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

Question 1

1.1 [3]

$x^2-3x-10=0$

$(x-5)(x+2)=0$

$x=5$ or $x=-2$

Answer:

$x=5$ or $x=-2$

1.2 [4]

$x=\frac{-3\pm \sqrt{9+56}}{4}$

$x\approx 1,27$

or $x\approx -2,77$

Answer:

$x\approx 1,27$ or $x\approx -2,77$

1.3 [3]

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

$x<-6$

or $x>2$

Answer:

$x<-6$ or $x>2$

1.4 [2]

$\frac{4\sqrt{3}-2\sqrt{3}}{\sqrt{3}}$

$=2$

Answer:

$2$

Question 2

2.1 [3]

$T_1=3$, $T_{20}=79$, $n=20$

$S_{20}=10(3+79)$

$=820$

Answer:

820

2.2.1 [3]

Second difference 2, $a=1$

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

$T_n=n^2+4$

Answer:

$T_n=n^2+4$

2.2.2 [2]

$n^2+4=404\Rightarrow n^2=400$

$n=20$, so $T_{20}$

Answer:

$T_{20}$

2.3 [3]

Write $S_n$ forwards and backwards

Add: each of the $n$ pairs is $2a+(n-1)d$

$2S_n=n[2a+(n-1)d]$

Answer:

Proof

Question 3

3.1 [3]

$p=2$, $q=1$

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

$a=4$

Answer:

$a=4$, $p=2$, $q=1$

3.2 [2]

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

$x=-6$

Answer:

$(-6;0)$

3.3 [2]

Through $(-2;1)$: $1=2+c$

$y=-x-1$

Answer:

$y=-x-1$

3.4 [3]

$\frac{4}{x+2}\ge 2$

This needs $x+2>0$ and $x+2\le 2$

$-2<x\le 0$

Answer:

$-2<x\le 0$

3.5 [2]

Replace $x$ with $-x$

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

Answer:

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

Question 4

4.1 [5]

Definition

$f(x+h)=x^2+2xh+h^2-4x-4h+1$

$f(x+h)-f(x)=2xh+h^2-4h$

$\underset{h\to 0}{\lim }(2x+h-4)$

$f^{\prime }(x)=2x-4$

Answer:

$f^{\prime }(x)=2x-4$

4.2 [4]

$\frac{4x^2-4x+1}{x}=4x-4+x^{-1}$

$4-x^{-2}$

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

Answer:

$4-\frac{1}{x^2}$

4.3.1 [2]

$h^{\prime }(t)=25-10t$

$h^{\prime }(1)=15$ m/s

Answer:

15 m/s

4.3.2 [3]

$25t-5t^2=0\Rightarrow t=5$ ($t=0$ is the kick)

$h^{\prime }(5)=25-50$

$=-25$ m/s (25 m/s downwards)

Answer:

$-25$ m/s

Question 5

5.1 [1]

$f^{-1}(x)=3^x$

Answer:

$f^{-1}(x)=3^x$

5.2 [1]

Domain: $x>2$

Answer:

$x>2$

5.3 [3]

${\log }_3(x-2)+1=0\Rightarrow {\log }_3(x-2)=-1$

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

$x=\frac{7}{3}$

Answer:

$\left(\frac{7}{3};0\right)$

5.4 [4]

${\log }_3[x(x-2)]=1$

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

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

$x=-1$ is outside the domain, so $x=3$

Answer:

$x=3$

5.5 [2]

$x<3^2=9$

and $x>0$: $0<x<9$

Answer:

$0<x<9$

Question 6

6.1 [3]

$f(x)=a(x+1{)}^2+8$

$0=4a+8\Rightarrow a=-2$

$f(x)=-2(x^2+2x+1)+8=-2x^2-4x+6$

Answer:

$f(x)=-2x^2-4x+6$

6.2 [3]

$-2x^2-4x+6=-2x+2$

$x^2+x-2=0\Rightarrow x=1$ or $x=-2$

$(-2;6)$

Answer:

$(-2;6)$

6.3 [3]

$y\le 8$

$(1;0)$

$h$ touches the $x$-axis at one point

Answer:

$y\le 8$; $(1;0)$, so $h$ touches the $x$-axis once

6.4 [3]

The line $y=k$ must cut the parabola on both sides of the $y$-axis

It must lie below the $y$-intercept, $f(0)=6$

$k<6$

Answer:

$k<6$

Question 7

7.1 [4]

$P(\text{boy})=\frac{70}{150}$ and $P(\text{gym})=\frac{60}{150}=0,4$

$P(\text{boy})\times P(\text{gym})\approx 0,187$

$P(\text{boy and gym})=\frac{28}{150}\approx 0,187$

Equal, so independent

Answer:

Independent

7.2 [4]

$P(\text{no green})=\frac{5}{8}\times \frac{4}{7}$

$=\frac{20}{56}$

$1-\frac{20}{56}$

$=\frac{9}{14}$

Answer:

$\frac{9}{14}\approx 0,643$

7.3.1 [1]

$7!=5040$

Answer:

5 040

7.3.2 [3]

Treat the three as one block: 5 units

$5!\times 3!$

$=720$

Answer:

720

7.3.3 [2]

$\frac{720}{5040}$

$=\frac{1}{7}\approx 0,143$

Answer:

$\frac{1}{7}\approx 0,143$

Question 8

8.1 [3]

${\left(1+\frac{i}{4}\right)}^4=1,12$

$i=4\left(1,12^{\frac{1}{4}}-1\right)$

$\approx 11,49\%$

Answer:

11,49% p.a. compounded quarterly

8.2 [5]

New oven: $420000(1,065{)}^5\approx 575436,40$

Old oven: $420000(0,85{)}^5\approx 186356,23$

Target $\approx 389080,17$

$x=\frac{389080,17\times \frac{0,09}{12}}{{\left(1+\frac{0,09}{12}\right)}^{60}-1}$

$\approx$ R5 158,56

Answer:

R5 158,56 per month

8.3 [4]

$2500000=\frac{25000\left[1-(1,006{)}^{-n}\right]}{0,006}$

$(1,006{)}^{-n}=0,4$

$n=-\frac{\log 0,4}{\log 1,006}\approx 153,17$

153 full withdrawals

Answer:

153 full withdrawals ($n\approx 153,17$)

8.4 [3]

$x=\frac{180000\times \frac{0,135}{12}}{1-{\left(1+\frac{0,135}{12}\right)}^{-48}}\approx 4873,74$

Balance $=\frac{4873,738\dots \left[1-{\left(1+\frac{0,135}{12}\right)}^{-28}\right]}{\frac{0,135}{12}}$

$\approx$ R116 504,66

Answer:

R116 504,66

Question 9

9.1 [4]

$f^{\prime }(x)=-3x^2+6x+9=0$

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

$(3;22)$

$(-1;-10)$

Answer:

$(3;22)$ and $(-1;-10)$

9.2 [2]

$f^{\prime \prime }(x)=-6x+6=0\Rightarrow x=1$

$(1;6)$

Answer:

$(1;6)$

9.3 [3]

$f^{\prime }(1)=-3+6+9=12$

$y-6=12(x-1)$

$y=12x-6$

Answer:

$y=12x-6$

9.4 [3]

Increasing: $-1<x<3$

$-3x^2+6x<0\Rightarrow x(x-2)>0$

$x<0$ or $x>2$

Answer:

Increasing for $-1<x<3$; $f^{\prime }(x)<9$ for $x<0$ or $x>2$

9.5 [3]

Exactly two $x$-intercepts when one turning point lies on the $x$-axis

Maximum $(3;22+k)$: $22+k=0\Rightarrow k=-22$

Minimum $(-1;-10+k)$: $-10+k=0\Rightarrow k=10$

Answer:

$k=-22$ or $k=10$

Question 10

10.1 [5]

Let $k=x^2-x$: $(k-2)(k-6)=0$

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

$x=2$ or $x=-1$

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

$x=3$ or $x=-2$

Answer:

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

10.2 [4]

Let $k=3^x$: $k^2-4k+3=0$

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

$3^x=1\Rightarrow x=0$

$3^x=3\Rightarrow x=1$

Answer:

$x=0$ or $x=1$

10.3 [4]

$\Delta =(k+3{)}^2-4(3k)=k^2-6k+9$

$=(k-3{)}^2$

$(k-3{)}^2\ge 0$ for all real $k$, so the roots are real

When $k$ is rational, $a$, $b$ and $c$ are rational and $\sqrt{\Delta }=|k-3|$ is rational, so the roots are rational

Answer:

$\Delta =(k-3{)}^2\ge 0$, so the roots are real; $\sqrt{\Delta }=|k-3|$ is rational when $k$ is rational, so the roots are rational (they are $3$ and $k$)

Question 11

11.1 [4]

$\frac{n}{2}[4+5(n-1)]=1428$

$5n^2-n-2856=0$

$n=\frac{1+\sqrt{1+57120}}{10}$

$n=24$

Answer:

24 terms

11.2 [6]

$ar=12$ and $\frac{a}{1-r}=64$

$\frac{12}{r(1-r)}=64$

$64r^2-64r+12=0\Rightarrow 16r^2-16r+3=0$

$(4r-1)(4r-3)=0$

$r=\frac{1}{4}$, $a=48$

or $r=\frac{3}{4}$, $a=16$

Answer:

$a=48$, $r=\frac{1}{4}$ or $a=16$, $r=\frac{3}{4}$

11.3 [4]

$\frac{3(2^n-1)}{1}>10000$

$2^n>3334,33$

$n>\frac{\log 3334,33}{\log 2}\approx 11,70$

12 terms

Answer:

12 terms ($n>11,70$)

Question 12

12.1 [7]

$\pi r^{2}h=1000\Rightarrow h=\frac{1000}{\pi r^2}$

$S=2\pi r^2+2\pi rh=2\pi r^2+\frac{2000}{r}$

$S^{\prime }(r)=4\pi r-\frac{2000}{r^2}=0$

$r^3=\frac{500}{\pi }$

$r\approx 5,42$ cm

$S^{\prime \prime }(r)>0$, so a minimum

$S\approx 553,58$ cm²

Answer:

$r\approx 5,42$ cm; $S\approx 553,58$ cm²