IEB-style Grade 12 Mathematics Paper 2: Practice Set 2 memo

Question 1

1.1 [2]

4 ; 15 ; 35

50 ; 57 ; 60

Answer:

4 ; 15 ; 35 ; 50 ; 57 ; 60

1.2 [4]

Start at $(15;0)$

Points at upper boundaries: $(20;4)$, $(25;15)$, $(30;35)$

$(35;50)$, $(40;57)$, $(45;60)$

Smooth curve, labelled axes

Answer:

Ogive

1.3 [2]

Read at cumulative frequency 30

Median $\approx 28,75$ minutes (accept 28 to 29,5)

Answer:

About 28,75 minutes

1.4 [3]

Midpoints 17,5 ; 22,5 ; 27,5 ; 32,5 ; 37,5 ; 42,5

$\sum fx=1745$

$\bar{x}\approx 29,08$ minutes

Answer:

About 29,08 minutes

1.5 [2]

$60-50=10$ runners

$\frac{10}{60}\approx 16,7\%$

Answer:

About 16,7%

Question 2

2.1 [5]

Construction: diameter through the point of contact; join to the other end of the chord

tan $\perp$ radius gives $90°$

$\angle$ in semi-circle gives $90°$

Both angles equal $90°$ minus the same angle

$\angle$s in the same seg complete the proof

Answer:

Proof

2.2 [2]

$\widehat{ACB}=90°$ ($\angle$ in semi-circle)

$\widehat{ABC}=65°$ (sum of $\angle$s in $△$)

Answer:

$90°$ and $65°$

2.3 [2]

$\widehat{BCT}=\widehat{CAB}$ (tan chord theorem)

$=25°$

Answer:

25°

2.4 [3]

$\widehat{ABC}$ is an exterior angle of $△BCT$

$\hat{T}=\widehat{ABC}-\widehat{BCT}$ (ext $\angle$ of $△$)

$=65°-25°=40°$

Answer:

40°

2.5 [2]

$\widehat{ACO}=25°$ ($OA=OC$, radii; $\angle$s opp equal sides)

$\widehat{OCT}=90°$ (tan $\perp$ radius)

Answer:

$25°$ and $90°$

Question 3

3.1 [3]

$a\approx 299,76$

$b\approx -23,05$

$\hat{y}=299,76-23,05x$

Answer:

$\hat{y}=299,76-23,05x$

3.2 [2]

$r\approx -0,996$

Very strong negative correlation

Answer:

$r\approx -0,996$; very strong negative

3.3 [2]

Each extra year of age

is linked to a drop in price of about R23 050

Answer:

About R23 050 less per year

Question 4

4.1 [3]

$M_{PR}=(1;0)$

$M_{QS}=(1;0)$

The diagonals bisect each other, so $PQRS$ is a parallelogram

Answer:

Diagonals share midpoint $(1;0)$

4.2 [3]

$PQ=\sqrt{36+16}=\sqrt{52}$

$QR=\sqrt{16+36}=\sqrt{52}$

A parallelogram with adjacent sides equal is a rhombus

Answer:

$P{Q}^2=Q{R}^2=52$, so $PQ=QR=\sqrt{52}$

4.3 [3]

$m_{PQ}=\frac{4}{6}=\frac{2}{3}$

$m_{QR}=\frac{-6}{4}=-\frac{3}{2}$

Product $-1$, so $\hat{Q}=90°$; a rhombus with a right angle is a square

Answer:

$m_{PQ}\times m_{QR}=-1$

4.4 [2]

Area $=P{Q}^2$

$=52$ square units

Answer:

52 square units

4.5 [2]

$\tan \theta =\frac{2}{3}$

$\theta \approx 33,69°$

Answer:

About 33,69°

4.6 [3]

$m_{PR}=\frac{-1-1}{6+4}=-\frac{1}{5}$, so the inclination of $PR$ is $180°-11,31°=168,69°$

Angle between the lines $=168,69°-33,69°=135°$, so the acute angle is $45°$

The diagonals of a square bisect its $90°$ angles

Answer:

45°

4.7 [3]

The centre is where the diagonals meet: $(1;0)$

$r^2=(-4-1{)}^2+(1-0{)}^2=26$

$(x-1{)}^2+y^2=26$

Answer:

$(x-1{)}^2+y^2=26$

4.8 [3]

$m_{\text{radius}}=\frac{5-0}{2-1}=5$

$m_{\text{tan}}=-\frac{1}{5}$

$y=-\frac{1}{5}x+\frac{27}{5}$

Answer:

$y=-\frac{1}{5}x+\frac{27}{5}$

Question 5

5.1 [4]

$\hat{A}$ common

$\widehat{ADB}=\widehat{ABC}=90°$

$\widehat{ABD}=\hat{C}$ (sum of $\angle$s in $△$)

$△ABD|||△ACB$ (equiangular $△$s)

Answer:

Proof

5.2 [3]

$△ADB|||△BDC$ (both equiangular to $△ABC$)

$\frac{AD}{BD}=\frac{BD}{DC}$

$B{D}^2=AD\cdot DC$

Answer:

Proof

5.3 [4]

$B{D}^2=9\times 16=144$

$BD=12$ cm

$A{B}^2=AD\cdot AC=9\times 25$

$AB=15$ cm

Answer:

$BD=12$ cm, $AB=15$ cm

5.4 [4]

Area $△ABD=\frac{1}{2}\times AD\times BD=\frac{1}{2}\times 9\times 12=54$ cm²

Also area $△ABD=\frac{1}{2}\times AB\times h=\frac{15}{2}h$

$\frac{15}{2}h=54$

$h=7,2$ cm

Answer:

7,2 cm

Question 6

6.1 [1]

$(3+1{)}^2+(5-2{)}^2=16+9=25$

Answer:

25 = 25

6.2 [4]

$m_{CA}=\frac{5-2}{3+1}=\frac{3}{4}$

$m_{\text{tan}}=-\frac{4}{3}$

$y-5=-\frac{4}{3}(x-3)$

$y=-\frac{4}{3}x+9$

Answer:

$y=-\frac{4}{3}x+9$

6.3 [3]

$B(0;9)$

$AB=\sqrt{3^2+4^2}$

$=5$

Answer:

$B(0;9)$, $AB=5$

6.4 [3]

$\widehat{CAB}=90°$ (tan $\perp$ radius)

Area $=\frac{1}{2}\times CA\times AB=\frac{1}{2}\times 5\times 5$

$=12,5$ square units

Answer:

12,5 square units

Question 7

7.1 [3]

$\cos 3x\cos x+\sin 3x\sin x=\cos (3x-x)$

$=\cos 2x$

At $x=30°$: $\cos 60°=\frac{1}{2}$

Answer:

$\cos 2x$; $\frac{1}{2}$

7.2 [6]

$3\sin x+2=1-2{\sin }^{2}x$

$2{\sin }^{2}x+3\sin x+1=0$

$(2\sin x+1)(\sin x+1)=0$

$\sin x=-\frac{1}{2}$: $x=210°$

or $x=330°$

$\sin x=-1$: $x=270°$

Answer:

$x=210°$, $270°$ or $330°$

7.3 [5]

$\sin 2x=2\sin x\cos x$

$1-\cos 2x=2{\sin }^{2}x$

$\frac{2\sin x\cos x}{2{\sin }^{2}x}=\frac{\cos x}{\sin x}=\frac{1}{\tan x}$

Undefined when $\sin x=0$ or $\cos x=0$ (where $\tan x$ is 0 or undefined)

For example $x=90°$, $180°$ or $270°$

Answer:

Proof; undefined at $x=90°$, $180°$, $270°$

Question 8

8.1 [3]

$Q{R}^2=85^2+120^2-2(85)(120)\cos 48°$

$QR\approx 89,30$ m

Answer:

About 89,30 m

8.2 [3]

$\sin \hat{R}=\frac{85\sin 48°}{89,30}$, so $\hat{R}\approx 45,02°$

$\widehat{PQR}=180°-48°-45,02°\approx 86,98°$

$\hat{R}$ is opposite a shorter side, so it must be acute; using the sine rule for $\hat{Q}$ would be ambiguous

Answer:

$\widehat{PRQ}\approx 45,02°$, $\widehat{PQR}\approx 86,98°$

8.3 [3]

Area $=\frac{1}{2}(85)(120)\sin 48°$

$\approx 3790,04$ m²

Answer:

About 3 790,04 m²

8.4 [3]

Area $=\frac{1}{2}\times QR\times h$

$h=\frac{2\times 3790,04}{89,30}$

$\approx 84,88$ m

Answer:

About 84,88 m

8.5 [6]

(a) $\sin C=\frac{10\sin 35°}{7}$

$\approx 0,819$

$\hat{C}\approx 55,02°$

or $\hat{C}\approx 124,98°$

(b) $35°+124,98°<180°$

so the obtuse angle also leaves a positive third angle

Answer:

$\hat{C}\approx 55,02°$ or $124,98°$

Question 9

9.1 [6]

$\cos (360°-x)=\cos x$

$\sin (x-180°)=-\sin x$

$\tan (-x)=-\tan x$

${\sin }^2(90°+x)={\cos }^{2}x$

$\frac{-\cos x\sin x}{-\frac{\sin x}{\cos x}{\cos }^{2}x}$

$=1$

Answer:

$1$

9.2 [4]

$\cos (45°-30°)=\cos 45°\cos 30°+\sin 45°\sin 30°$

$=\frac{1}{\sqrt{2}}\cdot \frac{\sqrt{3}}{2}+\frac{1}{\sqrt{2}}\cdot \frac{1}{2}$

$=\frac{\sqrt{3}+1}{2\sqrt{2}}$

$=\frac{\sqrt{6}+\sqrt{2}}{4}$

Answer:

$\frac{\sqrt{6}+\sqrt{2}}{4}\approx 0,9659$

9.3 [6]

(a) $\cos 2A=1-2{\sin }^{2}A$

$=1-\frac{50}{169}$

$=\frac{119}{169}$

(b) $\cos A=-\frac{12}{13}$ (quadrant 3)

$\sin A\cos 45°+\cos A\sin 45°=\frac{1}{\sqrt{2}}\left(-\frac{5}{13}-\frac{12}{13}\right)$

$=-\frac{17}{13\sqrt{2}}=-\frac{17\sqrt{2}}{26}$

Answer:

(a) $\frac{119}{169}\approx 0,704$ (b) $-\frac{17\sqrt{2}}{26}\approx -0,925$

Question 10

10.1 [5]

$\widehat{BAC}=\widehat{ACD}$ (alt $\angle$s; $AB\parallel DC$)

$\widehat{ACD}$ is subtended by chord $AD$ and $\widehat{BAC}$ by chord $BC$

Equal angles at the circumference stand on equal chords

$AD=BC$

Answer:

Proof

10.2 [8]

(a) $PQ=\frac{1}{2}BC$ (midpt theorem)

$=9$ cm

(b) $PQ\parallel BC$ (midpt theorem), so in $△ABR$, $PS\parallel BR$ through the midpoint $P$

$S$ is the midpoint of $AR$ and $PS=\frac{1}{2}BR$

$BR=6$, so $PS=3$ cm

$SQ=9-3=6$ cm

(c) $\frac{\text{area }△APS}{\text{area }△ABR}=\frac{\frac{1}{2}\cdot AP\cdot AS\sin \hat{A}}{\frac{1}{2}\cdot AB\cdot AR\sin \hat{A}}=\frac{AP}{AB}\cdot \frac{AS}{AR}$

$=\frac{1}{2}\times \frac{1}{2}=\frac{1}{4}$

Answer:

(a) 9 cm (b) $PS=3$ cm, $SQ=6$ cm (c) $\frac{1}{4}$

Question 11

11.1 [7]

$(x+1{)}^2+(x+k-2{)}^2=25$

$2x^2+2(k-1)x+(k-2{)}^2-24=0$

Two points: $\Delta >0$: $4(k-1{)}^2-8[(k-2{)}^2-24]>0$

$k^2-6k-41<0$

Critical values $k=3\pm 5\sqrt{2}$

$3-5\sqrt{2}<k<3+5\sqrt{2}$

Answer:

$3-5\sqrt{2}<k<3+5\sqrt{2}$