TMUA-P2-PRF1-012 — Paper P2 · Direct proof · Level 3

Options

1. A
   Squaring preserves order for all real numbers, so $x < y$ gives $x^2 < y^2$ immediately.
2. B
   Take $x = 2$, $y = 3$: $4 < 9$, so the claim holds generally.
3. C
   Then $y^2-x^2 = (y-x)(y+x)$. Since $y > x$ we have $y-x > 0$, and since $y > x > 0$ we have $y+x > 0$. Hence $y^2-x^2 > 0$, i.e. $x^2 < y^2$.
4. D
   Since $y-x > 0$, adding $y^2-x^2$ to both sides gives the result.

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

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