Why isn't the answer to TMUA-P2Y-PRF-012 shown upfront?

The options are lettered without marking the correct one so you attempt TMUA-P2Y-PRF-012 first — TMUA questions reward spotting the trap before you check the worked solution below.

Do I need a calculator for TMUA-P2Y-PRF-012?

No — calculators are banned in the TMUA. TMUA-P2Y-PRF-012 is set for a no-calculator target of about 200 seconds; the worked solution shows the fastest by-hand route.

Which TMUA specification point is TMUA-P2Y-PRF-012?

TMUA-P2Y-PRF-012 is Paper P2 · Disproof by counterexample (specification code PRF4), level 2. Drill more of the same topic from this page.

TMUA-P2Y-PRF-012 — Paper P2 · Disproof by counterexample · Level 2

P2PRF4Level 2~200s target

Spec: PRF4 · No calculator · ~200s

Disprove the claim: for all real numbers

x

and

y

(x + y)^2 = x^2 + y^2

Options

$x = 0$ , $y = 0$ : both sides are $0$ , so the claim holds.
$x = 1$ , $y = 1$ : left side $(2)^2 = 4$ but right side $1 + 1 = 2$ , so the equality fails.
The claim is true because squaring distributes over addition.
$x = 1$ , $y = 0$ : both sides equal $1$ , so the claim holds.

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

Hint (check trope)

No written hint yet — check the trope: single-counterexample, square-number-trap. Name the trap first, work the stem by hand, then open the solution below.

Show solution

Need one pair where the sides differ:

(1 + 1)^2 = 4 \ne 2 = 1^2 + 1^2

; the missing cross term is

2xy

. Chain: pick numbers

\Rightarrow

evaluate both sides

\Rightarrow

exhibit inequality. Trap: zero or single-zero pairs satisfy the equation since

2xy = 0

and therefore confirm rather than refute; distribution of squaring over addition is the misconception under test.

Keep practising

Same topic · same level
P2PRF4Level 2
TMUA-P2-PRF4-003
Disprove the claim: “For every real number x, |x| > x.”
Disproof by counterexample
Open TMUA-P2-PRF4-003 →
Same trap: square-number-trap
P1MM2Level 2
TMUA-P1-MM2-011
The sequence 2, 6, 12, 20, 30, continues in the same pattern. What is its 8th term?
Sequences and series
Open TMUA-P1-MM2-011 →
Step up a level
P2PRF4Level 3
TMUA-P2-PRF4-007
An olympiad-style training sheet invites students to test the sweeping divisibility conjecture that for all integers m a…
Disproof by counterexample
Open TMUA-P2-PRF4-007 →

Browse more

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Options

x = 0

y = 0

: both sides are

0

, so the claim holds.

x = 1

y = 1

: left side

(2)^2 = 4

but right side

1 + 1 = 2

, so the equality fails.

The claim is true because squaring distributes over addition.

x = 1

y = 0

: both sides equal

1

, so the claim holds.

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

Keep practising

Same topic · same level

P2PRF4Level 2

TMUA-P2-PRF4-003

Disprove the claim: “For every real number x, |x| > x.”

Disproof by counterexample

Open TMUA-P2-PRF4-003 →

Same trap: square-number-trap

P1MM2Level 2

TMUA-P1-MM2-011

The sequence 2, 6, 12, 20, 30, continues in the same pattern. What is its 8th term?

Sequences and series

Open TMUA-P1-MM2-011 →

Step up a level

P2PRF4Level 3

TMUA-P2-PRF4-007

An olympiad-style training sheet invites students to test the sweeping divisibility conjecture that for all integers m a…

Disproof by counterexample

Open TMUA-P2-PRF4-007 →

TMUA-P2Y-PRF-012 — Paper P2 · Disproof by counterexample · Level 2

Options

Keep practising

Master Disproof by counterexample

TMUA-P2Y-PRF-012 — Paper P2 · Disproof by counterexample · Level 2

Options

Keep practising

Master Disproof by counterexample