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Let f(x) = [x2]-[x]2, where [.] denotes the greatest integer function. Then
  • a)
    f (x) is discontinuous for all integral values of x
  • b)
    f (x) is discontinuous only at x = 0, 1
  • c)
    f (x) is continuous only at x = 1
  • d)
    none of these
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
Let f(x) = [x2]-[x]2, where [.] denotes the greatest integer function....

Also f(1)= 0. So, f(x) is continuous at x = 1.
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Most Upvoted Answer
Let f(x) = [x2]-[x]2, where [.] denotes the greatest integer function....
Solution:
Given, f(x) = [x^2] - [x]^2
Let's consider the left-hand limit and the right-hand limit at x = 1

Left-hand limit:
lim f(x) as x approaches 1 from the left side
= lim ([x^2] - [x]^2) as x approaches 1 from the left side
= [(1-ε)^2] - [1-ε]^2 (where ε is a small positive number)
= [1 - 2ε + ε^2] - [1 - 2ε + ε^2]
= 0

Right-hand limit:
lim f(x) as x approaches 1 from the right side
= lim ([x^2] - [x]^2) as x approaches 1 from the right side
= [1^2] - [1]^2
= 0

Since the left-hand limit and the right-hand limit are equal, f(x) is continuous at x = 1.

Now, let's consider the intervals between integers:

For x ∈ [n, n+1), where n is an integer, we have:
[x^2] = n^2 (since n^2 ≤ x^2 < />
[x]^2 = n^2
Therefore, f(x) = n^2 - n^2 = 0 for all x in the interval [n, n+1).

For x = n, we have:
[x^2] = n^2
[x]^2 = n^2
Therefore, f(x) = n^2 - n^2 = 0 at x = n.

Hence, f(x) is continuous at x = 1 and discontinuous at all integral values of x except x = 1. Therefore, the correct option is (c).
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Let f(x) = [x2]-[x]2, where [.] denotes the greatest integer function. Thena)f (x) is discontinuous for all integral values of xb)f (x) is discontinuous only at x = 0, 1c)f (x) is continuous only at x = 1d)none of theseCorrect answer is option 'C'. Can you explain this answer?
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