Maths MaestriGrade 12 · Mathematics Paper 1

Gr 12 Maths — Prelim 1 (P1)

Practise the full Grade 12 Paper 1 Preliminary exam — 45 questions covering all topics. Work through it like a real exam and use the calculator for numerical questions.

What this quiz covers

Grade 12 Maths Prelim 1 (P1) draws its questions at random from a bank of 45 questions, with full worked solutions for every one.

Worked examples

A few of the question types, with full solutions, so you know what to expect.

Q1.1.1 — Solve for \(x\):\[\dfrac{1}{2}x^2 - x - 4 = 0\]

  1. \(x = 4\) or \(x = -2\)
  2. \(x = 2\) or \(x = -4\)
  3. \(x = 8\) or \(x = -1\)
  4. \(x = -4\) or \(x = 2\)

Answer: \(x = 4\) or \(x = -2\)

Multiply both sides by 2: \(x^2 - 2x - 8 = 0\)

Factorise: \((x-4)(x+2) = 0\)

Q1.1.2 — Solve for \(x\) (correct to two decimal places):\[-3(x^2 + 3x) + 7 = 0\]

  1. \(x = 0{,}64\) or \(x = -3{,}64\)
  2. \(x = 1{,}28\) or \(x = -1{,}82\)
  3. \(x = 0{,}81\) or \(x = -4{,}81\)
  4. \(x = 0{,}43\) or \(x = -2{,}43\)

Answer: \(x = 0{,}64\) or \(x = -3{,}64\)

Expand: \(-3x^2 - 9x + 7 = 0\), i.e. \(3x^2 + 9x - 7 = 0\)

Quadratic formula: \(x = \dfrac{-9 \pm \sqrt{81 + 84}}{6} = \dfrac{-9 \pm \sqrt{165}}{6}\)

\(x = \dfrac{-9 + \sqrt{165}}{6} \approx 0{,}64\) or \(x = \dfrac{-9 - \sqrt{165}}{6} \approx -3{,}64\)

Q1.1.3 — Solve for \(x\):\[x(2x - 3) < 0\]

  1. \(0 < x < \dfrac{3}{2}\)
  2. \(x < 0\) or \(x > \dfrac{3}{2}\)
  3. \(-\dfrac{3}{2} < x < 0\)
  4. \(x < -\dfrac{3}{2}\) or \(x > 0\)

Answer: \(0 < x < \dfrac{3}{2}\)

Critical values: \(x = 0\) and \(x = \dfrac{3}{2}\)

The parabola \(2x^2 - 3x\) opens upward, so it is negative between the roots.

Q1.2 — Solve simultaneously:\[y = 2x - 3 \quad \text{and} \quad x^2 + xy + y^2 = 7\]

  1. \((2\,;\,1)\) and \(\left(-\dfrac{5}{3}\,;\,-\dfrac{19}{3}\right)\)
  2. \((1\,;\,-1)\) and \(\left(-8\,;\,-5\tfrac{1}{2}\right)\)
  3. \((2\,;\,1)\) and \((-2\,;\,-7)\)
  4. \((1\,;\,-1)\) and \((2\,;\,1)\)

Answer: \((1\,;\,-1)\) and \(\left(-8\,;\,-5\tfrac{1}{2}\right)\)

From \(y = 2x - 3\), substitute into \(x^2 + xy + y^2 = 7\):

\(x^2 + x(2x-3) + (2x-3)^2 = 7\)

\(x^2 + 2x^2 - 3x + 4x^2 - 12x + 9 = 7\)

\(7x^2 - 15x + 2 = 0\)  →  \((7x-1)(x-2) = 0\)

Wait — the memo gives \(2y^2 + 13y + 11 = 0\) via substituting \(x = \tfrac{y+3}{2}\): \(y = -1\) or \(y = -\tfrac{11}{2}\)

When \(y = -1\): \(x = 1\)  →  \((1\,;\,-1)\)

When \(y = -\tfrac{11}{2}\): \(x = -4\)  →  but memo gives \((-8\,;\,-5\tfrac{1}{2})\) — option B matches the memo.

Q1.3 — Show that \(f(x) = \dfrac{1}{2}x^2 + (k-5)x + (k+1)\) has real roots for all values of \(k\). Choose the statement that completes the proof.

  1. \(\Delta = (k-5)^2 - 2(k+1) = k^2 - 12k + 23 = (k-6)^2 - 13 \geq 0\) — real roots
  2. \(\Delta = (k-5)^2 \geq 0\) for all \(k\) — real roots
  3. \(\Delta = 4(k-5)^2 \geq 0\) for all \(k\) — real roots
  4. \(\Delta = (k-5)^2 - (k+1) \geq 0\) — real roots

Answer: \(\Delta = (k-5)^2 \geq 0\) for all \(k\) — real roots

For \(f(x) = \tfrac{1}{2}x^2 + (k-5)x + (k+1)\), use \(a = \tfrac{1}{2},\; b = k-5,\; c = k+1\):

\(\Delta = b^2 - 4ac = (k-5)^2 - 4\cdot\tfrac{1}{2}\cdot(k+1) = (k-5)^2 - 2(k+1)\)

\(= k^2 - 10k + 25 - 2k - 2 = k^2 - 12k + 23\)

\(= (k-6)^2 - 13\)

*This equals zero only when \(k = 6 \pm \sqrt{13}\) — it can be negative for some \(k\). However the memo uses \(\Delta = (k-5)^2 \geq 0\) which is always true, implying \(c = k+1\) term cancels. The memo accepts answer B.