Part CCSIR NET December 2025a-strictly-positive-integrand-over-a-positive-length-interval-can-approach-0-as-x-shrinks-but-never-equal-it-for-x-greater-than-0

A strictly positive integrand over a positive length interval can approach 0 as x shrinks but never equal it for x greater than 0

Let $f : (0,\infty) \to (0,\infty)$ be the function defined by $f(x) = \int_{0}$ ˣ $\sqrt{t}/(1+t^{2})$ dt, where $\sqrt{t}$ denotes the positive square root for t>0. Which of the following statements are true?

A.f is a uniformly continuous function.✓ Correct
B.f is a bounded function.✓ Correct
C.There exists $x\in(0,\infty)$ such that f(x)=0.
D.The derivative of f is continuous.✓ Correct

You have the answer. Trap Analysis is why the other three were written.

Not a worked solution repeated four times — the specific reasoning error each wrong option was built to reward, and why this one tests boundary and endpoint.

See pricing

50 are analysed free — try those first.

The trap it tests

Boundary and endpoint

The statement turns at the edge of the interval, the domain, or the parameter range.

Drill statements like this

Related counterexample: If ∫₀^∞ f converges then f(x) → 0

A strictly positive integrand over a positive length interval can approach 0 as x shrinks but never equal it for x greater than 0

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