Let \(f(x) = [x]\) and \(g(x) = x - [x]\), where [x] denotes the greatest integer less than or equal to x. Consider the following statements:
(A) f is discontinuous at \(x = 1\).
(B) f + g is continuous at \(x = 1\).
(C) fg is continuous at \(x = 1\).
(D) f − g is continuous at \(x = 1\).
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A), (B) and (C) only
Near x = 1 the two functions are:
\(0 \lt x \lt 1: \ f(x) = 0, \ g(x) = x\)
\(1 \le x \lt 2: \ f(x) = 1, \ g(x) = x - 1\)
As x → 1 from the left, f → 0 and g → 1. As x → 1 from the right, f → 1 and g → 0. At x = 1, f(1) = 1 and g(1) = 0.
\(f(x) + g(x) = x\) for all x
\(fg\): left limit \(0 \times 1 = 0\), right limit \(1 \times 0 = 0\), value \(1 \times 0 = 0\)
\(f - g\): left limit \(0 - 1 = -1\), right limit \(1 - 0 = 1\)
So f jumps at 1, f + g is the identity function x, fg has limit 0 equal to its value 0, and f − g has one-sided limits −1 and 1.
Checking the statements
(A) f is discontinuous at \(x = 1\). — Correct. Left limit 0, right limit 1.
(B) f + g is continuous at \(x = 1\). — Correct. f + g = x, which is continuous everywhere.
(C) fg is continuous at \(x = 1\). — Correct. Both one-sided limits and the value are 0.
(D) f − g is continuous at \(x = 1\). — Incorrect. One-sided limits −1 and 1 are unequal.
This shows that a sum or product of two discontinuous functions can still be continuous.