Question 647930
|x-3| ≦ k. Find a value of k for which the inequality has no solution. 
<pre>
An absolute value can NEVER equal to a negative number, so this
inequality:

|x-3| &#8806; -1

has no solution.  (Any negative number on the right will do)

Answer: when k = -1 or any other negative number.

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</pre>Find a value for which the inequality has exactly one solution.
Find a value of k for which a solution exists but for which the solution set does not include 5.
<pre>
Both those problems involve cases where a solution exists, which will be
when k is a non-negative number. 

Let k be a non-negative number, that is, k &#8807; 0, so that the
inequality will have a solution.

Then |x-3| &#8806; k will become this three sided inequality:

      -k &#8806; x-3 &#8806; k  where   x &#8807; 0

Add 3 to all 3 sides:

     3-k &#8806; x &#8806; k+3

    The solution is the closed interval [3-k, k+3]

-------------------------------------------

Find a value for which the inequality has exactly one solution.

The closed interval [3-k, k+3] will have infinitely many solutions
except when the interval shrinks to just one point, and that will be 
when its endpoints are equal, so we set them equal:

     3-k = k+3
     -2k = 0
       k = 0

So |x-3| &#8806; 0 has exactly one solution, when k=0

    3-0 &#8806; x &#8806; 0+3
      3 &#8806; x &#8806; 3

Which means that solution is x = 3.

Answer: when k=0, there is exactly one solution, x = 3.

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Find a value of k for which a solution exists but for which the solution set does not include 5.

    The solution is the closed interval [3-k, k+3]

It will not contain k if k is either 

(a) less than the left endpoint

      k < 3-k
     2k < 3
      k < {{{3/2}}}

That is 0 &#8806; k < {{{3/2}}}

So we can pick any value for k in the interval  [0, {{{3/2}}}), say k=1

[actually the value k=0 that we used above would be a possible answer
here because the one-point interval [3-0,0+3] or [3,3} does not contain 
k=0.]

or

(b) greater than the right endpoint

    k > k+3 
    0 > 3

That is a contradiction so we discard this case.

So k = 1 is a good answer for this.
 
Edwin</pre>