Question

The function ƒ(x) = [x]² – [x²] (where [y] is the greatest integer less than or equal to y), is discontinuous at –

• A. All integers
• B. All integers except 0 and 1
• C. All integers except 0
• D. All integers except 1

Answer 1

Answer: (D)

All integers except 1

Answer 2

Note that ƒ(x) = 0 for each integral value of x.

Also, if 0 ≤ x < 1, then 0 ≤ x² < 1

∴ [x] = 0 and [x²] = 0 ⇒ ƒ(x) = 0 for 0 ≤ x < 1.

Next, if 1 ≤ x <$\sqrt { 2 }$, then

1 ≤ x² < 2 ⇒   [x] = 1 and [x²] = 1

Thus, ƒ(x) = [x]² – [x²] = 0 if 1 ≤ x < $\sqrt { 2 }$.

It follows that ƒ(x) = 0, if 0 ≤ x <$\sqrt { 2 }$.

This shows that ƒ(x) must be continuous at x = 1.

However, at points x other than integers and not lying between 0 and$\sqrt { 2 }$, ƒ(x) ≠ 0.

Thus, ƒ is discontinuous at all integers except 1.

Hence (4) is the correct answer.