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Part CCSIR NET December 2025f2-is-continuous-everywhere-so-it-cannot-mask-f1-s-jump-at-x-equals-1-only-right-continuity-survives

F2 is continuous everywhere so it cannot mask f1 s jump at x equals 1 only right continuity survives

Consider the following real-valued functions F1F_{1} and F2F_{2} defined on R\mathbb{R}, given by F1(x)=1F_{1}(x)=1 if x≥1, 0 otherwise, and F2(x)=∫0F_{2}(x)=\int_{0}ˣexp(−t)dt if x≥0, 0 otherwise. Define another function F:R→[0,1]F:\mathbb{R}\to[0,1] by F(x)=(2/3)F1(x)+(1/3)F2(x)F(x)=(2/3)F_{1}(x)+(1/3)F_{2}(x) for all x∈Rx\in\mathbb{R}. Which of the following statements are true?

  1. A.F is non-decreasing on R\mathbb{R}✓ Correct
  2. B.lim⁡(x→∞)F(x)=1\lim(x\to\infty) F(x) = 1✓ Correct
  3. C.F is left-continuous on R\mathbb{R}
  4. D.F is right-continuous on R\mathbb{R}✓ 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: Continuous on a bounded interval ⇒ bounded

More on this topic

The chapter behind this: Continuity vs uniform continuity vs Lipschitz — free to read

From Continuity and Differentiation › Continuity, uniform continuity, Lipschitz

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