Let f : [0, ∞) → ℝ be the periodic function of period 1 given by f(x) = 1 − |2x − 1| for x ∈ [0, 1], and define g : [0, ∞) → ℝ by g(x) = f(x²). Which of the following statements are true?

A, B, C. f is a continuous tent function with f(0) = f(1) = 0, so its periodic extension is continuous; continuous periodic ⇒ uniformly continuous. g = f∘x² is continuous but its oscillations speed up as x → ∞ (the tents get compressed), so it is not uniformly continuous.

Periodic uniform continuity — CSIR NET December 2023 Part C, solved

Periodic uniform continuity

Let $f : [0, \infty) \to \mathbb{R}$ be the periodic function of period 1 given by f(x) = 1 − |2x − 1| for x ∈ [0, 1], and define $g : [0, \infty) \to \mathbb{R}$ by $g(x) = f(x^{2})$ . Which of the following statements are true?

A.f is continuous on $[0, \infty)$ .✓ Correct
B.f is uniformly continuous on $[0, \infty)$ .✓ Correct
C.g is continuous on $[0, \infty)$ .✓ Correct
D.g is uniformly continuous on $[0, \infty)$ .

The idea, in plain words

The zigzag itself is perfectly well behaved, repeating the same shape at the same width forever. But feeding in the square of the input speeds things up as you travel out, squashing each zigzag into a narrower and narrower space, so the graph gets steeper without limit.

Solution

f is a continuous tent function with f(0) = f(1) = 0

so its periodic extension is continuous; continuous periodic ⇒ uniformly continuous. g = f∘ $x^{2}$ is continuous but its oscillations speed up as $x \to \infty ($ the tents get compressed)

so it is not uniformly continuous.

Why each option is right or wrong

Checked against 4 symbolic computations

A.Correct
True.

On [0, 1] the tent 1 − |2x − 1| is continuous, with f(0) = 0, f(1/2) = 1 and f(1) = 0 — all three verified.

Matching values at the two endpoints is exactly what lets the period-1 extension glue together continuously.
B.Correct
True. f is continuous and periodic on $[0, \infty)$ .

A continuous periodic function is uniformly continuous: it is uniformly continuous on one closed period by Heine–Cantor, and periodicity carries that same modulus to the whole line.
C.Correct
True. g is the composition of the continuous map $x \mapsto x^{2}$ with the continuous f, and compositions of continuous functions are continuous.
D.WrongPointwise vs uniform
False, and this is the point of the question.

Uniform continuity is not preserved under composition

because $x \mapsto x^{2}$ is not uniformly continuous on $[0, \infty)$ .

Concretely, take $x_{n} = \sqrt{n + 1/2}$ and $y_{n} = \sqrt{n}$ : the gap $x_{n} - y_{n} \to 0$

yet $g(x_{n}) - g(y_{n})$ stays at 1

because squaring compresses each successive tent into a shorter interval.

SymPy makes the mechanism explicit: $(t + 1/t)^{2} - t^{2} = 2 + 1/t^{2}$

which stays bounded away from zero even as the input gap 1/t → 0.

1 of these 1 wrong options correspond to a named reasoning error — the same errors recur across subjects.

The trap it tests

Pointwise vs uniform

The mode of convergence — or of continuity — was the wrong one.

Drill statements like this

Last revised 12 Sept 2026. Found a mistake? Tell us — corrections are the fastest thing we act on.

Periodic uniform continuity

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