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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 xRx\in\mathbb{R}. Which of the following statements are true?

  1. A.F is non-decreasing on R\mathbb{R}
  2. B.lim(x)F(x)=1\lim(x\to\infty) F(x) = 1
  3. C.F is left-continuous on R\mathbb{R}
  4. D.F is right-continuous on R\mathbb{R}

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.

See pricing

50 are analysed free — try those first.

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 DifferentiationContinuity, uniform continuity, Lipschitz

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