The function ƒ(x) = [x] 2 – [x 2 ] (where [y] is the greatest integer less than or equal to y), is discontinuous at -
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Note that ƒ(x) = 0 for each integral value of x.
Also, if 0 ≤ x < 1, then 0 ≤ x 2 < 1
∴ [x] = 0 and [x 2 ] = 0 ⇒ ƒ(x) = 0 for 0 ≤ x < 1.
Next, if 1 ≤ x <
, then
1 ≤ x 2 < 2 ⇒ [x] = 1 and [x 2 ] = 1
Thus, ƒ(x) = [x] 2 – [x 2 ] = 0 if 1 ≤ x <
.
It follows that ƒ(x) = 0, if 0 ≤ x <
.
This shows that ƒ(x) must be continuous at x = 1.
However, at points x other than integers and not lying between 0 and
, ƒ(x) ≠ 0.
Thus, ƒ is discontinuous at all integers except 1.
Hence is the correct answer.
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