50 questions answered · 150 marks · DBE
Mathematics Grade 12 NSC May/June 2026 P2 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
Eleven questions, all compulsory. Show all calculations, diagrams and graphs you use, since answers alone may not earn full marks. An approved non-programmable calculator may be used unless stated otherwise; round to two decimal places unless told otherwise. Diagrams are not drawn to scale, and reasons must be given for statements in Questions 9, 10 and 11.
Question 1
4 questionsThere are 12 players in a cricket team. The number of balls each player faced in a tournament and the number of runs each scored are given below.
| Balls faced (\(x\)) | 62 | 111 | 22 | 48 | 124 | 180 | 89 | 96 | 95 | 260 | 98 | 115 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|
| Runs scored (\(y\)) | 124 | 209 | 41 | 88 | 208 | 320 | 148 | 158 | 350 | 388 | 145 | 165 |
A scatter plot of the data is also given (number of balls faced on the horizontal axis from 0 to 300, number of runs scored on the vertical axis from 0 to 400). The points rise steadily from left to right, except for the point \((95\,;\,350)\), which sits far above the rest.
Enter the \(x\) and \(y\) values into the calculator (statistics, linear regression mode).
\[a = 39{,}47\]\[b = 1{,}44\]\(\hat{y} = 39{,}47 + 1{,}44x\)
About 255 runs (\(255{,}28\) if the unrounded calculator values are used)
Mean of \(y\):
\[\bar{y} = \frac{2344}{12} = 195{,}33\]Standard deviation of \(y\):
\[\sigma_y = 101{,}84\]One standard deviation above the mean:
\[195{,}33 + 101{,}84 = 297{,}17\]Runs above \(297{,}17\): 320, 350 and 388
3 players
This player scored far more runs than the other players who faced a similar number of balls (about 90 to 100 balls gave about 145 to 160 runs).
Question 2
5 questions2.1 Marks (as a %) of Grade 12 learners at a school in the June 2025 Mathematics examination:
| Class interval (%) | Frequency |
|---|---|
| \(20 \lt x \leq 30\) | 8 |
| \(30 \lt x \leq 40\) | 14 |
| \(40 \lt x \leq 50\) | 16 |
| \(50 \lt x \leq 60\) | 21 |
| \(60 \lt x \leq 70\) | 16 |
| \(70 \lt x \leq 80\) | 9 |
| \(80 \lt x \leq 90\) | 6 |
| \(90 \lt x \leq 100\) | 2 |
2.2 A box and whisker diagram shows the marks (as a %) of learners at another school in the same examination. The five values labelled from left to right are: minimum \((x - 2)\%\), lower quartile \(\left(\frac{3x - 3}{2}\right)\%\), median \((2x + 5)\%\), upper quartile \((3x)\%\) and maximum \(91\%\).
Add each frequency to the running total:
| Class interval (%) | Frequency | Cumulative frequency |
|---|---|---|
| \(20 \lt x \leq 30\) | 8 | 8 |
| \(30 \lt x \leq 40\) | 14 | 22 |
| \(40 \lt x \leq 50\) | 16 | 38 |
| \(50 \lt x \leq 60\) | 21 | 59 |
| \(60 \lt x \leq 70\) | 16 | 75 |
| \(70 \lt x \leq 80\) | 9 | 84 |
| \(80 \lt x \leq 90\) | 6 | 90 |
| \(90 \lt x \leq 100\) | 2 | 92 |
Cumulative frequencies: 8; 22; 38; 59; 75; 84; 90; 92
Ground the curve at \((20\,;\,0)\), the lower boundary of the first class.
Plot each cumulative frequency at the upper boundary of its class: \((30\,;\,8)\), \((40\,;\,22)\), \((50\,;\,38)\), \((60\,;\,59)\), \((70\,;\,75)\), \((80\,;\,84)\), \((90\,;\,90)\), \((100\,;\,92)\).
Join the points with a smooth, increasing S-shaped curve.
Ogive from \((20\,;\,0)\) rising through the plotted points to \((100\,;\,92)\).
\(n = 92\), so the median is at a cumulative frequency of:
\[\frac{92}{2} = 46\]Read across from 46 to the curve and down to the \(x\)-axis.
\(\text{Median} \approx 54\%\)
\(27\%\)
Question 3
5 questionsAS intersects the \(y\)-axis at \(C(0\,;\,6)\). \(B(-4\,;\,8)\) is the midpoint of AC. F is the point \((0\,;\,-6)\). LF is drawn parallel to AS and cuts the \(x\)-axis at G. E is a point on LF such that the equation of AE is \(y = -\frac{7}{2}x - 18\). AE cuts the \(x\)-axis at T. SF is drawn.
Diagram: A lies in the second quadrant above B, with line AS sloping down to the right through B and C to S in the first quadrant. Line LF runs parallel to AS below it, from L (upper left, beyond G) through G on the negative \(x\)-axis and E (third quadrant) to F on the negative \(y\)-axis. Line AE runs steeply down from A through T on the negative \(x\)-axis (between G and O) to E. Segment SF joins S to F.
B is the midpoint of AC:
\[\frac{x_A + 0}{2} = -4 \quad\therefore x_A = -8\]\[\frac{y_A + 6}{2} = 8 \quad\therefore y_A = 10\]\(A(-8\,;\,10)\)
(\(LF \parallel AS\))
\(F(0\,;\,-6)\) is the \(y\)-intercept, so \(c = -6\)
\(y = -\frac{1}{2}x - 6\)
E lies on AE and LF, so equate:
\[-\frac{1}{2}x - 6 = -\frac{7}{2}x - 18\]\[-x - 12 = -7x - 36\]\[6x = -24\]\[x = -4\]\[y = -\frac{1}{2}(-4) - 6 = -4\]\(E(-4\,;\,-4)\)
so \(C\hat{S}F = 90^\circ\)
\(\therefore\) CF is a diameter (chord subtends a right angle at the circle)
Centre = midpoint of CF = \(O(0\,;\,0)\) and \(r = 6\)
\(x^2 + y^2 = 36\)
so the inclination of AE is \(105{,}95^\circ\)
\(\therefore G\hat{T}E = 105{,}95^\circ\) (vertically opposite the inclination angle at T)
\[\tan(L\hat{G}X) = m_{LF} = -\frac{1}{2}\]so the inclination of LF is \(L\hat{G}T = 153{,}43^\circ\)
In \(\triangle GTE\): \(L\hat{G}T = G\hat{T}E + G\hat{E}T\) (ext \(\angle\) of \(\triangle\))
\[G\hat{E}T = 153{,}43^\circ - 105{,}95^\circ\]\(A\hat{E}G = 47{,}49^\circ\)
Question 4
7 questionsThe circle with centre M has equation \((x - 2)^2 + (y - 4)^2 = 25\). Diameter TMR and tangent SR are drawn. The equation of line ST is \(y = 2x + 5\). ST cuts the circle again at V. VR is drawn.
Diagram: T is on the circle at the lower left, R at the upper right end of the diameter through M. Line ST rises steeply from T through V (top of the circle) to S, where it meets the tangent at R.
Substitute \(y = 2x + 5\) into the circle:
\[(x - 2)^2 + (2x + 5 - 4)^2 = 25\]\[x^2 - 4x + 4 + 4x^2 + 4x + 1 - 25 = 0\]\[5x^2 - 20 = 0\]\[x^2 - 4 = 0\]\[(x - 2)(x + 2) = 0\]\[x = 2 \text{ or } x = -2\]At T, \(x = -2\):
\[y = 2(-2) + 5 = 1\]\(\therefore T(-2\,;\,1)\)
From 4.2 the other intersection has \(x = 2\):
\[y = 2(2) + 5 = 9\]\(V(2\,;\,9)\)
\(m_{TR} = \frac{3}{4}\)
M is the midpoint of TR:
\[x_R = 2(2) - (-2) = 6\]\[y_R = 2(4) - 1 = 7\]so \(R(6\,;\,7)\)
\[m_{SR} = -\frac{4}{3}\](tangent \(\perp\) radius)
\[y - 7 = -\frac{4}{3}(x - 6)\]\[y = -\frac{4}{3}x + 8 + 7\]\(y = -\frac{4}{3}x + 15\)
Centre N = midpoint of AC:
\[N = \left(\frac{-3 + 2}{2}\,;\,\frac{-1 + 3}{2}\right) = \left(-\frac{1}{2}\,;\,1\right)\]\[AC = \sqrt{(2 - (-3))^2 + (3 - (-1))^2} = \sqrt{41}\]so \(r = \frac{\sqrt{41}}{2}\) and \(r^2 = \frac{41}{4}\)
\[\left(x + \frac{1}{2}\right)^2 + (y - 1)^2 = \frac{41}{4}\]\[x^2 + x + \frac{1}{4} + y^2 - 2y + 1 = \frac{41}{4}\]\[x^2 + x + y^2 - 2y - 9 = 0\]At the intersection points both equations hold, so equate:
\[x^2 + x + y^2 - 2y - 9 = x^2 - 4x + y^2 - 8y - 5\]\[5x + 6y - 4 = 0\]\(6y + 5x - 4 = 0\), i.e. \(y = -\frac{5}{6}x + \frac{2}{3}\)
The radius becomes smaller (the circle shrinks; the centre stays at M).
Question 5
4 questions\(\sin(A + B) = \sin A\cos B + \cos A\sin B\)
Reference angle \(= 60^\circ\); \(\cos\) is positive in quadrants 1 and 4
\(x = 60^\circ\) or \(x = 300^\circ\)
LHS = RHS
(\(\cos 10^\circ \gt 0\))
LHS = RHS
Question 6
3 questions\(\cos 330^\circ = \cos 30^\circ\), \(\cos 365^\circ = \cos 5^\circ\), \(\sin 175^\circ = \sin 5^\circ\), \(\sin 210^\circ = -\sin 30^\circ\)
\[\text{Expression} = \cos 30^\circ\cos 5^\circ + \sin 5^\circ(-\sin 30^\circ)\]\[= \cos 30^\circ\cos 5^\circ - \sin 30^\circ\sin 5^\circ\]\[= \cos(30^\circ + 5^\circ)\]\(= \cos 35^\circ\)
\(\cos 35^\circ = p\), so draw a right triangle with hypotenuse 1 and adjacent side \(p\); the third side is \(\sqrt{1 - p^2}\), so:
\[\sin 35^\circ = \sqrt{1 - p^2}\]\[\sin 70^\circ = \sin 2(35^\circ) = 2\sin 35^\circ\cos 35^\circ\]\(\sin 70^\circ = 2p\sqrt{1 - p^2}\)
\(\sin x = 1\) is rejected, since \(\sin x \lt 0\)
Reference angle \(= 41{,}81^\circ\); \(\sin\) is negative in quadrants 3 and 4
\(x = 221{,}81^\circ + k\cdot 360^\circ\) or \(x = 318{,}19^\circ + k\cdot 360^\circ\), \(k \in \mathbb{Z}\)
Question 7
6 questionsThe graph of \(f(x) = 2\sin(x - 45^\circ)\) is drawn for \(x \in [-180^\circ\,;\,180^\circ]\). The \(y\)-axis is marked from \(-3\) to \(3\) and the \(x\)-axis at \(-180^\circ\), \(-90^\circ\), \(90^\circ\) and \(180^\circ\).
Diagram: \(f\) has a minimum value of \(-2\) at \(x = -45^\circ\) and a maximum value of \(2\) at \(x = 135^\circ\). It cuts the \(x\)-axis at \(x = -135^\circ\) and \(x = 45^\circ\), and the \(y\)-axis at \(-\sqrt{2}\).
\(f'(x) \geq 0\) where \(f\) is increasing (or stationary): from the minimum at \(x = -45^\circ\) to the maximum at \(x = 135^\circ\).
\(-45^\circ \leq x \leq 135^\circ\)
Asymptotes: \(x = -90^\circ\) and \(x = 90^\circ\) (dotted vertical lines).
\(x\)-intercepts: \((-180^\circ\,;\,0)\), \((0^\circ\,;\,0)\) and \((180^\circ\,;\,0)\).
Points: \(g(-135^\circ) = -1\), \(g(-45^\circ) = 1\), \(g(45^\circ) = -1\), \(g(135^\circ) = 1\).
Shape: the tan graph reflected in the \(x\)-axis, so each branch decreases from left to right: one branch from \((-180^\circ\,;\,0)\) down towards \(x = -90^\circ\), a middle branch from top left near \(x = -90^\circ\) through \((0^\circ\,;\,0)\) to bottom right near \(x = 90^\circ\), and a branch from the top near \(x = 90^\circ\) down to \((180^\circ\,;\,0)\).
Three decreasing branches of \(-\tan x\) with asymptotes at \(x = \pm 90^\circ\) and \(x\)-intercepts at \(-180^\circ\), \(0^\circ\) and \(180^\circ\).
\(3\)
Count the points where the graphs of \(g\) and \(f\) intersect.
2
The asymptotes of \(g\) at \(x = -90^\circ\) and \(x = 90^\circ\) move \(45^\circ\) to the right (\(x = -270^\circ\) moves to \(-225^\circ\), outside the interval).
\(x = -45^\circ\) and \(x = 135^\circ\)
Question 8
3 questions8.1 \(\triangle RTP\) is drawn with T at the origin and P on the positive \(x\)-axis; R lies above the \(x\)-axis. \(RT = p\) and \(RP = t\).
8.2 D, B and C are points in the same horizontal plane. AB is a vertical flagpole with \(AB = 4\) units. A rope is tied from A, the top of the flagpole, to C. \(B\hat{A}C = B\hat{C}D = \theta\) and \(B\hat{D}C = \alpha\).
Diagram: A is directly above B. Triangle BDC lies flat on the ground, and the rope AC joins the top of the pole to C.
Construction: draw \(RK \perp TP\) with K on TP, and let \(RK = h\).
In \(\triangle RKT\):
\[\sin\hat{T} = \frac{h}{p} \quad\therefore h = p\sin\hat{T}\]In \(\triangle RKP\):
\[\sin\hat{P} = \frac{h}{t} \quad\therefore h = t\sin\hat{P}\]\[\therefore p\sin\hat{T} = t\sin\hat{P}\]Divide both sides by \(pt\):
\(\dfrac{\sin\hat{T}}{t} = \dfrac{\sin\hat{P}}{p}\)
In right-angled \(\triangle ABC\) (AB vertical):
\[\tan\theta = \frac{BC}{AB} = \frac{BC}{4}\]\(BC = 4\tan\theta\)
In \(\triangle BDC\): \(D\hat{B}C = 180^\circ - (\alpha + \theta)\) (sum of \(\angle\)s in \(\triangle\))
\[\frac{DC}{\sin[180^\circ - (\alpha + \theta)]} = \frac{BC}{\sin\alpha}\](sine rule)
\[DC = \frac{4\tan\theta\sin(\alpha + \theta)}{\sin\alpha}\]\[= \frac{4\tan\theta(\sin\alpha\cos\theta + \cos\alpha\sin\theta)}{\sin\alpha}\]\[= \frac{4\sin\theta(\sin\alpha\cos\theta + \cos\alpha\sin\theta)}{\cos\theta\sin\alpha}\]\[= 4\sin\theta\left[\frac{\sin\alpha\cos\theta}{\cos\theta\sin\alpha} + \frac{\cos\alpha\sin\theta}{\cos\theta\sin\alpha}\right]\]\[= 4\sin\theta\left(1 + \frac{\tan\theta}{\tan\alpha}\right)\]\(DC = 4\sin\theta\left(1 + \dfrac{\tan\theta}{\tan\alpha}\right)\)
Question 9
4 questions9.1 O is the centre of the circle. RSTV is a cyclic quadrilateral, with R and T opposite each other.
9.2 A, B, C, D and E lie on a circle. BA is produced to F. \(\hat{B}_2 = 27^\circ\) and \(\hat{C} = 88^\circ\). AE, ED, DC, BC, BE and BD are drawn.
Angle labels in 9.2: at B, \(\hat{B}_1 = \angle ABE\), \(\hat{B}_2 = \angle EBD\), \(\hat{B}_3 = \angle DBC\). At E, \(\hat{E}_1 = \angle AEB\) and \(\hat{E}_2 = \angle BED\). At D, \(\hat{D}_1 = \angle EDB\) and \(\hat{D}_2 = \angle BDC\). At A, \(\hat{A}_1 = \angle BAE\) and \(\hat{A}_2 = \angle FAE\) (outside the circle). \(\hat{C} = \angle BCD = 88^\circ\).
Construction: draw radii OS and OV. Let \(\hat{O}_1\) be the angle at O facing T and \(\hat{O}_2\) the reflex angle at O facing R.
- \(\hat{O}_2 = 2\hat{R}\) (\(\angle\) at centre = 2 \(\times\) \(\angle\) at circumference)
- \(\hat{O}_1 = 2\hat{T}\) (\(\angle\) at centre = 2 \(\times\) \(\angle\) at circumference)
- \(\hat{O}_1 + \hat{O}_2 = 360^\circ\) (\(\angle\)s round a pt)
- \(2\hat{R} + 2\hat{T} = 360^\circ\)
- \(\therefore \hat{R} + \hat{T} = 180^\circ\)
\(\hat{R} + \hat{T} = 180^\circ\)
- \(\hat{E}_2 + \hat{C} = 180^\circ\) (opp \(\angle\)s of cyclic quad BCDE)
- \(\hat{E}_2 = 180^\circ - 88^\circ\)
\(\hat{E}_2 = 92^\circ\)
- \(\hat{B}_3 = \hat{B}_2\) (equal chords; equal \(\angle\)s)
\(\hat{B}_3 = 27^\circ\)
- \(\hat{D}_1 = 180^\circ - 27^\circ - 92^\circ = 61^\circ\) (sum of \(\angle\)s in \(\triangle BED\))
- \(\hat{A}_2 = \hat{D}_1\) (ext \(\angle\) of cyclic quad ABDE)
\(\hat{A}_2 = 61^\circ\)
Question 10
5 questionsO is the centre of the circle. Chord AS is perpendicular to diameter BOD at K. OR is drawn with R on BS. AB, AD and OS are drawn. \(\hat{B}_1 = x\).
Angle labels: \(\hat{B}_1 = \angle ABD\), \(\hat{B}_2 = \angle DBS\). \(\hat{A}_1 = \angle BAS\), \(\hat{A}_2 = \angle SAD\). \(\hat{K}_3 = \angle OKS\) (\(= 90^\circ\)). \(\hat{O}_1 = \angle BOR\), \(\hat{O}_2 = \angle ROS\), \(\hat{O}_3 = \angle SOK\). \(\hat{S}_1 = \angle OSK\), \(\hat{S}_2 = \angle OSR\). \(\hat{R}_1 = \angle ORB\), \(\hat{R}_2 = \angle ORS\).
- \(D\hat{A}B = 90^\circ\) (\(\angle\) in semi-circle)
- \(\hat{D} = 180^\circ - 90^\circ - x\) (sum of \(\angle\)s in \(\triangle ABD\))
\(\hat{D} = 90^\circ - x\)
- \(\hat{S}_1 + \hat{S}_2 = \hat{D} = 90^\circ - x\) (\(\angle\)s in the same segment)
- In \(\triangle BKS\): \(\hat{K}_3 = 90^\circ\) (given \(AS \perp BD\))
- \(\hat{B}_2 = 180^\circ - 90^\circ - (90^\circ - x)\) (sum of \(\angle\)s in \(\triangle\))
\(\hat{B}_2 = x\)
- \(\hat{S}_2 = \hat{B}_2 = x\) (\(\angle\)s opp equal sides; \(OB = OS\))
- \(\hat{S}_1 = (\hat{S}_1 + \hat{S}_2) - \hat{S}_2 = 90^\circ - x - x\)
\(\hat{S}_1 = 90^\circ - 2x\)
- \(\hat{R}_2 = 90^\circ\) (line from centre to midpt of chord)
- \(\hat{K}_3 = 90^\circ\) (given \(AS \perp BD\))
- \(\hat{R}_2 + \hat{K}_3 = 180^\circ\)
- \(\therefore\) OKSR is a cyclic quadrilateral (converse opp \(\angle\)s of cyclic quad)
OKSR is cyclic.
Question 11
4 questionsFA is a common tangent at A to the larger circle ABC and the smaller circle ADE. Chords AB and AC of the larger circle cut the smaller circle at D and E respectively. BC is drawn. Another tangent from C meets FA produced at G. DE produced meets tangent GC at H.
Angle labels: \(\hat{A}_1 = \angle FAD\), \(\hat{A}_2 = \angle DAE\) (\(= \angle BAC\)), \(\hat{A}_3 = \angle EAG\). \(\hat{D}_1 = \angle ADE\), \(\hat{D}_2 = \angle EDB\). \(\hat{E}_1 = \angle AED\), \(\hat{E}_2 = \angle DEC\), \(\hat{E}_3 = \angle CEH\), \(\hat{E}_4 = \angle HEA\). \(\hat{C}_1 = \angle BCA\), \(\hat{C}_2 = \angle ACH\). \(\hat{H}_1 = \angle CHE\), \(\hat{H}_2 = \angle EHG\).
In \(\triangle ADE\) and \(\triangle ABC\):
- \(\hat{A}_2 = \hat{A}_2\) (common)
- \(\hat{E}_1 = \hat{A}_1\) (tan chord theorem, small circle)
- \(\hat{C}_1 = \hat{A}_1\) (tan chord theorem, large circle)
- \(\therefore \hat{E}_1 = \hat{C}_1\)
- \(\hat{D}_1 = \hat{B}\) (third \(\angle\) of \(\triangle\), or tan chord theorem using \(\hat{A}_3\))
- \(\therefore \triangle ADE \,|||\, \triangle ABC\) (\(\angle\angle\angle\))
\(\triangle ADE \,|||\, \triangle ABC\)
- \(\hat{D}_1 = \hat{B}\) (proved in 11.1), so \(DE \parallel BC\) (corresp \(\angle\)s =)
- \(\dfrac{AD}{DB} = \dfrac{AE}{EC}\) (line \(\parallel\) one side of \(\triangle\))
- \(AD = k\), \(AB = 3k\), so \(DB = 2k\)
- \(\dfrac{k}{2k} = \dfrac{4}{EC}\)
- \(EC = 8\)
\(EC = 8\) cm
In \(\triangle ECH\) and \(\triangle CBA\):
- \(\hat{E}_3 = \hat{C}_1\) (alt \(\angle\)s; \(BC \parallel DH\))
- \(\hat{C}_2 = \hat{B}\) (tan chord theorem)
- \(\hat{H}_1 = \hat{A}_2\) (sum of \(\angle\)s in \(\triangle\))
- \(\therefore \triangle ECH \,|||\, \triangle CBA\) (\(\angle\angle\angle\))
- \(\therefore \dfrac{EH}{CA} = \dfrac{EC}{CB}\) (\(|||\) \(\triangle\)s)
\(\therefore \dfrac{EH}{EC} = \dfrac{CA}{CB}\)
- \(\dfrac{EH}{EC} = \dfrac{CA}{CB}\) (proved in 11.3), so \(EC = \dfrac{EH\cdot CB}{CA}\)
- \(\dfrac{EC}{AC} = \dfrac{BD}{AB}\) (line \(\parallel\) one side of \(\triangle\); \(DE \parallel BC\)), so \(EC = \dfrac{AC\cdot BD}{AB}\)
- \(EC \times EC = \dfrac{EH\cdot CB}{AC} \times \dfrac{AC\cdot BD}{AB}\)
- \(EC^2 = \dfrac{EH\cdot CB\cdot DB}{AB}\)
- \(EC^2 = \dfrac{EH\cdot CB\cdot DB}{3AD}\) (\(AD : AB = 1 : 3\), so \(AB = 3AD\))
\(\therefore EC = \sqrt{\dfrac{EH\cdot CB\cdot DB}{3AD}}\)