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

Question 1

1.1.1 [3]

$\frac{3^{n+2}\cdot 3^{2n}}{3^{3n}}$

$=3^{n+2+2n-3n}=3^2$

$=9$

Answer:

$9$

1.1.2 [2]

$3+4\sqrt{3}+4$

$=7+4\sqrt{3}$

Answer:

$7+4\sqrt{3}\approx 13,93$

1.1.3 [2]

$\frac{6\sqrt{3}}{3}$

$=2\sqrt{3}$

Answer:

$2\sqrt{3}\approx 3,46$

1.1.4 [3]

$\frac{2\sqrt{3}+5\sqrt{3}}{3\sqrt{3}}$

$=\frac{7}{3}$

Answer:

$\frac{7}{3}\approx 2,33$

1.2.1 [2]

$x(x-7)=0$

$x=0$ or $x=7$

Answer:

$x=0$ or $x=7$

1.2.2 [3]

$x^2+3x-18=0$

$(x+6)(x-3)=0$

$x=-6$ or $x=3$

Answer:

$x=-6$ or $x=3$

1.2.3 [4]

$x^2+6x=-\frac{5}{2}$

$(x+3{)}^2=\frac{13}{2}$

$x=-3\pm \sqrt{6,5}$

$x\approx -0,45$ or $x\approx -5,55$

Answer:

$x\approx -0,45$ or $x\approx -5,55$

1.3 [4]

$x^2-2x=2x-3\Rightarrow x^2-4x+3=0$

$(x-1)(x-3)=0$

$(1;-1)$

$(3;3)$

Answer:

$(1;-1)$ and $(3;3)$

Question 2

2.1 [4]

First differences $4;6;8$; second difference 2

$a=1$

$3+b=4\Rightarrow b=1$; $c=1$

$T_n=n^2+n+1$

Answer:

$T_n=n^2+n+1$

2.2 [1]

$T_{30}=900+30+1=931$

Answer:

931

2.3 [3]

Second difference 2, $a=1$

$3+b=4\Rightarrow b=1$; $c=0$

$T_n=n^2+n$

Answer:

$T_n=n^2+n$

2.4 [2]

$n^2+n-420=0\Rightarrow (n+21)(n-20)=0$

Figure 20

Answer:

Figure 20

Question 3

3.1 [3]

$f(x)=a(x+1)(x-5)$

$-5=a(1)(-5)\Rightarrow a=1$

$f(x)=x^2-4x-5$

Answer:

$f(x)=x^2-4x-5$

3.2 [3]

$(x-2{)}^2-4-5$

$=(x-2{)}^2-9$

Turning point $(2;-9)$

Answer:

$(x-2{)}^2-9$; $(2;-9)$

3.3 [2]

$\frac{0-(-9)}{5-2}$

$=3$

Answer:

$3$

3.4 [3]

Decreasing for $x<2$

$x^2-4x-12\ge 0\Rightarrow (x-6)(x+2)\ge 0$

$x\le -2$ or $x\ge 6$

Answer:

$x<2$; $x\le -2$ or $x\ge 6$

Question 4

4.1 [4]

$q=-4$

$-2=a-4\Rightarrow a=2$

$0=2b-4$

$b=2$

Answer:

$a=2$, $b=2$, $q=-4$

4.2 [3]

$y>-4$

Reflection: $y=-g(x)$

$y=-2\cdot 2^x+4$

Answer:

$y>-4$; $y=-2\cdot 2^x+4$

4.3 [3]

$2\cdot 2^x>16$

$2^x>8=2^3$

$x>3$

Answer:

$x>3$

4.4 [2]

$g(2)=4$, $g(0)=-2$

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

Answer:

$3$

4.5 [3]

Asymptote $y=-4$ dotted

Intercepts $(0;-2)$ and $(1;0)$

Increasing curve approaching the asymptote on the left

Answer:

Sketch

Question 5

5.1 [3]

36 equally likely outcomes

Totals of 8: $(2,6),(3,5),(4,4),(5,3),(6,2)$

$P=\frac{5}{36}$

Answer:

$\frac{5}{36}\approx 0,139$

5.2 [4]

$P(B)=0,55$

$P(A\text{ or }B)=0,35+0,55$

$=0,9$

$P(\text{neither})=0,1$

Answer:

0,9 and 0,1

5.3 [6]

(a) Tree: study 0,7 / not 0,3; pass 0,8 or 0,4

$0,7\times 0,8=0,56$

$0,3\times 0,4=0,12$

$P(\text{pass})=0,68$

(b) $0,7\times 0,2$

$=0,14$

Answer:

(a) 0,68 (b) 0,14

Question 6

6.1 [4]

$p=-1$

$q=-2$

$0=\frac{a}{2}-2$

$a=4$

Answer:

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

6.2 [1]

$y=-4-2=-6$

Answer:

$(0;-6)$

6.3 [3]

Through $(1;-2)$

$y=x-3$

$y=-x-1$

Answer:

$y=x-3$ and $y=-x-1$

6.4 [4]

$\frac{4}{x-1}-2=x-3\Rightarrow \frac{4}{x-1}=x-1$

$(x-1{)}^2=4$

$x=3$ or $x=-1$

$(3;0)$ and $(-1;-4)$

Answer:

$(3;0)$ and $(-1;-4)$

6.5 [3]

The graph is above the $x$-axis between the vertical asymptote and the $x$-intercept

$1<x$

$<3$

Answer:

$1<x<3$

Question 7

7.1 [3]

$A=120000(1-0,015{)}^8$

$\approx 106334$ people

Answer:

About 106 334 people

7.2 [3]

$140000=350000(1-6i)$

$1-6i=0,4$

$i=10\%$

Answer:

10% p.a.

7.3 [3]

${\left(1+\frac{i}{2}\right)}^2=1,11$

$i=2\left(1,11^{\frac{1}{2}}-1\right)$

$\approx 10,71\%$

Answer:

10,71% p.a. compounded half-yearly

7.4 [6]

$i=0,02$ per quarter

First deposit grows 16 quarters: $x(1,02{)}^{16}$

Second deposit grows 8 quarters: $x(1,02{)}^8$

$x\left[(1,02{)}^{16}+(1,02{)}^8\right]=80000$

$x\approx \frac{80000}{2,5444}$

$x\approx$ R31 441,04

Answer:

R31 441,04

Question 8

8.1 [3]

$2x+(x+10)-6+8=60$

$3x=48$

$x=16$

Answer:

$x=16$

8.2 [2]

$A$ only $=32-6=26$

$P=\frac{26}{60}\approx 0,43$

Answer:

$\frac{26}{60}\approx 0,43$

8.3 [2]

$P(A)P(B)=\frac{32}{60}\times \frac{26}{60}\approx 0,23$ and $P(A\text{ and }B)=\frac{6}{60}=0,1$

Not equal, so not independent

Answer:

Not independent

Question 9

9.1 [5]

Let $k=x^2+2x$: $(k-3)(k-8)=0$

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

$x=1$ or $x=-3$

$x^2+2x-8=0\Rightarrow (x+4)(x-2)=0$

$x=2$ or $x=-4$

Answer:

$x=-4$, $-3$, $1$ or $2$

9.2 [4]

Let $k=\sqrt{x}\ge 0$: $k^2+k-6=0$

$(k+3)(k-2)=0$

$k=-3$ is rejected

$\sqrt{x}=2\Rightarrow x=4$

Answer:

$x=4$

9.3 [4]

$x^2-mx+4=0$

A tangent meets the curve once, so $\Delta =0$

$m^2-16=0$

$m=\pm 4$

Answer:

$m=4$ or $m=-4$

9.4 [4]

$3^x\cdot 3^{2y}=3^4$

$x+2y=4$

Substitute $x=y+1$: $3y+1=4$

$y=1$ and $x=2$

Answer:

$x=2$, $y=1$

9.5 [5]

$x^2-6x+11=(x-3{)}^2+2$

$(x-3{)}^2\ge 0$, so $x^2-6x+11\ge 2$

The fraction is largest when the denominator is smallest

Maximum $=\frac{4}{2}=2$

at $x=3$

Answer:

Maximum 2 at $x=3$

Question 10

10.1 [3]

First differences: $4+2(n-1)=2n+2$

$2n+2=60\Rightarrow n=29$

Between $T_{29}$ and $T_{30}$

Answer:

Between $T_{29}$ and $T_{30}$

10.2 [3]

$n^2+n-100=0$

$n=\frac{-1+\sqrt{401}}{2}\approx 9,51$

Not a natural number, so no figure has 100 dots ($T_9=90$ and $T_{10}=110$)

Answer:

$n\approx 9,51$ is not a natural number

10.3 [6]

Plan A total: $50n$

Plan B totals $2;6;12;20;\dots$

$T_n=n^2+n$

$n^2+n>50n\Rightarrow n(n-49)>0$

$n>49$

From week 50

Answer:

From week 50

10.4 [3]

$T_n=n(n+1)+1$

$n$ and $n+1$ are consecutive integers, so one of them is even and $n(n+1)$ is even

An even number plus 1 is odd, so every $T_n$ is odd

Answer:

$T_n=n(n+1)+1$ is odd

Question 11

11.1 [4]

$-x+k=\frac{4}{x-1}-2\Rightarrow (-x+k+2)(x-1)=4$

$x^2-(k+3)x+(k+6)=0$

No intersection: $\Delta =(k+3{)}^2-4(k+6)=k^2+2k-15<0$

$(k+5)(k-3)<0\Rightarrow -5<k<3$

Answer:

$-5<k<3$