IEB-style Grade 12 Mathematics Paper 2: Practice Set 1 memo

Question 1

1.1 [2]

$\bar{x}=\frac{48}{12}=4$ hours

Median $=\frac{3,5+4}{2}=3,75$ hours

Answer:

Mean 4 hours; median 3,75 hours

1.2 [2]

Calculator: $\sigma \approx 1,94$

Answer:

$\sigma \approx 1,94$ hours

1.3 [3]

Minimum 1,5 and maximum 9

$Q_1=2,75$

$Q_3=4,75$ (median 3,75)

Answer:

$(1,5;2,75;3,75;4,75;9)$

1.4 [3]

IQR $=4,75-2,75=2$

Upper fence $=4,75+1,5\times 2=7,75$

$9>7,75$, so 9 is an outlier

Answer:

Upper fence 7,75

1.5 [2]

$\frac{48+2t}{14}=4\Rightarrow t=4$ hours

Both new values equal the mean, so the sum of squared deviations is unchanged while $n$ increases from 12 to 14: the standard deviation decreases

Answer:

4 hours each; the standard deviation decreases

Question 2

2.1 [3]

$a\approx -34,74$

$b\approx 8,53$

$\hat{y}=-34,74+8,53x$

Answer:

$\hat{y}=-34,74+8,53x$

2.2 [2]

$r\approx 0,999$

Very strong positive correlation

Answer:

$r\approx 0,999$; very strong positive

2.3 [2]

$\hat{y}=-34,74+8,53(30)$

$\approx 221$ cold drinks

Answer:

About 221

2.4 [1]

45 °C is far outside the data (extrapolation), so the trend may not continue

Answer:

Extrapolation

Question 3

3.1 [2]

$\widehat{BOD}=2\times 64°$ ($\angle$ at centre $=2\times \angle$ at circumference)

$=128°$

Answer:

128°

3.2 [2]

$\widehat{BCD}=180°-64°$ (opp $\angle$s of cyclic quad)

$=116°$

Answer:

116°

3.3 [2]

$OB=OD$ (radii), so $\widehat{OBD}=\widehat{ODB}$

$\widehat{OBD}=\frac{1}{2}(180°-128°)=26°$

Answer:

26°

3.4 [3]

$AB=AD$, so $\widehat{ABD}=\widehat{ADB}$ ($\angle$s opp equal sides)

$\widehat{ABD}=\frac{1}{2}(180°-64°)=58°$

$\widehat{ACB}=\widehat{ADB}=58°$ ($\angle$s in the same seg)

Answer:

$\widehat{ABD}=58°$, $\widehat{ACB}=58°$

3.5 [5]

Construction: join $B$ and $D$ to the centre

${\hat{O}}_1=2\hat{A}$

${\hat{O}}_2=2\hat{C}$ ($\angle$ at centre $=2\times \angle$ at circumference)

${\hat{O}}_1+{\hat{O}}_2=360°$ ($\angle$s round a pt)

$\hat{A}+\hat{C}=180°$

Answer:

Proof

Question 4

4.1 [2]

$AB=\sqrt{8^2+4^2}=\sqrt{80}$

$=4\sqrt{5}$

Answer:

$4\sqrt{5}\approx 8,94$

4.2 [2]

$\left(\frac{-3+8}{2};\frac{4+2}{2}\right)$

$M(2,5;3)$

Answer:

$M(2,5;3)$

4.3 [3]

$m_{AB}=\frac{8-4}{5+3}=\frac{1}{2}$

$m_{BC}=\frac{2-8}{8-5}=-2$

$m_{AB}\times m_{BC}=-1$, so $AB\perp BC$

Answer:

$m_{AB}\times m_{BC}=-1$

4.4 [4]

$\hat{B}=90°$, so $AC$ is a diameter (chord subtends $90°$)

Centre $M(2,5;3)$

$r^2=(8-2,5{)}^2+(2-3{)}^2=31,25$

$(x-2,5{)}^2+(y-3{)}^2=31,25$

Answer:

$(x-2,5{)}^2+(y-3{)}^2=31,25$

4.5 [3]

The diagonals of a rectangle bisect each other, so $M$ is also the midpoint of $BD$

$\frac{5+x}{2}=2,5\Rightarrow x=0$

$\frac{8+y}{2}=3\Rightarrow y=-2$: $D(0;-2)$

Answer:

$D(0;-2)$

4.6 [3]

$BC=\sqrt{9+36}=\sqrt{45}$

Area $=\sqrt{80}\times \sqrt{45}$

$=60$ square units

Answer:

60 square units

Question 5

5.1 [2]

$\widehat{ACB}=x$

(tan chord theorem)

Answer:

$\widehat{ACB}=x$

5.2 [3]

$\hat{P}$ is common

$\widehat{PAB}=\widehat{PCA}$ (tan chord theorem)

$△PAB|||△PCA$ (equiangular $△$s)

Answer:

Proof

5.3 [2]

$\frac{PA}{PC}=\frac{PB}{PA}$ ($|||$ $△$s)

$P{A}^2=PB\cdot PC$

Answer:

Proof

5.4 [3]

$P{A}^2=4\times 9=36$, so $PA=6$ cm

$\frac{AB}{CA}=\frac{PA}{PC}$ ($|||$ $△$s)

$AB:AC=6:9=2:3$

Answer:

$PA=6$ cm; $AB:AC=2:3$

5.5 [4]

In $△PAC$: $\widehat{PCA}=x=40°$

$\widehat{PAC}=180°-30°-40°=110°$ (sum of $\angle$s in $△$)

$\widehat{BAC}=\widehat{PAC}-\widehat{PAB}$

$=110°-40°=70°$

Answer:

$\widehat{PAC}=110°$, $\widehat{BAC}=70°$

Question 6

6.1 [1]

$180°-50°-70°=60°$

Answer:

60°

6.2 [3]

$\frac{BF}{\sin 50°}=\frac{60}{\sin 60°}$

$BF=\frac{60\sin 50°}{\sin 60°}$

$\approx 53,07$ m

Answer:

About 53,07 m

6.3 [3]

$\tan 32°=\frac{TF}{BF}$

$TF=53,07\tan 32°$

$\approx 33,16$ m

Answer:

About 33,16 m

6.4 [3]

The closest point is the foot of the perpendicular from $F$ to $AB$; its distance from $F$ is $BF\sin 70°$

$=53,07\sin 70°\approx 49,87$ m

$\tan \theta =\frac{33,16}{49,87}\Rightarrow \theta \approx 33,62°$

Answer:

About 33,62°

6.5 [3]

$AF=\frac{60\sin 70°}{\sin 60°}\approx 65,10$ m

$T{A}^2=A{F}^2+T{F}^2$

$TA\approx 73,06$ m

Answer:

About 73,06 m

Question 7

7.1 [4]

$\sin 210°=-\sin 30°=-\frac{1}{2}$

$\tan 315°=-\tan 45°=-1$

$\cos 120°=-\cos 60°=-\frac{1}{2}$

$\frac{\left(-\frac{1}{2}\right)(-1)}{-\frac{1}{2}}=-1$

Answer:

$-1$

7.2 [4]

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

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

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

$=\frac{\sin x}{\cos x}=\tan x$

Answer:

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

7.3 [2]

$\sin 2x=0$ (this also covers $\cos x=0$, where $\tan x$ is undefined)

$x=k\cdot 90°$, $k\in \mathbb{Z}$

Answer:

$x=k\cdot 90°$, $k\in \mathbb{Z}$

7.4 [4]

$\sin (180°+x)=-\sin x$ and $\cos (90°-x)=\sin x$

${\cos }^2(-x)={\cos }^{2}x$

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

$=\cos 2x$

Answer:

$\cos 2x$

7.5 [4]

$\sin 74°=2\sin 37°\cos 37°$

$\sin 37°=\sqrt{1-p^2}$ (from a sketch or the square identity)

$\sin 74°=2p\sqrt{1-p^2}$

Answer:

$2p\sqrt{1-p^2}$

Question 8

8.1 [3]

$\frac{AD}{DB}=\frac{AE}{EC}$ (line $\parallel$ one side of $△$; $DE\parallel BC$)

$\frac{6}{4}=\frac{7,5}{EC}$

$EC=5$ cm

Answer:

5 cm

8.2 [3]

$\frac{BF}{FC}=\frac{AE}{EC}=\frac{3}{2}$ (line $\parallel$ one side of $△$; $EF\parallel AB$)

$BF=\frac{3}{5}\times 20$

$=12$ cm

Answer:

12 cm

8.3 [4]

$\hat{A}$ is common

$\widehat{ADE}=\widehat{ABC}$ (corresp $\angle$s; $DE\parallel BC$)

$\widehat{AED}=\widehat{ACB}$ (corresp $\angle$s; $DE\parallel BC$)

$△ADE|||△ABC$ (equiangular $△$s)

Answer:

Proof

8.4 [3]

$\frac{DE}{BC}=\frac{AD}{AB}=\frac{6}{10}$ ($|||$ $△$s)

Area $=\frac{1}{2}ab\sin A$, and both sides scale by $\frac{3}{5}$

Ratio $={\left(\frac{3}{5}\right)}^2=\frac{9}{25}$

Answer:

$\frac{9}{25}=0,36$

Question 9

9.1 [4]

$x^2-4x+4+y^2+6y+9=12+13$

$(x-2{)}^2+(y+3{)}^2=25$

$N(2;-3)$, radius 5

Answer:

$N(2;-3)$, $r=5$

9.2 [1]

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

Answer:

25 = 25

9.3 [4]

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

$m_{\text{tan}}=-\frac{3}{4}$ (tan $\perp$ radius)

$y-1=-\frac{3}{4}(x-5)$

$y=-\frac{3}{4}x+\frac{19}{4}$

Answer:

$y=-\frac{3}{4}x+\frac{19}{4}$

9.4 [4]

$Q{N}^2=(10-2{)}^2+(3+3{)}^2=100$

Tangent $\perp$ radius, so tangent${}^2=Q{N}^2-r^2$

$=75$

Length $=5\sqrt{3}$

Answer:

$5\sqrt{3}\approx 8,66$

9.5 [3]

Horizontal tangents touch at the top and bottom of the circle

$k=-3+5=2$

or $k=-3-5=-8$

Answer:

$k=2$ or $k=-8$

9.6 [4]

The line cuts the circle twice when its distance from $N(2;-3)$ is less than the radius 5, or substitute and use $\Delta >0$

$(x-2{)}^2+(x+c+3{)}^2=25\Rightarrow 2x^2+(2c+2)x+(c+3{)}^2-21=0$

$\Delta =(2c+2{)}^2-8\left[(c+3{)}^2-21\right]=-4(c^2+10c-25)>0$

$c^2+10c-25<0\Rightarrow -5-5\sqrt{2}<c<-5+5\sqrt{2}$

Answer:

$-5-5\sqrt{2}<c<-5+5\sqrt{2}$ (about $-12,07<c<2,07$)

9.7 [3]

The centres and $P$ are collinear and $P$ is the midpoint of $N{N}^{\prime }$ (both radii are 5)

$N^{\prime }=(2\times 5-2;2\times 1+3)=(8;5)$

$(x-8{)}^2+(y-5{)}^2=25$

Answer:

$(x-8{)}^2+(y-5{)}^2=25$

Question 10

10.1 [2]

Period 180°

Amplitude 1

Answer:

Period 180°, amplitude 1

10.2 [4]

$f$: period 180°, amplitude 1

$f$ through $(0;1)$, $(\pm 90°;-1)$, $(\pm 180°;1)$

$g$: standard sine curve

Intersections at $-90°$, $30°$, $150°$

Answer:

Sketch

10.3 [6]

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

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

$\sin x=\frac{1}{2}$: $x=30°+k\cdot 360°$

or $x=150°+k\cdot 360°$

$\sin x=-1$: $x=270°+k\cdot 360°$

$k\in \mathbb{Z}$

Answer:

$x=30°+k\cdot 360°$, $150°+k\cdot 360°$ or $270°+k\cdot 360°$, $k\in \mathbb{Z}$

10.4 [3]

$-180°\le x\le 30°$

or $150°\le x\le 180°$

(at $x=-90°$ the graphs touch, so it is included)

Answer:

$-180°\le x\le 30°$ or $150°\le x\le 180°$

10.5 [3]

$f$ completes two full periods on $[-180°;180°]$

$y=k$ cuts $f$ four times for $-1<k<1$ and three times for $k=1$ (at $-180°$, $0°$ and $180°$)

Exactly two solutions only when $k=-1$ ($x=\pm 90°$)

Answer:

$k=-1$