = |10 – x| + |x – 2| – |4 – x|$$ is attained at x = ?

A. 2
B. 4
C. 8
D. 10
E. 12

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)
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Re: The minimum value of f(x) = |10 – x| + |x – 2| – |4 – x| is attained  [#permalink]

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30 May 2020, 08:07
Is there any logic for similar questions?
|x| + |y| - |z|

I also solved using the substitution method.
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Re: The minimum value of f(x) = |10 – x| + |x – 2| – |4 – x| is attained   [#permalink] 30 May 2020, 08:07

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

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