**Answer**

(1) INSUFFICIENT: Statement (1) is insufficient because y is unbounded when both x and k can vary. Therefore y has no definite maximum.

To show that y is unbounded, let's calculate y for a special sequence of (x, k) pairs. The sequence starts at (-2, 1) and doubles both values to get the next (x, k) pair in the sequence.

y1 = | -2 – 1 | – | -2 + 1 | = 3 – 1 = 2

y2 = | -4 – 2 | – | -4 + 2 | = 6 – 2 = 4

y3 = | -8 – 4 | – | -8 + 4 | = –12 + 4 = 8

etc.

In this sequence y doubles each time so it has no definite maximum, so statement (1) is insufficient.

(2) SUFFICIENT: Statement (2) says that k = 3, so y = | x – 3 | – | x + 3 |. Therefore to maximize y we must maximize | x – 3 | while simultaneously trying to minimize | x + 3 |. This state holds for very large negative x. Let's try two different large negative values for x and see what happens:

If x = -100 then:

y = |-100 – 3| – |-100 + 3|

y = 103 – 97 = 6

If x = -101 then:

y = |-101 – 3| – |-101 + 3|

y = 104 – 98 = 6

We see that the two expressions increase at the same rate, so their difference remains the same. As x decreases from 0, y increases until it reaches 6 when x = –3. As x decreases further, y remains at 6 which is its maximum value.

The correct answer is B.

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