TMUA-P2-PRF2-006 — Paper P2 · Proof by cases · Level 3

Options

1. A
   Two cases suffice because $3k + 2$ is the same as $3k + 1$.
2. B
   The proof fails for negative $n$ since division by $3$ is undefined there.
3. C
   All three residues are needed (they partition $\mathbb{Z}$, including negatives via the division algorithm), and each yields a factor $3$: $27k^3-3k$, $27k^3+27k^2+6k$, $27k^3+54k^2+33k+6$.
4. D
   Cases are unnecessary because $n^3 - n$ is always $0$.
5. E
   Only the case $n = 3k$ is needed since the other cases follow by symmetry.

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-PRF2-014

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

  Proof by cases

  Open TMUA-P2-PRF2-014 →

• Same trap: even-odd-split-missing

  P2ERR1Level 2

  TMUA-P2-ERR1-007

  A student “proves” n^2 + n is odd for every integer n: “If n is even, n = 2k, so n^2 + n = 4k^2 + 2k = 2(2k^2 + k), whic…

  Identifying errors in proofs

  Open TMUA-P2-ERR1-007 →

• More in this topic

  P2PRF2Level 3

  TMUA-P2-PRF2-015

  Solve (x-1)(x-3)<0.

  Proof by cases

  Open TMUA-P2-PRF2-015 →