Skip to content
Part BCSIR NET December 2023

Sequences: convergence, monotone, Bolzano–Weierstrass, Cauchy

Consider the following subset of R:U=\mathbb{R}: U = {x∈R:x2−9x+18≤0,x2−7x+12≤0x \in \mathbb{R} : x^{2} - 9x + 18 \le 0, x^{2} - 7x + 12 \le 0}. Which one of the following statements is true?

  1. A.inf U = 5.
  2. B.inf U = 4.
  3. C.inf U = 3.✓ Correct
  4. D.inf U = 2.

The idea, in plain words

Two conditions have to hold at once, so solve each one separately and keep only the overlap. Each gives a stretch of the line, and the answer is the left-hand end of the part they share.

Solution

x2−9x+18≤0⇔x∈[3,6];x2−7x+12≤0⇔x∈[3,4]x^{2} - 9x + 18 \le 0 \Leftrightarrow x \in [3, 6]; x^{2} - 7x + 12 \le 0 \Leftrightarrow x \in [3, 4].

So U = [3, 4] and inf U = 3.

Why each option is right or wrong

Checked against 2 symbolic computations
  • False.

    5 lies in [3, 6] but not in [3, 4]

    so it is not even an element of U.

    Choosing 5 almost always means solving x2−9x+18≤0x^{2} - 9x + 18 \le 0

    reading off its interval, and never imposing the second constraint.

  • False — 4 is the supremum of U, not the infimum.

    SymPy gives the roots as {3, 6} and {3, 4}

    so U = [3, 4]: 3 is the left endpoint and 4 the right.

    This option is what correct algebra plus a misread of which endpoint was asked for produces

    which is why it is the most-picked wrong answer.

  • C.Correct

    True. x2−9x+18≤0x^{2} - 9x + 18 \le 0 has roots 3 and 6

    so it holds exactly on [3,6].x2−7x+12≤0[3, 6]. x^{2} - 7x + 12 \le 0 has roots 3 and 4

    so it holds on [3, 4].

    Both factorisations verified by SymPy.

    The intersection is [3, 4]

    so inf U = 3.

  • False.

    2 satisfies neither inequality.

    This is the factorisation-slip option: reading x2−7x+12x^{2} - 7x + 12 as (x − 2)(x − 5) — which is actually x2−7x+10x^{2} - 7x + 10 — plants a spurious root at 2.

3 of these 3 wrong options correspond to a named reasoning error — the same errors recur across subjects.

Related counterexample: aₙ^{1/n} → L ⇒ aₙ₊₁/aₙ → L

More on this topic

The chapter behind this: Sequences in ℝ — the implication map — free to read

From The Real Line › Sequences: convergence, monotone, Bolzano–Weierstrass, Cauchy

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

ShareWhatsAppTelegram