IEB-style Grade 11 Mathematics Paper 2: Practice Set 1 memo

Question 1

1.1 [2]

$\frac{620}{10}$

$=62\%$

Answer:

62%

1.2 [2]

Calculator ${\sigma }_x$

$\approx 11,88$

Answer:

$\sigma \approx 11,88$

1.3 [3]

New mean $=62+5=67\%$

Standard deviation unchanged: $\sigma \approx 11,88$

Every mark and the mean move up by 5, so each deviation $x-\bar{x}$ stays the same

Answer:

Mean 67%; $\sigma \approx 11,88$ (unchanged)

1.4 [3]

Minimum 42, maximum 85

Median $=\frac{60+63}{2}=61,5$

$Q_1=55$, $Q_3=70$

Answer:

$(42;55;61,5;70;85)$

1.5 [2]

Box from 55 to 70 with median at 61,5

Whiskers to 42 and 85, on a labelled scale

Answer:

Box plot

1.6 [2]

Slightly positively skewed

Mean (62) $>$ median (61,5); $Q_3-$ median $=8,5>$ median $-Q_1=6,5$

Answer:

Slightly positively skewed

Question 2

2.1 [3]

Midpoints 1 ; 3 ; 5 ; 7 ; 9

$\sum fx=128$

$\bar{x}\approx 4,27$ kg

Answer:

About 4,27 kg

2.2 [1]

$4<m\le 6$

Answer:

$4<m\le 6$

2.3 [2]

Cumulative frequencies 5 ; 14 ; 24

The 15th and 16th values lie in $4<m\le 6$

Answer:

$4<m\le 6$

Question 3

3.1 [1]

$m_{AB}=\frac{2+1}{4+2}=\frac{1}{2}$

Answer:

$\frac{1}{2}=0,5$

3.2 [3]

$m_{AC}=\frac{-5+1}{0+2}=-2$

$m_{AB}\times m_{AC}=\frac{1}{2}\times (-2)=-1$

So $AB\perp AC$

Answer:

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

3.3 [4]

$AB=\sqrt{36+9}=3\sqrt{5}$

$AC=\sqrt{4+16}=2\sqrt{5}$

Area $=\frac{1}{2}\times 3\sqrt{5}\times 2\sqrt{5}$

$=15$ square units

Answer:

15 square units

3.4 [2]

$\left(\frac{4+0}{2};\frac{2-5}{2}\right)$

$M(2;-1,5)$

Answer:

$M(2;-1,5)$

3.5 [3]

$m=\frac{-1,5+1}{2+2}=-\frac{1}{8}$

$y+1=-\frac{1}{8}(x+2)$

$y=-\frac{1}{8}x-\frac{5}{4}$

Answer:

$y=-\frac{1}{8}x-\frac{5}{4}$

3.6 [2]

$\tan \theta =\frac{1}{2}$

$\theta \approx 26,57°$

Answer:

About 26,57°

Question 4

4.1 [5]

$\cos 300°=\cos 60°=\frac{1}{2}$

$\tan 135°=-\tan 45°=-1$

$\sin (-30°)=-\frac{1}{2}$

$\frac{\frac{1}{2}(-1)}{-\frac{1}{2}}$

$=1$

Answer:

$1$

4.2 [4]

$\cos (90°-x)=\sin x$

$\cos (180°+x)=-\cos x$

$\sin (-x)=-\sin x$

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

Answer:

$\cos x$

4.3 [5]

(a) $\sin \theta =-\frac{4}{5}$, quadrant 3: $x=-3$, $r=5$

$\cos \theta =-\frac{3}{5}$

(b) $\tan \theta =\frac{4}{3}$

$\cos (180°-\theta )=-\cos \theta =\frac{3}{5}$

$\frac{4}{3}+\frac{3}{5}=\frac{29}{15}$

Answer:

(a) $-\frac{3}{5}$ (b) $\frac{29}{15}\approx 1,93$

4.4 [4]

$\frac{(1-\sin x)(1+\sin x)}{1-\sin x}$

$=1+\sin x$

Undefined when $\sin x=1$

$x=90°+k\cdot 360°$, $k\in \mathbb{Z}$

Answer:

Undefined for $x=90°+k\cdot 360°$

Question 5

5.1 [5]

Given $M$ the midpoint of $AB$; join $OA$, $OB$ and $OM$

$OA=OB$ (radii); $AM=MB$ (given); $OM$ common

$△OMA\equiv △OMB$ (SSS)

${\hat{M}}_1={\hat{M}}_2$ ($\equiv$ triangles)

${\hat{M}}_1+{\hat{M}}_2=180°$ ($\angle$s on a str line), so ${\hat{M}}_1={\hat{M}}_2=90°$

Answer:

Proof

5.2 [2]

$OM\perp AB$ (line from centre to midpt of chord); $AM=12$

$OM=\sqrt{169-144}=5$ cm

Answer:

5 cm

5.3 [5]

$ON=\sqrt{169-25}$

$=12$ cm

$O$, $M$ and $N$ lie on one line perpendicular to both chords

Same side of $O$, so distance $=ON-OM$

$=7$ cm

Answer:

7 cm

Question 6

6.1 [3]

$\hat{A}=65°$

$b=\frac{12\sin 40°}{\sin 65°}$

$\approx 8,51$ cm

Answer:

About 8,51 cm

6.2 [4]

Opposite the 11 cm side

$\cos \theta =\frac{36+64-121}{2(6)(8)}$

$=-\frac{21}{96}$

$\theta \approx 102,64°$

Answer:

About 102,64°

6.3 [2]

$\frac{1}{2}(6)(8)\sin 102,64°$

$\approx 23,42$ cm²

Answer:

About 23,42 cm²

6.4 [5]

$\sin C=\frac{12\sin 42°}{9}\approx 0,892$

$\hat{C}\approx 63,15°$

or $\hat{C}\approx 116,85°$

$42°+116,85°<180°$

so both triangles exist

Answer:

$\hat{C}\approx 63,15°$ or $116,85°$

Question 7

7.1 [3]

$M_{PR}=(2;2)$

$M_{QS}=(2;2)$

The diagonals bisect each other

Answer:

Common midpoint $(2;2)$

7.2 [2]

$P{Q}^2=16+4=20$ and $Q{R}^2=4+16=20$

Adjacent sides equal, so a rhombus

Answer:

$P{Q}^2=Q{R}^2=20$

7.3 [2]

$m_{PQ}=\frac{1}{2}$, $m_{QR}=-2$; product $-1$

A rhombus with a right angle is a square

Answer:

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

7.4 [3]

Line $PQ$: $y-3=\frac{1}{2}(x+1)$

$7-3=\frac{1}{2}(k+1)$

$k=7$

Answer:

$k=7$

7.5 [2]

$\tan \theta =-2$, reference angle $63,43°$

$\theta \approx 116,57°$

Answer:

About 116,57°

7.6 [3]

$m_{PR}=\frac{1-3}{5+1}=-\frac{1}{3}$, so the inclination of $PR$ is $180°-18,43°=161,57°$

Inclination of $PQ$ is $26,57°$; $161,57°-26,57°=135°$, so the acute angle is $45°$

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

Answer:

45°

Question 8

8.1 [2]

$\widehat{ABC}=180°-112°$ (opp $\angle$s of cyclic quad)

$=68°$

Answer:

68°

8.2 [2]

$\widehat{AOC}=2\times 68°$ ($\angle$ at centre $=2\times \angle$ at circumference)

$=136°$

Answer:

136°

8.3 [2]

$OA=OC$ (radii)

$\widehat{OAC}=\frac{1}{2}(180°-136°)=22°$

Answer:

22°

8.4 [3]

$\widehat{ACT}=\widehat{ABC}$ (tan chord theorem)

$=68°$

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

Answer:

$68°$ and $90°$

8.5 [5]

$\widehat{DAC}=\widehat{DCA}$ ($\angle$s opp equal sides)

$\widehat{DAC}=\frac{1}{2}(180°-112°)=34°$ (sum of $\angle$s in $△$)

$\widehat{DCT}=\widehat{DAC}=34°$ (tan chord theorem)

$\widehat{DCA}=34°$

$\widehat{DCT}=\widehat{DCA}$, so $CD$ bisects $\widehat{ACT}$

Answer:

$\widehat{DAC}=34°$; $\widehat{DCT}=\widehat{DCA}=34°$

Question 9

9.1 [4]

Cylinder: $\pi (3{)}^2(10)=90\pi$

Hemisphere: $\frac{1}{2}\cdot \frac{4}{3}\pi (3{)}^3=18\pi$

Total $=108\pi$

$\approx 339,29$ m³

Answer:

About 339,29 m³

9.2 [4]

Curved surface of the cylinder: $2\pi (3)(10)=60\pi$

Hemisphere: $\frac{1}{2}\cdot 4\pi (3{)}^2=18\pi$

Total $=78\pi$

$\approx 245,04$ m²

Answer:

About 245,04 m²

9.3 [2]

1 m³ = 1 000 litres

$\approx 339292$ litres

Answer:

About 339 292 litres

9.4 [2]

Volume scales by $k^3$

$2^3=8$

Answer:

8 times larger

Question 10

10.1 [3]

$\widehat{OAP}=90°$ and $\widehat{OBP}=90°$ (tan $\perp$ radius)

$\widehat{OAP}+\widehat{OBP}=180°$

$PAOB$ is cyclic (converse opp $\angle$s of cyclic quad)

Answer:

Proof

10.2 [6]

$\widehat{AOB}=180°-54°$ (opp $\angle$s of cyclic quad)

$=126°$

$\widehat{ACB}=\frac{1}{2}\times 126°$ ($\angle$ at centre $=2\times \angle$ at circumference)

$=63°$

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

$=63°$

Answer:

$126°$, $63°$, $63°$

10.3 [3]

$PA=PB$ (tans from common pt) and $OA=OB$ (radii)

So both $P$ and $O$ are equidistant from $A$ and $B$, and lie on the perpendicular bisector of $AB$

Therefore $OP\perp AB$

Answer:

Proof

Question 11

11.1 [6]

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

$\cos x=\frac{1}{2}$: $x=\pm 60°+k\cdot 360°$

$\cos x=-1$: $x=180°+k\cdot 360°$

$k\in \mathbb{Z}$

Answer:

$x=\pm 60°+k\cdot 360°$ or $x=180°+k\cdot 360°$, $k\in \mathbb{Z}$

11.2 [4]

$f$: amplitude 2, period 360°

$g$: period 180°, range $0\le y\le 2$

$g$ has maximums of 2 at $0°$, $180°$, $360°$ and minimums of 0 at $90°$, $270°$

Both graphs labelled

Answer:

Sketch

11.3 [4]

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

${\sin }^{2}x+\sin x-1=0\Rightarrow \sin x=\frac{-1\pm \sqrt{5}}{2}$

$\sin x=\frac{-1-\sqrt{5}}{2}<-1$ has no solution, so $\sin x=\frac{\sqrt{5}-1}{2}\approx 0,618$

$x\approx 38,17°$ or $x\approx 141,83°$

Answer:

$x\approx 38,17°$ or $x\approx 141,83°$

11.4 [4]

The largest angle is opposite the longest side: $(x+2{)}^2=x^2+(x+1{)}^2-2x(x+1)\cos 120°$

$x^2+4x+4=2x^2+2x+1+x^2+x=3x^2+3x+1$

$2x^2-x-3=0\Rightarrow (2x-3)(x+1)=0$

$x=1,5$ cm ($x=-1$ is not a length)

Answer:

$x=1,5$ cm