52 questions answered · 150 marks · DBE
Mathematics Grade 12 NSC May/June 2026 P1 Questions and Answers
Every question from the NSC Examination (May/June) (DBE, May/June 2026), with a full answer written from the Grade 12 Mathematics textbook and the official marking approach. Open a question to see its answer, or use Hide answers to practise first.
Instructions
Answer all ten questions. Show all your working, since answers alone may not get full marks. An approved non-graphical calculator may be used, and answers should be rounded to two decimal places unless stated otherwise. Diagrams are not drawn to scale.
Question 1
6 questions\(x = 3 \text{ or } x = -3\)
\(x = 1{,}27 \text{ or } x = -2{,}77\)
Multiply by \(-1\) and reverse the sign:
\[x^2 - 3x - 4 \lt 0\]\[(x - 4)(x + 1) \lt 0\]Critical values: \(x = 4\) and \(x = -1\)
The parabola \(y = x^2 - 3x - 4\) is below the \(x\)-axis between its roots.
\(-1 \lt x \lt 4\), i.e. \(x \in (-1\,;\,4)\)
Check \(x = \frac{31}{4}\): \(8 - 2\left(\frac{31}{4}\right) = -7{,}5 \lt 0\), but a square root cannot be negative, so reject it.
Check \(x = 2\): \(\sqrt{16} + 4 = 8\) ✓
\(x = 2\)
Substitute (1) into \(-2x^2 + 3xy = y^2\):
\[-2x^2 + 3x(2 - x) = (2 - x)^2\]\[-2x^2 + 6x - 3x^2 = 4 - 4x + x^2\]\[-6x^2 + 10x - 4 = 0\]\[3x^2 - 5x + 2 = 0\]\[(3x - 2)(x - 1) = 0\]\[x = \frac{2}{3} \text{ or } x = 1\]\[y = 2 - \frac{2}{3} = \frac{4}{3} \text{ or } y = 2 - 1 = 1\]\(x = \frac{2}{3}\,;\, y = \frac{4}{3}\) or \(x = 1\,;\, y = 1\)
Difference of two squares:
\[\left(\sqrt{6p}\right)^2 - \left(\sqrt[4]{128}\right)^2 = 54 + a\sqrt{2}\]\[6p - \sqrt{128} = 54 + a\sqrt{2}\]\[6p - 8\sqrt{2} = 54 + a\sqrt{2}\]Equate the rational parts and the \(\sqrt{2}\) parts:
\[6p = 54, \text{ so } p = 9\]\[a = -8\]\[a - p = -8 - 9\]\(a - p = -17\)
Question 2
6 questionsGiven the arithmetic series: \(6 + 10 + 14 + \dots\)
\(n = 150\), so the 150th term
\(S_n = 2n^2 + 4n\)
\(4n + 2\) is the general term of \(6 + 10 + 14 + \dots\), so the sum is the 100th to the 150th terms.
\[\sum_{n=100}^{150}(4n + 2) = S_{150} - S_{99}\]\[= \left[2(150)^2 + 4(150)\right] - \left[2(99)^2 + 4(99)\right]\]\[= 45\,600 - 19\,998\]\(= 25\,602\)
The first four terms of a quadratic number pattern are: \(-1\,;\,-4x\,;\,11\,;\,2x + 12\)
Terms: \(-1\,;\,-4x\,;\,11\,;\,2x + 12\)
First differences: \(-4x + 1\,;\,11 + 4x\,;\,2x + 1\)
Second differences:
\[(11 + 4x) - (-4x + 1) = 8x + 10\]\[(2x + 1) - (11 + 4x) = -2x - 10\]The second difference is constant:
\[8x + 10 = -2x - 10\]\[10x = -20\]\(x = -2\)
Terms: \(-1\,;\,8\,;\,11\,;\,8\)
First differences: \(9\,;\,3\,;\,-3\)
Second difference: \(-6\)
\[2a = -6, \text{ so } a = -3\]\[3a + b = 9:\; 3(-3) + b = 9, \text{ so } b = 18\]\[a + b + c = -1:\; -3 + 18 + c = -1, \text{ so } c = -16\]\(T_n = -3n^2 + 18n - 16\)
The first differences \(9\,;\,3\,;\,-3\,;\,\dots\) form an arithmetic sequence with \(a = 9\) and \(d = -6\).
\[\frac{n}{2}\left[2(9) + (n - 1)(-6)\right] \lt -1\,380\]\[n(18 - 6n + 6) \lt -2\,760\]\[-6n^2 + 24n + 2\,760 \lt 0\]\[n^2 - 4n - 460 \gt 0\]\[n = \frac{4 \pm \sqrt{16 + 1\,840}}{2}\]\[n \gt 23{,}54 \text{ or } n \lt -19{,}54 \text{ (not applicable)}\]\(n = 24\): from 24 first differences onwards the sum is less than \(-1\,380\)
Question 3
2 questionsTwo clocks, A and B, were observed from 09:00 to 10:00. The following observations were made:
- The time on both clocks was exactly 09:00.
- Clock A measured time correctly from 09:00 to 10:00.
- Clock B measured half the time that clock A measured in the first five minutes.
- Thereafter, for every five minutes, clock B measured \(50\%\) of the time it measured in the previous 5 minutes.
From 09:00 to 10:00 there are 12 five-minute intervals, so \(n = 12\).
Clock B measured: \(2{,}5\,;\,1{,}25\,;\,0{,}625\,;\,\dots\) minutes
This is geometric with \(a = 2{,}5 = \frac{5}{2}\) and \(r = 0{,}5 = \frac{1}{2}\).
\[S_{12} = \frac{a(1 - r^n)}{1 - r} = \frac{\frac{5}{2}\left(1 - \left(\frac{1}{2}\right)^{12}\right)}{1 - \frac{1}{2}}\]\[S_{12} = 4{,}998\ldots \text{ minutes}\]Clock B shows 09:05 (just under 5 minutes past 9)
The total time clock B can ever measure is 5 minutes, which is less than 6 minutes.
Clock B will never reach 09:06.
Question 4
9 questionsSketched below are the graphs of \(f(x) = -a^x + q\) and \(g(x) = -x^2 + 2x + 3\). B and C are the \(x\)-intercepts of \(g\). D, the turning point of \(g\), is also a point on the asymptote of \(f\). T is a point on \(f\) such that \(DT\) is parallel to the \(y\)-axis.
Diagram: \(g\) is a parabola opening downwards, cutting the \(x\)-axis at B (left of O) and C (right of O), with its turning point D in the first quadrant. A horizontal dashed line through D is the asymptote of \(f\). \(f\) is a decreasing exponential curve that lies below this asymptote: it approaches the asymptote far to the left, passes through the \(y\)-axis at the same point as \(g\), crosses \(g\) again between D and C, and cuts the \(x\)-axis to the right of C. A vertical dashed line from D down to \(f\) meets \(f\) at T.
\(D(1\,;\,4)\)
\(g\) opens downwards with its turning point at \(x = 1\), so it rises to the left of the turning point.
\(x \lt 1\), i.e. \(x \in (-\infty\,;\,1)\)
\(x = 3 \text{ or } x = -1\), so \(B(-1\,;\,0)\) and \(C(3\,;\,0)\)
The horizontal line \(y = k\) must cut \(g\) twice to the right of the \(y\)-axis.
\(g(0) = 3\) (\(y\)-intercept) and the maximum value is 4.
The line must lie above the \(y\)-intercept and below the turning point.
\(3 \lt k \lt 4\)
The asymptote of \(f\) passes through \(D(1\,;\,4)\), so it is \(y = 4\), and \(f\) lies below it.
\(y \lt 4\), i.e. \(y \in (-\infty\,;\,4)\)
\(q = 4\) (asymptote \(y = 4\))
T is directly below D:
\[T\left(1\,;\,4 - \frac{3}{2}\right) = T\left(1\,;\,\frac{5}{2}\right)\]Substitute T into \(f(x) = -a^x + 4\):
\[\frac{5}{2} = -a^1 + 4\]\(a = \frac{3}{2}\) and \(q = 4\)
For every \(x\)-value there is exactly one corresponding \(y\)-value (\(f\) is a one-to-one relation; any vertical line cuts it only once).
For the inverse swop \(x\) and \(y\):
\[x = \left(\frac{3}{2}\right)^y\]\(y = \log_{\frac{3}{2}} x\)
\(T\left(1\,;\,\frac{5}{2}\right)\) on \(f\). \(h(x) = -f(x) + 4\), so
\[T' = \left(1\,;\,-\frac{5}{2} + 4\right) = \left(1\,;\,\frac{3}{2}\right)\]Reflecting in \(y = x\) gives \(T''\left(\frac{3}{2}\,;\,1\right)\).
Sketch: an increasing logarithmic curve \(y = \log_{\frac{3}{2}} x\) that lies to the right of the \(y\)-axis, with the \(y\)-axis (\(x = 0\)) as its vertical asymptote, cutting the \(x\)-axis at \((1\,;\,0)\) and passing through \(T''\left(\frac{3}{2}\,;\,1\right)\). There is no \(y\)-intercept.
Question 5
7 questionsThe graphs of \(f(x) = \frac{-6}{x + 2} - 1\) and \(g(x) = \frac{1}{2}x + 4\) are drawn below. A and B are the \(x\)- and \(y\)-intercepts of \(f\) respectively. A is also a point of intersection of \(f\) and \(g\).
Diagram: \(f\) is a hyperbola with a vertical dashed asymptote to the left of the \(y\)-axis and a horizontal dashed asymptote just below the \(x\)-axis. The left branch of \(f\) rises steeply as it nears the vertical asymptote and cuts the \(x\)-axis at A. The right branch lies below the \(x\)-axis, cutting the \(y\)-axis at B and flattening out towards the horizontal asymptote. \(g\) is a straight line with a positive gradient through A, crossing the left branch of \(f\) again and cutting the positive \(y\)-axis.
The vertical asymptote is \(x = -2\).
\(x \in \mathbb{R},\; x \neq -2\)
Let \(f(x) = 0\):
\[\frac{-6}{x + 2} - 1 = 0\]\[-6 = x + 2\]\[x = -8\]\(A(-8\,;\,0)\)
The axes of symmetry pass through the point where the asymptotes meet, \((-2\,;\,-1)\).
With gradient 1:
\[y = x + c\]\[-1 = -2 + c, \text{ so } c = 1\]\(h(x) = x + 1\)
\((6\,;\,7)\)
\(f(x) + g(x) \geq 0\) means \(f(x) \geq -g(x)\), where \(-g(x) = -\frac{1}{2}x - 4\) (the reflection of \(g\) in the \(x\)-axis).
\(-g\) also passes through \(A(-8\,;\,0)\) and through \((0\,;\,-4) = B\), so \(f\) and \(-g\) meet at \(x = -8\) and \(x = 0\).
\(f\) lies on or above \(-g\) from A up to the vertical asymptote, and from B onwards.
\(-8 \leq x \lt -2 \text{ or } x \geq 0\)
Take the right angle at D, with the third vertex on \(g\) directly above or below D.
Base \(AD = k + 8\) and height \(= g(k) = \frac{1}{2}k + 4 = \frac{1}{2}(k + 8)\)
\[\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}\]\[64 = \frac{1}{2}(k + 8)\left(\frac{1}{2}k + 4\right)\]\[\left(\frac{1}{2}k + 4\right)^2 = 64\]\[\frac{1}{2}k + 4 = \pm 8\]\(k = 8 \text{ or } k = -24\)
Question 6
4 questions\(A = P(1 + i)^n\), with \(n = 2 \times 2 = 4\) half-years
\[85\,000 = 75\,000\left(1 + \frac{r}{200}\right)^4\]\[\frac{17}{15} = \left(1 + \frac{r}{200}\right)^4\]\[1 + \frac{r}{200} = \sqrt[4]{\frac{17}{15}}\]\[\frac{r}{200} = 0{,}03178\ldots\]\(r = 6{,}36\%\)
William wants to buy a house. The bank requires William to pay a deposit of \(10\%\) of the purchase price of the house. The bank will grant William a loan to cover the outstanding amount, subject to the following conditions:
- The loan is to be repaid over 20 years.
- Interest on the loan is fixed at \(11{,}5\%\) p.a., compounded monthly.
- An instalment of R10 665 is payable at the end of each month after the loan has been granted, starting one month after the loan has been granted.
Deposit = R111 118,44
About 16,80 years (16 years and 10 months)
Instalments in 16 years:
\[11\,000 \times 16 \times 12 = R\,2\,112\,000\]Add the deposit:
\[2\,112\,000 + 120\,000\]R2 232 000
Question 7
6 questions\(f'(x) = 2x + 2\)
\(\sqrt[3]{27x^4} = 3x^{\frac{4}{3}}\) and \(\frac{2}{x^3} = 2x^{-3}\)
\[\frac{d}{dx}\left(3x^{\frac{4}{3}} + 2x^{-3}\right)\]\[= 3 \times \frac{4}{3}x^{\frac{1}{3}} + 2 \times (-3)x^{-4}\]\(= 4x^{\frac{1}{3}} - 6x^{-4}\) (or \(4\sqrt[3]{x} - \frac{6}{x^4}\))
The graph of \(k(x) = 2x^3 + bx^2 + cx + d\) has turning points at \(x = -2\) and \(x = 0\). Determine the values of \(x\) for which:
The derivative is zero at the turning points.
\(x = -2 \text{ or } x = 0\)
The point of inflection is halfway between the turning points:
\[x = \frac{-2 + 0}{2} = -1\]\(k\) has a positive leading coefficient, so it is concave down to the left of the point of inflection.
\(x \lt -1\)
\(k'(x) \gt 0\) (\(k\) increasing) when \(x \lt -2\) or \(x \gt 0\).
Replace \(x\) by \(x - 3\):
\[x - 3 \lt -2 \text{ or } x - 3 \gt 0\]\(x \lt 1 \text{ or } x \gt 3\)
Question 8
4 questionsThe graph of \(h(x) = -x^3 + px^2 + qx + r\) is drawn below. \(C(2\,;\,0)\) and \(D(5\,;\,0)\) are the \(x\)-intercepts and B is the \(y\)-intercept of \(h\). A and C are the turning points of \(h\).
Diagram: a cubic graph that comes down from the top left, cuts the positive \(y\)-axis at B, decreases to touch the \(x\)-axis at \(C(2\,;\,0)\) (a local minimum), rises to a local maximum A above the \(x\)-axis, then falls and cuts the \(x\)-axis at \(D(5\,;\,0)\) and continues downwards to the right.
\(C(2\,;\,0)\) is a turning point on the \(x\)-axis, so \((x - 2)\) is a repeated factor; D gives the factor \((x - 5)\).
\[h(x) = -1(x - 2)^2(x - 5)\]\[h(x) = -(x^2 - 4x + 4)(x - 5)\]\[h(x) = -(x^3 - 9x^2 + 24x - 20)\]\[h(x) = -x^3 + 9x^2 - 24x + 20\]\(\therefore p = 9,\; q = -24,\; r = 20\)
B is the \(y\)-intercept: \(h(0) = r = 20\), so \(B(0\,;\,20)\).
\(OB = 20\) units
\((3\,;\,2)\)
\(x = 1\) (between B and C) or \(x = 5\) (not between B and C)
\[h(1) = -1 + 9 - 24 + 20 = 4\]So the point of contact is \((1\,;\,4)\).
\[y - 4 = -9(x - 1)\]\(y = -9x + 13\)
Question 9
3 questionsThe oxygen content in the water, \(t\) days after organic waste was dumped into a lake, can be represented by \(C(t) = \frac{1}{1\,200}\left(t^3 - 15t^2 + 1\,200\right)\) for \(0 \leq t \leq 15\).
Let \(t = 0\):
\[C(0) = \frac{1}{1\,200}(0 - 0 + 1\,200)\]\(C(0) = 1\)
So \(t = 10\) gives a minimum.
After 10 days
The rate of change is \(C'(t) = \frac{1}{400}t^2 - \frac{1}{40}t\).
Its turning point:
\[C''(t) = \frac{1}{200}t - \frac{1}{40} = 0\]\[t - 5 = 0\]\(t = 5\) days
Question 10
5 questionsDr Calculus's wardrobe has:
| TIES | TROUSERS | SHIRTS |
|---|---|---|
| 6 blue | 1 blue | 3 red |
| 4 red | 1 black | 2 pink |
| 5 green | 1 brown | 4 white |
| 1 grey | 3 blue | |
| 4 green |
Totals: 15 ties, 4 trousers, 16 shirts.
Ties: 15, trousers: 4, shirts: 16
\(15 \times 4 \times 16 = 960\) outfits
Only a red tie with a pink shirt is not allowed (there are no red or pink trousers).
Outfits with a red tie and pink shirt:
\[4 \times 4 \times 2 = 32\]\[960 - 32\]\(= 928\) outfits
Blue ties: 6; blue or black trousers: 2; blue shirts: 3
\[\text{Number of such outfits} = 6 \times 2 \times 3 = 36\]\[P = \frac{36}{960}\]\(= \frac{3}{80} = 0{,}0375\)
A survey was conducted among 54 people to determine which of the three types of tea (A, B or C) they drank. The results are summarised below.
- Some people did not drink any tea at all.
- \(P(\text{drinking types A and B and C}) = \frac{7}{54}\)
- 9 people drank types B and C.
- 2 people drank types A and B but not type C.
- 18 people drank type B.
- No-one drank types A and C but not type B.
- Drinking type A is independent from drinking type B.
- \(P(\text{not drinking type (A or B)}) = 2P(\text{not drinking types (A or B or C)})\)
A and B are independent:
\[P(A \text{ and } B) = P(A) \times P(B)\]\[\frac{9}{54} = \frac{x}{54} \times \frac{18}{54}\]\[x = \frac{9 \times 54}{18}\]27 people drank type A
A and B and C \(= 7\) (from \(\frac{7}{54} \times 54\))
Only B and C \(= 9 - 7 = 2\)
Only A and B \(= 2\)
Only A and C \(= 0\)
Only B \(= 18 - 7 - 2 - 2 = 7\)
Only A \(= 27 - 7 - 2 - 0 = 18\)
\[n(A \text{ or } B) = 27 + 18 - 9 = 36\]\[\text{not } (A \text{ or } B) = 54 - 36 = 18\]Not (A or B) = only C + none, and it is twice none:
\[18 = 2 \times \text{none}, \text{ so none} = 9 \text{ and only C} = 9\]Venn diagram: only A = 18, only B = 7, only C = 9, A and B only = 2, B and C only = 2, A and C only = 0, all three = 7, outside all circles = 9 (total 54).