IEB-style Grade 11 Mathematics Paper 2: Practice Set 2 memo

Question 1

1.1 [2]

3 ; 12 ; 30

44 ; 50

Answer:

3 ; 12 ; 30 ; 44 ; 50

1.2 [3]

Start at $(140;0)$

Points at upper boundaries: $(150;3)$, $(160;12)$, $(170;30)$, $(180;44)$, $(190;50)$

Smooth curve, labelled axes

Answer:

Ogive

1.3 [2]

Read at cumulative frequency 25

$\approx 167,2$ cm

Answer:

About 167,2 cm

1.4 [3]

$Q_1$ at 12,5: $\approx 160,3$ cm

$Q_3$ at 37,5

$\approx 175,4$ cm

Answer:

About 160,3 cm and 175,4 cm

1.5 [2]

At 175 cm, cumulative frequency $\approx 37$, so about 13 learners are taller

$\frac{13}{50}=26\%$

Answer:

About 26%

1.6 [3]

$Q_1=5,5$ and $Q_3=8,5$, so IQR $=3$

Upper fence $=8,5+1,5\times 3=13$

$20>13$, so 20 is an outlier

Answer:

Upper fence 13

1.7 [2]

$\bar{x}=8$

$\sigma \approx 4,85$

Answer:

$\sigma \approx 4,85$

1.8 [2]

New mean $=\frac{44}{7}\approx 6,29$

The standard deviation decreases (to about 1,83), because the value furthest from the mean has been removed

Answer:

Mean about 6,29; the standard deviation decreases

Question 2

2.1 [2]

$\left(\frac{1-3}{2};\frac{6-2}{2}\right)$

$M(-1;2)$

Answer:

$M(-1;2)$

2.2 [4]

$m_{AB}=\frac{6+2}{1+3}=2$

$m_{\perp }=-\frac{1}{2}$

$y-2=-\frac{1}{2}(x+1)$

$y=-\frac{1}{2}x+\frac{3}{2}$

Answer:

$y=-\frac{1}{2}x+\frac{3}{2}$

2.3 [3]

$-\frac{1}{2}(5)+\frac{3}{2}=-1$ ✓

Every point on the perpendicular bisector is equidistant from $A$ and $B$

$CA=\sqrt{16+49}=\sqrt{65}$ and $CB=\sqrt{64+1}=\sqrt{65}$

Answer:

$CA=CB=\sqrt{65}$

2.4 [2]

$\tan \theta =2$

$\theta \approx 63,43°$

Answer:

About 63,43°

2.5 [4]

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

Inclination of $BC\approx 7,13°$

$\widehat{ABC}=63,43°-7,13°$

$\approx 56,31°$

Answer:

About 56,31°

Question 3

3.1 [2]

$\widehat{ADC}=\widehat{CBE}$ (ext $\angle$ of cyclic quad)

$=80°$

Answer:

80°

3.2 [2]

$\widehat{ABD}=\widehat{ACD}$ ($\angle$s in the same seg)

$=30°$

Answer:

30°

3.3 [2]

$\widehat{ABD}+\widehat{DBC}+\widehat{CBE}=180°$ ($\angle$s on a str line)

$\widehat{DBC}=70°$

Answer:

70°

3.4 [2]

$\widehat{DAC}=\widehat{DBC}$ ($\angle$s in the same seg)

$=70°$

Answer:

70°

3.5 [5]

Construction: draw diameter $AOD$ and join $DB$

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

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

$\widehat{TAB}=90°-\widehat{BAD}$ and $\widehat{ADB}=90°-\widehat{BAD}$ (sum of $\angle$s in $△ABD$), so $\widehat{TAB}=\widehat{ADB}$

$\widehat{ADB}=\widehat{ACB}$ ($\angle$s in the same seg), so $\widehat{TAB}=\widehat{ACB}$

Answer:

$\widehat{TAB}=\widehat{ADB}=\widehat{ACB}$

Question 4

4.1 [5]

$\sin 135°=\frac{1}{\sqrt{2}}$, so ${\sin }^{2}135°=\frac{1}{2}$

$\cos 240°=-\frac{1}{2}$

$\tan 300°=-\sqrt{3}$

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

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

Answer:

$\frac{1+\sqrt{3}}{2}\approx 1,37$

4.2 [5]

$\sin (360°-x)=-\sin x$; $\tan (-x)=-\tan x$

$\cos (180°+x)=-\cos x$; $\tan (180°-x)=-\tan x$

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

$=\tan x$

Answer:

$\tan x$

4.3 [3]

$\tan 65°=\frac{\sin 65°}{\cos 65°}=\frac{\cos 25°}{\sin 25°}$ (co-functions)

$\cos 25°=\sqrt{1-t^2}$ (square identity; $25°$ is acute, so the positive root)

$\tan 65°=\frac{\sqrt{1-t^2}}{t}$

Answer:

$\frac{\sqrt{1-t^2}}{t}$

Question 5

5.1 [2]

$\frac{1}{3}(9)(2)$

$=6$ m³

Answer:

6 m³

5.2 [3]

From the centre of the base to the middle of a side is 1,5 m

$s=\sqrt{2^2+1,5^2}$

$=2,5$ m

Answer:

2,5 m

5.3 [3]

Four triangles, no floor

$4\times \frac{1}{2}(3)(2,5)$

$=15$ m²

Answer:

15 m²

5.4 [2]

$\frac{1}{3}\pi (1,5{)}^2(2)$

$\approx 4,71$ m³

Answer:

About 4,71 m³

5.5 [2]

Area scales by $k^2$

$1,5^2=2,25$

Answer:

2,25

Question 6

6.1 [4]

$m_{EF}=\frac{8-4}{7-1}=\frac{2}{3}$

$m_{DE}=\frac{4-k}{3}$

$\frac{4-k}{3}=\frac{2}{3}$

$k=2$

Answer:

$k=2$

6.2 [3]

$m=\frac{2}{3}$

$y-8=\frac{2}{3}(x-7)$

$y=\frac{2}{3}x+\frac{10}{3}$

Answer:

$y=\frac{2}{3}x+\frac{10}{3}$

6.3 [4]

$m_{EG}=-\frac{3}{2}$

$y-4=-\frac{3}{2}(x-1)$

At $x=0$: $y=4+\frac{3}{2}$

$G(0;5,5)$

Answer:

$G(0;5,5)$

6.4 [4]

$EG=\sqrt{1+2,25}\approx 1,80$

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

$\widehat{GEF}=90°$, so area $=\frac{1}{2}\times EG\times EF$

$=\frac{1}{2}\sqrt{3,25\times 52}=6,5$ square units

Answer:

$EG\approx 1,80$; area 6,5 square units

Question 7

7.1 [5]

$PM=8$ cm (line from centre $\perp$ chord)

Let the radius be $r$: $OM=r-4$

$r^2=8^2+(r-4{)}^2$

$0=64-8r+16$

$r=10$ cm

Answer:

10 cm

7.2 [1]

$90°$ ($\angle$ in semi-circle)

Answer:

90°

7.3 [5]

$R{P}^2=R{M}^2+P{M}^2=16+64$

$RP=\sqrt{80}=4\sqrt{5}$

$RS=20$

$P{S}^2=400-80=320$

$PS=8\sqrt{5}$

Answer:

$RP=4\sqrt{5}\approx 8,94$ cm, $PS=8\sqrt{5}$ cm

Question 8

8.1 [3]

$A{C}^2=49+81-2(7)(9)\cos 120°$

$=193$

$AC\approx 13,89$ m

Answer:

About 13,89 m

8.2 [3]

$\frac{\sin \widehat{BAC}}{9}=\frac{\sin 120°}{13,89}$

$\sin \widehat{BAC}\approx 0,561$

$\widehat{BAC}\approx 34,13°$

Answer:

About 34,13°

8.3 [3]

$A{D}^2=13,89^2+6^2-2(13,89)(6)\cos 50°$

$AD\approx 11,04$ m

Answer:

About 11,04 m

8.4 [5]

Area $△ABC=\frac{1}{2}(7)(9)\sin 120°$

$\approx 27,28$ m²

Area $△ACD=\frac{1}{2}(13,89)(6)\sin 50°$

$\approx 31,93$ m²

Total $\approx 59,21$ m²

Answer:

About 59,21 m²

8.5 [5]

$\widehat{BCA}=180°-120°-34,13°=25,87°$ (sum of $\angle$s in $△$)

$\widehat{BCD}=25,87°+50°=75,87°$

$B{D}^2=9^2+6^2-2(9)(6)\cos 75,87°$

$BD\approx 9,52$ m

Answer:

About 9,52 m

Question 9

9.1 [2]

$\frac{180°}{3}$

$=60°$

Answer:

60°

9.2 [4]

Reference angle $50,19°$

$x-20°=-50,19°+k\cdot 180°$

$x=149,81°$

or $x=329,81°$

Answer:

$x\approx 149,81°$ or $x\approx 329,81°$

9.3 [4]

$f$: sine curve shifted 30° right; $x$-intercepts $-150°$ and $30°$

$f$: maximum $(120°;1)$, minimum $(-60°;-1)$

$g$: standard cosine curve

Labels and endpoints

Answer:

Sketch

9.4 [4]

$\cos x=\sin (90°-x)$, so $\sin (x-30°)=\sin (90°-x)$

$x-30°=90°-x+k\cdot 360°\Rightarrow x=60°+k\cdot 180°$

The other case gives $x-30°=90°+x$, which has no solution

$x=60°$ or $x=-120°$

Answer:

$x=-120°$ or $x=60°$

9.5 [5]

LHS $=\frac{\sin x}{\cos x}+\frac{\cos x}{1+\sin x}$

$=\frac{\sin x(1+\sin x)+{\cos }^{2}x}{\cos x(1+\sin x)}$

$=\frac{\sin x+{\sin }^{2}x+{\cos }^{2}x}{\cos x(1+\sin x)}$

$=\frac{1+\sin x}{\cos x(1+\sin x)}$

$=\frac{1}{\cos x}=$ RHS

Answer:

Proof

Question 10

10.1 [6]

$\widehat{TAB}=\widehat{ACB}$ (tan chord theorem)

$\widehat{TAB}=\widehat{ADE}$ (alt $\angle$s; $TA\parallel DE$)

So $\widehat{ADE}=\widehat{ECB}$

Exterior angle equals interior opposite angle: $BDEC$ is cyclic (converse ext $\angle$ of cyclic quad)

Answer:

Proof

10.2 [8]

$\widehat{ACB}=65°$ (tan chord theorem)

$\widehat{ABC}=180°-50°-65°=65°$

$\widehat{ABC}=\widehat{ACB}$

so $AB=AC$ (sides opp equal $\angle$s)

$\widehat{ADE}=\widehat{TAB}=65°$ (alt $\angle$s)

$\widehat{AED}=\widehat{ABC}=65°$ (ext $\angle$ of cyclic quad $BDEC$)

Answer:

$65°$, $65°$, $65°$, $65°$; $AB=AC$