TMUA-P2-PRF2-017 — Paper P2 · Proof by cases · Level 3

Options

1. A
   Assume $a,b \ge 0$ only.
2. B
   Square both sides without any sign discussion.
3. C
   Exhaust signs: if $a,b \ge 0$ or $a,b \le 0$ then equality up to sign holds; if they have opposite signs then $|a+b|=||a|-|b|| \le |a|+|b|$. All sign cases are covered.
4. D
   Assume $a+b \ge 0$ only.

Letters only — the answer key is hidden. Work it out, then check the solution below.

Keep practising

• Same topic · same level

  P2PRF2Level 3

  TMUA-P2-PRF2-006

  A lecturer proves that n^3 - n is divisible by 3 for every integer n by splitting into remainders modulo 3, writing n =…

  Proof by cases

  Open TMUA-P2-PRF2-006 →

• Same trap: case-split-modulus

  P2PRF2Level 1

  TMUA-P2-PRF2-007

  To solve |x-3|=5 by cases, which split is complete and correct?

  Proof by cases

  Open TMUA-P2-PRF2-007 →

• More in this topic

  P2PRF2Level 3

  TMUA-P2-PRF2-014

  Solve |x-1|+|x-3|=4.

  Proof by cases

  Open TMUA-P2-PRF2-014 →