50 questions answered · 150 marks · DBE
Mathematics Grade 12 NSC November 2024 P2 Questions and Answers
Every question from the NSC Examination (DBE, November 2024), 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 11 questions in the answer book. Show all working, since answers alone may not earn full marks. A non-programmable calculator may be used unless a question says otherwise. Round to two decimal places unless told otherwise, and give reasons for statements in Questions 9 to 11.
Question 1
6 questionsAt the start of a season a rugby coach recorded the weight (in kg) of the 15 players in a junior boys' team and the number of push-ups each player could do in one minute. The least squares regression line is drawn on the scatter plot.
| Weight in kg (\(x\)) | 34 | 32 | 40 | 27 | 33 | 28 | 27 | 55 | 39 | 44 | 30 | 57 | 28 | 32 | 31 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| Push-ups per minute (\(y\)) | 26 | 21 | 6 | 20 | 16 | 26 | 23 | 7 | 18 | 14 | 17 | 8 | 28 | 25 | 20 |
Diagram: a scatter plot with weight (0 to 60 kg) on the horizontal axis and push-ups per minute (0 to 30) on the vertical axis. The points slope downwards from left to right and a straight regression line runs through them. One point, \((40\,;\,6)\), lies far below the line.
Enter the data in the calculator in statistics (linear regression) mode.
\[a = 39{,}456\ldots \approx 39{,}46\]\[b = -0{,}590\ldots \approx -0{,}59\]\(\hat{y} = 39{,}46 - 0{,}59x\)
About \(22{,}35\) push-ups (so about 22 push-ups)
Adding the same amount to every value moves the whole data set up but does not change how spread out it is.
It has no influence: the standard deviation stays the same.
The point furthest below the regression line is \((40\,;\,6)\), with \(y = 6\).
Target at 40 kg:
\[\hat{y} = 39{,}46 - 0{,}59(40) = 15{,}86 \approx 16\]Increase needed:
\[16 - 6 = 10\]10 push-ups
Question 2
6 questionsThe ogive (cumulative frequency graph) shows the time taken (in minutes) by 60 employees to travel to work each morning.
Diagram: time in minutes (0 to 140) on the horizontal axis and cumulative frequency (0 to 65) on the vertical axis. The curve starts at \((0\,;\,0)\), passes through about \((20\,;\,5)\), \((40\,;\,13)\), \((60\,;\,26)\), \((80\,;\,41)\), \((100\,;\,56)\) and levels off at \((120\,;\,60)\).
Median position:
\[\frac{1}{2} \times 60 = 30\text{th value}\]Reading across from 30 to the ogive and down: median \(\approx 65\) minutes
\(Q_1\) position:
\[\frac{1}{4} \times 60 = 15\text{th value}\]\(Q_1 \approx 44\) minutes
\(Q_3\) position:
\[\frac{3}{4} \times 60 = 45\text{th value, so } Q_3 \approx 84 \text{ minutes}\]\[\text{IQR} = Q_3 - Q_1 = 84 - 44\]\(\text{IQR} = 40\) minutes
Five-number summary: minimum 5 ; \(Q_1\) 44 ; median 65 ; \(Q_3\) 84 ; maximum 120.
Sketch: draw a box from 44 to 84 with a vertical line inside it at 65. Draw a whisker from the left side of the box to 5 and another from the right side of the box to 120.
Box: 44 to 84 with median at 65; whiskers to 5 and 120.
From the ogive, 26 employees travel less than 60 minutes.
Employees who qualify:
\[60 - 26 = 34\]Percentage:
\[\frac{34}{60} \times 100\]\(\approx 56{,}67\%\)
Time above an hour:
\[110 - 60 = 50 \text{ minutes}\]50 minutes = two full 20-minute intervals plus part of a third, so 3 intervals.
Time at home:
\[3 \times 30 \text{ minutes}\]90 minutes (\(1{,}5\) hours)
Question 3
7 questions\(\triangle ACD\) has vertices \(A\), \(D(3\,;\,-9)\) and \(C(-9\,;\,0)\), where \(A\) is a point in the second quadrant. \(B(-1\,;\,k)\) lies on side \(DC\).
Diagram: \(C\) is on the negative \(x\)-axis, \(D\) is in the fourth quadrant and \(A\) is high up in the second quadrant. Side \(AD\) crosses the \(y\)-axis above the origin. \(B\) lies on \(DC\) between \(C\) and \(D\), just left of the \(y\)-axis.
\(m_{DC} = -\frac{3}{4}\)
\(y = -\frac{3}{4}x - \frac{27}{4}\)
\(B(-1\,;\,k)\) lies on \(DC\), so substitute \(x = -1\):
\[k = -\frac{3}{4}(-1) - \frac{27}{4}\]\[k = \frac{3}{4} - \frac{27}{4} = -\frac{24}{4}\]\(k = -6\)
\(DC = 15\) units
\(\dfrac{DB}{DC} = \dfrac{1}{3}\)
In \(\triangle ACD\), \(MB \parallel AC\), so \(\dfrac{DM}{DA} = \dfrac{DB}{DC} = \dfrac{1}{3}\) (line \(\parallel\) one side of \(\triangle\))
\[\frac{\text{Area } \triangle MBD}{\text{Area } \triangle ACD} = \frac{\frac{1}{2} \cdot DM \cdot DB \cdot \sin\hat{D}}{\frac{1}{2} \cdot DA \cdot DC \cdot \sin\hat{D}}\]\[= \frac{DM}{DA} \times \frac{DB}{DC}\]\[= \frac{1}{3} \times \frac{1}{3}\]\(= \dfrac{1}{9}\)
Equation of \(AD\) through \(D(3\,;\,-9)\) with \(m = -4\):
\[-9 = -4(3) + c, \text{ so } c = 3 \text{ and } y = -4x + 3\]\(AD^2 = 612\):
\[(x - 3)^2 + (y + 9)^2 = 612\]\[(x - 3)^2 + (-4x + 12)^2 = 612\]\[x^2 - 6x + 9 + 16x^2 - 96x + 144 = 612\]\[17x^2 - 102x - 459 = 0\]\[x^2 - 6x - 27 = 0\]\[(x - 9)(x + 3) = 0\]\(x = 9\) (not valid, \(A\) is in the second quadrant) or \(x = -3\)
\[y = -4(-3) + 3 = 15\]\(A(-3\,;\,15)\)
Question 4
6 questions\(M(1\,;\,3)\) is the centre of the circle. The circle cuts the \(x\)-axis at \(N\). \(ST\) is a tangent to the circle at \(P(3\,;\,9)\). \(R(d\,;\,1)\), with \(d \gt 0\), and \(L\) lie on the circle. \(O\) and \(V\) are the \(x\)-intercepts of \(PL\) and \(RL\) respectively.
Diagram: \(PL\) is a diameter passing through \(M\) and through the origin \(O\), with \(L\) in the third quadrant. Chords \(PR\) and \(RL\) are drawn, \(R\) lying just above the \(x\)-axis on the right. \(RL\) cuts the \(x\)-axis at \(V\), between \(O\) and \(N\). The tangent \(ST\) touches the circle at \(P\), with \(S\) near the \(y\)-axis and \(T\) to the right.
\(PL\) is a diameter, so \(M\) is the midpoint of \(PL\).
\[x_L = 2(1) - 3 = -1\,;\quad y_L = 2(3) - 9 = -3\]\(L(-1\,;\,-3)\)
\(y = -\frac{1}{3}x + 10\)
\(x^2 + y^2 - 2x - 6y - 30 = 0\)
\(R(d\,;\,1)\) is on the circle:
\[d^2 + 1^2 - 2d - 6(1) - 30 = 0\]\[d^2 - 2d - 35 = 0\]\[(d - 7)(d + 5) = 0\]\(d = 7\) or \(d = -5\), but \(d \gt 0\)
\(d = 7\)
\(L\hat{V}O = R\hat{V}N = 26{,}57^\circ\) (vert opp \(\angle\)s \(=\))
\(P\hat{O}V = \hat{L} + L\hat{V}O\) (ext \(\angle\) of \(\triangle\))
\[\hat{L} = 71{,}57^\circ - 26{,}57^\circ\]\(\hat{L} = 45^\circ\)
\(\therefore PT \perp RT\)
Question 5
4 questionsLine \(OP\) is drawn with \(P(-3\,;\,-4)\). \(K\) lies on the positive \(x\)-axis and the reflex angle \(K\hat{O}P = A\), measured anticlockwise from \(OK\) to \(OP\).
\(P(-3\,;\,-4)\):
\[r = \sqrt{(-3)^2 + (-4)^2} = 5\]\[\cos A = \frac{x}{r} = -\frac{3}{5}\]\(\cos A = -\dfrac{3}{5}\)
\(\cos 2A = -\dfrac{7}{25}\)
\(\sin B\) is positive and \(90^\circ \lt B \lt 360^\circ\), so \(B\) is in the second quadrant: point \((x\,;\,4)\) with \(r = 5\).
\[x = -\sqrt{25 - 16} = -3, \text{ so } \cos B = -\frac{3}{5}\]\[\sin A = -\frac{4}{5} \text{ and } \cos A = -\frac{3}{5}\]\[\sin(A - B) = \sin A\cos B - \cos A\sin B\]\[= \left(-\frac{4}{5}\right)\left(-\frac{3}{5}\right) - \left(-\frac{3}{5}\right)\left(\frac{4}{5}\right)\]\[= \frac{12}{25} + \frac{12}{25}\]\(\sin(A - B) = \dfrac{24}{25}\)
\(= -\dfrac{p}{4}\)
Question 6
5 questionsGiven the identity: \(\cos(x - y) = \cos x\cos y + \sin x\sin y\)
\(= \cos x\cos y - \sin x\sin y\)
\(= \text{RHS}\)
Reference angle \(= 19{,}47^\circ\); \(\sin\) is negative in quadrants 3 and 4
\(x = 199{,}47^\circ\) or \(x = 340{,}53^\circ\)
Consider the function \(g(x) = \dfrac{4 - 8\sin^2 x}{3}\)
Maximum value of \(\cos 2x\) is 1
Maximum value of \(g = \dfrac{4}{3}\)
\(g\) is a maximum when \(\sin x = 0\). In \((0^\circ\,;\,360^\circ]\), \(0^\circ\) is excluded, so the first value is \(180^\circ\).
\(x = 180^\circ\)
Question 7
5 questionsThe graph of \(f(x) = \tan x\) is drawn for the interval \(x \in [-180^\circ\,;\,180^\circ]\). A second function is given in 7.3: \(g(x) = \cos 2x + 1\).
Diagram: the tan graph passes through \((-180^\circ\,;\,0)\), \((0^\circ\,;\,0)\) and \((180^\circ\,;\,0)\), with vertical dotted asymptotes at \(x = -90^\circ\) and \(x = 90^\circ\). The \(y\)-axis is marked from \(-3\) to \(3\).
\(\tan x \leq 0\) where the graph is on or below the \(x\)-axis.
\(x = -180^\circ\) or \(-90^\circ \lt x \leq 0^\circ\)
The graph is \(\cos 2x\) shifted up by 1, so it ranges from 0 to 2.
Maximum turning points: \((-180^\circ\,;\,2)\), \((0^\circ\,;\,2)\) and \((180^\circ\,;\,2)\). The \(y\)-intercept is \((0\,;\,2)\).
Minimum turning points touch the \(x\)-axis: \((-90^\circ\,;\,0)\) and \((90^\circ\,;\,0)\). These are the only \(x\)-intercepts.
Smooth wave from \((-180^\circ\,;\,2)\) down to \((-90^\circ\,;\,0)\), up to \((0^\circ\,;\,2)\), down to \((90^\circ\,;\,0)\) and up to \((180^\circ\,;\,2)\).
Divide by \(\cos x\) (\(\cos x \neq 0\), since \(\cos x = 0\) gives \(\sin x = \pm 1\), which is not a solution):
\[2\cos^2 x = \tan x\]\[2\cos^2 x - 1 + 1 = \tan x\]\[\cos 2x + 1 = \tan x, \text{ so } g(x) = f(x)\]The graphs intersect at \(x = 45^\circ\) and \(x = -135^\circ\) (\(180^\circ\) apart).
\(x = 45^\circ + k \cdot 180^\circ,\ k \in \mathbb{Z}\)
Question 8
2 questions\(C\) is the foot of a vertical building and \(D\) is the top. The height of the building, \(CD\), is 16 m. Two observers stand 19 m apart at points \(A\) and \(B\), where \(A\), \(B\) and \(C\) lie in the same horizontal plane. A painter works at point \(E\) on the building (on \(CD\)). The angle of elevation of \(D\) from \(A\) is \(46{,}85^\circ\). \(D\hat{E}B = 122^\circ\) and \(B\hat{C}A = 105{,}61^\circ\).
Diagram: \(\triangle ABC\) lies on the ground with \(AB = 19\) m and angle \(105{,}61^\circ\) at \(C\). \(DC\) is vertical with \(CD = 16\). Lines \(DA\) and \(EB\) are drawn. The obtuse angle \(122^\circ\) is at \(E\) between \(ED\) and \(EB\).
In right-angled \(\triangle DCA\):
\[\tan 46{,}85^\circ = \frac{DC}{AC} = \frac{16}{AC}\]\[AC = \frac{16}{\tan 46{,}85^\circ}\]\(AC = 15\) m
In \(\triangle ABC\), use the cosine rule with \(BC = x\):
\[19^2 = x^2 + 15^2 - 2(x)(15)\cos 105{,}61^\circ\]\[x^2 + 8{,}07x - 136 = 0\]\[x = \frac{-8{,}07 \pm \sqrt{(8{,}07)^2 - 4(1)(-136)}}{2(1)}\]\(x = 8{,}30\) m (\(x = -16{,}38\) is not valid)
\(B\hat{E}C = 180^\circ - 122^\circ = 58^\circ\) (\(\angle\)s on a str line)
In right-angled \(\triangle BCE\):
\[\tan 58^\circ = \frac{BC}{EC}\]\[EC = \frac{8{,}30}{\tan 58^\circ} = 5{,}19 \text{ m}\]\[DE = 16 - 5{,}19\]\(DE = 10{,}81\) m
Question 9
2 questions\(ABCD\) is a cyclic quadrilateral. \(BC\) is produced to \(E\). \(AC\) is drawn. \(\hat{A}_1 = \frac{1}{2}\hat{B}\), \(\hat{A}_2 = 46^\circ\) and \(\hat{C}_1 = 86^\circ\).
Diagram: \(\hat{A}_1 = \angle DAC\) and \(\hat{A}_2 = \angle CAB\). At \(C\), \(\hat{C}_1 = \angle DCE\) (the exterior angle between \(DC\) and \(CE\)), \(\hat{C}_2 = \angle DCA\) and \(\hat{C}_3 = \angle ACB\).
\(\hat{A}_1 + \hat{A}_2 = \hat{C}_1 = 86^\circ\) (ext \(\angle\) of cyclic quad)
\(\hat{A}_1 = 86^\circ - 46^\circ\)
\(\hat{A}_1 = 40^\circ\)
\(\hat{B} = 2\hat{A}_1 = 80^\circ\) (given \(\hat{A}_1 = \frac{1}{2}\hat{B}\))
\(\hat{D} = 180^\circ - 80^\circ = 100^\circ\) (opp \(\angle\)s of cyclic quad)
\(\hat{C}_2 = 180^\circ - 100^\circ - 40^\circ = 40^\circ\) (sum of \(\angle\)s in \(\triangle\))
\(\hat{C}_2 = \hat{A}_1 = 40^\circ\)
\(\therefore AD = DC\) (sides opp equal \(\angle\)s)
Question 10
3 questions\(\triangle RST\) is drawn. Line \(AB\) cuts \(RS\) at \(A\) and \(RT\) at \(B\) such that \(AB \parallel ST\).
Construction: join \(SB\) and \(TA\). Draw \(h_1\) from \(B \perp RA\) and \(h_2\) from \(A \perp RB\).
\(\dfrac{\text{Area } \triangle RAB}{\text{Area } \triangle ASB} = \dfrac{\frac{1}{2} \cdot RA \cdot h_1}{\frac{1}{2} \cdot AS \cdot h_1} = \dfrac{RA}{AS}\) (same height)
\(\dfrac{\text{Area } \triangle RAB}{\text{Area } \triangle ABT} = \dfrac{\frac{1}{2} \cdot RB \cdot h_2}{\frac{1}{2} \cdot BT \cdot h_2} = \dfrac{RB}{BT}\) (same height)
\(\text{Area } \triangle ASB = \text{Area } \triangle ABT\) (same base \(AB\) and same height; \(AB \parallel ST\))
\(\therefore \dfrac{\text{Area } \triangle RAB}{\text{Area } \triangle ASB} = \dfrac{\text{Area } \triangle RAB}{\text{Area } \triangle ABT}\)
\(\therefore \dfrac{RA}{AS} = \dfrac{RB}{BT}\)
\(O\) is the centre of the circle. \(\triangle PWS\) is drawn with \(P\), \(W\) and \(S\) on the circle, and \(PW\) passes through \(O\) (\(PW\) is a diameter). \(OR \perp PS\), with \(R\) on \(PS\). \(PRS\) is produced to \(T\). \(SW\) and \(OT\) intersect at \(V\). \(OV : OT = 1 : 4\).
\(PR = RS\) (line from centre \(\perp\) to chord)
\(PO = OW\) (radii)
\(\therefore OR = \frac{1}{2}WS\) (midpt theorem)
\(\therefore OR : WS = 1 : 2\)
\(OV : OT = 1 : 4\), so \(OV : VT = 1 : 3\)
\(RO \parallel SW\) (from 10.2.1), so in \(\triangle TRO\): \(\dfrac{RS}{ST} = \dfrac{OV}{VT} = \dfrac{1}{3}\) (line \(\parallel\) one side of \(\triangle\))
\(\dfrac{RS}{15} = \dfrac{1}{3}\), so \(RS = 5\)
\(PR = RS = 5\) (line from centre \(\perp\) to chord)
\(PT = 5 + 5 + 15\)
\(PT = 25\) units
Question 11
4 questions\(A\), \(B\), \(G\) and \(F\) lie on the larger circle. A smaller circle touches the larger circle internally at \(A\). \(EA\) is a common tangent to both circles. \(EBCF\) is a tangent to the smaller circle at \(C\). \(AC\) is produced to \(G\). \(AF\) cuts the smaller circle at \(D\). \(AB\), \(CD\) and \(GF\) are drawn.
Diagram labels: at \(A\), \(\hat{A}_1 = \angle EAB\), \(\hat{A}_2 = \angle BAC\), \(\hat{A}_3 = \angle CAD\). At \(C\), \(\hat{C}_1 = \angle ACB\), \(\hat{C}_2 = \angle ACD\), \(\hat{C}_3 = \angle DCF\), \(\hat{C}_4 = \angle FCG\). At \(D\), \(\hat{D}_1 = \angle ADC\), \(\hat{D}_2 = \angle CDF\). At \(F\), \(\hat{F}_1 = \angle AFC\), \(\hat{F}_2 = \angle CFG\). At \(B\), \(\hat{B}_1 = \angle ABE\), \(\hat{B}_2 = \angle ABC\).
\(\hat{D}_1 = E\hat{A}G = x\) (tan chord theorem, small circle)
\(\hat{C}_1 = \hat{D}_1 = x\) (tan chord theorem, small circle)
\(\hat{C}_4 = \hat{C}_1 = x\) (vert opp \(\angle\)s \(=\))
\(A\hat{F}G = E\hat{A}G = x\) (tan chord theorem, large circle)
\(\hat{D}_1\), \(\hat{C}_1\), \(\hat{C}_4\) and \(A\hat{F}G\) all equal \(x\)
\(\hat{D}_1 = A\hat{F}G = x\) (proved in 11.1)
\(\therefore DC \parallel FG\) (corresp \(\angle\)s \(=\))
\(\dfrac{AG}{AC} = \dfrac{AF}{AD}\) (line \(\parallel\) one side of \(\triangle\), \(DC \parallel FG\))
\(\therefore AG \cdot AD = AC \cdot AF\)
In \(\triangle AGF\) and \(\triangle ABC\):
\(A\hat{G}F = \hat{B}_2\) (\(\angle\)s in the same seg)
\(A\hat{F}G = \hat{C}_1 = x\) (proved in 11.1)
\(\hat{A}_3 = \hat{A}_2\) (sum of \(\angle\)s in \(\triangle\))
\(\therefore \triangle AGF \,|||\, \triangle ABC\) (\(\angle\angle\angle\))
\(\dfrac{GF}{BC} = \dfrac{AF}{AC}\) (\(|||\) \(\triangle\)s, \(\triangle AGF \,|||\, \triangle ABC\))
\(\therefore GF = \dfrac{BC \cdot AF}{AC}\) ... (1)
In \(\triangle ACD\) and \(\triangle FGC\):
\(\hat{D}_1 = \hat{C}_4 = x\) (proved in 11.1)
\(A\hat{C}D = A\hat{G}F\) (corresp \(\angle\)s, \(DC \parallel FG\))
\(\hat{A}_3 = \hat{F}_2\) (sum of \(\angle\)s in \(\triangle\))
\(\therefore \triangle ACD \,|||\, \triangle FGC\) (\(\angle\angle\angle\))
\(\dfrac{AC}{FG} = \dfrac{AD}{FC}\) (\(|||\) \(\triangle\)s), so \(AC = \dfrac{AD \cdot FG}{FC}\) ... (2)
Substitute (2) in (1): \(GF = \dfrac{BC \cdot AF \cdot FC}{AD \cdot FG}\)
\(\therefore GF^2 = \dfrac{BC \cdot FC \cdot AF}{AD}\)