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- \documentclass{article}
- \usepackage[utf8]{inputenc}
- \usepackage{graphicx}
- \title{MATHEMATICS: GRADE 12 PAPER 1 NOVEMBER 2017 MEMO}
- \author{Marks: 150}
- \date{Time: 3 hours}
- \begin{document}
- \maketitle
- \section*{Question 1}
- 1.1.1
- {Answer}{}
- $x=-7$ or $x=-2$\\
- {}{}
- {Memo}{}
- $x^{2}+9x+14=0$\\
- $(x+7)(x+2)=0$\\
- $x=-7$ or $x=-2$
- {}{}
- \\\\
- 1.1.2
- {Answer}{}
- $x=0,29$ or $x=-2,54$\\
- {}{}
- {Memo}{}
- $4x^{2}+9x-3=0$\\
- $x=\frac{-b \pm \sqrt{b^{2}-4ac}}{2a}$\\
- $=\frac{-9\pm\sqrt{9^{2}-4(4)(-3)}}{2(4)}$\\
- $=\frac{-9\pm\sqrt{129}}{8}$\\
- $x=0,29$ or $x=-2,54$\\
- OR \\
- $x^{2}+\frac{9}{4}x + \frac{81}{64} = \frac{3}{4} + \frac{81}{64}$\\
- $(x+\frac{9}{8})^{2}=\frac{129}{64}$\\
- $x+\frac{9}{8}=\pm\frac{\sqrt{129}}{8}$\\
- $x=\frac{-9\pm\sqrt{129}}{8}$\\
- $x = 0,29$ or $x=-2,54$
- {}{}
- \\\\
- 1.1.3
- {Answer}{}
- $x=5$\\
- {}{}
- {Memo}{}
- $\sqrt{x^{2}-5}=2\sqrt{x}$\\
- $x^{2}-5=4x$\\
- $x^{2}-4x-5=0$\\
- $(x-5)(x+1)=0$\\
- $x=5$ or $x=-1$\\
- $x=5$
- {}{}
- \\\\
- 1.2
- {Answer}{}
- $y=-1$ or $y=17$\\
- {}{}
- {Memo}{}
- $3x-y=4$\\
- $y = 3x-4$\\
- $x^{2}+2xy-y^{2}=-2$\\
- $x^{2}+2x(3x-4)-(3x-4)^{2}=-2$\\
- $x^{2}+6x^{2}-8x-(9x^{2}-24x+16)=-2$\\
- $7x^{2}-8x-9x^{2}+24x-16=-2$\\
- $-2x^{2}+16x-14=0$\\
- $x^{2}-8x+7=0$\\
- $(x-7)(x-1)=0$\\
- $x=1$ or $x=7$\\
- $y = 3(1)-4$ or $y=3(7)-4$\\
- $y=-1$ or $y=17$\\
- OR\\
- $3x-y=4$\\
- $x= \frac{y+4}{3}$\\
- $x^{2}+2xy-y^{2}=-2$\\
- $(\frac{y+4}{3})^{2} + 2(\frac{y+4}{3})y-y^{2}=-2$\\
- $y^{2}+8y+16+6y^{2}+24y-9y^{2}-18$\\
- $-2y^{2}+32y+34=0$\\
- $y^{2}-16y-17=0$\\
- $(y-17)(y+1)=0$\\
- $y = -1$ or $y = 17$\\
- $x= \frac{-1+4}{3}$ or $x= \frac{17+4}{3}$\\
- $x=1$ or $x= 7$
- {}{}
- \\\\
- 1.3.1
- {Answer}{}
- $x \in (-\infty; -4)$ or $x \in(-4; \infty)$\\
- {}{}
- {Memo}{}
- $x^2+8x+16>0$\\
- $(x+4)(x+4)>0$\\
- $x \in R, x\neq -4$\\
- OR\\
- $x \in (-\infty; -4)$ or $x \in(-4; \infty)$\\
- OR\\
- $x<-4 $ or $ x >-4$\\
- OR
- $x^2+8x+16>0$\\
- $(x+4)(x+4)>0$\\
- The function values remain positive\\
- $x \in R, x\neq -4$
- {}{}
- \includegraphics{graph8.png}
- \\\\
- 1.3.2
- {Answer}{}
- $0 < p < 16 $\\
- {}{}
- {Memo}{}
- For two negative unequal roots: $0 < p < 16$\\
- OR \\
- $x^{2} + 8x+16=p$\\
- $x^{2}+8x+16-p=0$\\
- $0<16-p<16$\\
- $-16<-p<0$\\
- $0<p<16$\\
- OR\\
- $x^{2}+8x+16-p=0$\\
- $x = \frac{-8\pm\sqrt{64-4(16-p)}}{2}$\\
- $0 < 64-4(16-p)<64$\\
- $0<4p<64$\\
- $0<p<16$\\
- OR\\
- $x^{2}+8x+16=p$\\
- $x^{2}+8x+16-p=0$\\
- Roots are real and unequal:\\
- $8^{2}-4(16-p)>0$\\
- $4p>0$\\
- $p>0$\\
- Roots are: $\frac{-8\pm\sqrt{4p}}{2}$\\
- For both roots to be negative:\\
- $\sqrt{4p}<8$\\
- $4p<64$\\
- $p<16$\\
- $0<p<16$\\
- {}{}
- \includegraphics{graph1.png}
- \\\\
- \section*{Question 2}
- 2.1.1
- {Answer}{}
- $-6$\\
- {}{}
- {Memo}{}
- First differences: $-9;-15;-21$\\
- Second difference: $-6$\\
- {}{}
- \includegraphics{graph2.png}
- \\\\
- 2.1.2
- {Answer}{}
- $T_{n}=-3n^{2}+8$\\
- {}{}
- {Memo}{}
- $T_{n}=an^{2}+bn+c$\\
- $a =\frac{second difference}{2} = -3$\\
- $3a + b = -9$\\
- $3(-3)+b=-9$\\
- $b=0$\\
- $a+b+c=5$\\
- $-3+0+c=5$\\
- $c=8$\\
- $T_{n}=-3n^{2}+8$\\
- OR\\
- $T_{n} = T_{1} + (n-1)d_{1} + \frac{(n-1)(n-2)d_{2}}{2}$\\
- $= 5+(n-1)(-9)+\frac{(n-1)(n-2)(-6)}{2}$\\
- $= 5-9n+9-3n^{2}+9n-6$\\
- $T_{n}=-3n^{2}+8$\\
- {}{}
- \\\\
- 2.1.3
- {Answer}{}
- The $93^{rd}$ term has a value of $-25939$.\\
- {}{}
- {Memo}{}
- $-3n^{2}+8=-25939$\\
- $-3n^{2}=-25947$\\
- $n^{2}=8649$\\
- $n=-93$ or $n=93$\\
- The $93^{rd}$ term has a value of $-25939$.
- {}{}
- \\\\
- 2.2.1
- {Answer}{}
- $T_{15} = 59$\\
- {}{}
- {Memo}{}
- $2k-7; k+8$ and $2k-1$\\
- $k+8-(2k-7)=2k-1-(k+8)$\\
- $-k+15=k-9$\\
- $2k=24$\\
- $k=12$\\
- $2k-7; k+8$ and $2k-1$\\
- $17;20;23.......$\\
- $d = 3$\\
- $T_{15}=17+14(3)$\\
- $=59$
- {}{}
- \\\\
- 2.2.2
- {Answer}{}
- $S_{30} = 3210$\\
- {}{}
- {Memo}{}
- Sequence is $17; 20; 23; 26; 29; 32 .....$\\
- Every alternate term of the sequence will be even. \\
- $20 + 26 + 32 + ....$\\
- $S_{30}=\frac{30}{2}[2(20)+(29)(6)]$\\
- $=15[40+174]$\\
- $= 3210$\\
- OR \\
- $T_{30}=20+29(6)$\\
- $=94$\\
- $S_{30}=\frac{30}{2}(20+194)$\\
- $=3210$
- {}{}
- \\\\
- \section*{Question 3}
- 3.1
- {Answer}{}
- $a=\frac{9-9r}{4}$ OR $a= \frac{2}{1+r}$ OR $a= \frac{1-r}{4r^{2}}$\\
- {}{}
- {Memo}{}
- $a+ar=2$\\
- $a(1+r)=2$\\
- $a=\frac{2}{1+r}$\\
- OR\\
- $\frac{a}{1-r} - 2 = \frac{1}{4}$\\
- $4a-8(1-r)=1-r$\\
- $4a-8+8r=1-r$\\
- $4a=9-9r$\\
- $a=\frac{9-9r}{4}$\\
- OR\\
- $S_{n}= \frac{a(r^{n}-1}{r-1}$\\
- $2= \frac{a(r^{2}-1}{r-1}$\\
- $2 = \frac{a(r-1)(r+1)}{r-1}$\\
- $2 = a(r+1)$\\
- $a = \frac{2}{r+1}$\\
- OR\\
- $\frac{ar^{2}}{1-r} = \frac{1}{4}$\\
- $a=\frac{1-r}{4r^{2}}$
- {}{}
- \\\\
- 3.2
- {Answer}{}
- $a = \frac{3}{2}$ and $r = \frac{1}{3}$\\
- {}{}
- {Memo}{}
- $S_{\infty} = T_{1}+T_{2}+\sum{\infty}{n=3}{T_{n}}$\\
- $S_{\infty}=2+\frac{1}{4}$\\
- $\frac{a}{1-r}=2+\frac{1}{4}$\\
- $\frac{a}{1-r}=\frac{9}{4}$\\
- $(\frac{2}{1+r})x(\frac{1}{1-r}) = \frac{9}{4}$\\
- $\frac{2}{1-r^{2}} = \frac{9}{4}$\\
- $8 = 9 - 9r^{2}$\\
- $9r^{2}=1$\\
- $r = \frac{1}{3}$\\
- $a = \frac{3}{2}$\\
- OR\\
- $S_{\infty} = T_{1}+T_{2}+\sum{\infty}{n=3}{T_{n}}$\\
- $S_{\infty} = 2+\frac{1}{4}$\\
- $\frac{a}{1-r} = 2 + \frac{1}{4}$\\
- $\frac{a}{1-r} = \frac{9}{4}$\\
- $4a=9-9r$\\
- $r=\frac{9-4a}{9}$\\
- $a+a(\frac{9-4a}{9})=2$\\
- $9a+9a-4a^{2} = 18$\\
- $2a^{2} - 9a+9 = 0$\\
- $(a-3)(2a-3)=0$\\
- $a=\frac{3}{2}$ or $a=3$\\
- $r=\frac{1}{3}$ or $r = -\frac{1}{3}$\\
- OR\\
- $r = \frac{2-a}{a}$\\
- $\frac{ar^{2}}{1-r} = \frac{1}{4}$\\
- $4ar^{2} = 1-r$\\
- $4a(\frac{2-a}{a})^{2} = 1-\frac{2-a}{a}$\\
- $16-16a+4a^{2}=2a+2$\\
- $2a^{2}-9a+9=0$\\
- $(2a-3)(a-3)=0$\\
- $a=\frac{3}{2}$, $a\neq3$\\
- $r = \frac{1}{3}$, $r\neq\frac{1}{3}$\\
- OR\\
- $S_{\infty} = T_{1}+T_{2}+\sum{\infty}{n=3}{T_{n}}$\\
- $S_{\infty} = 2+\frac{1}{4}$\\
- $\frac{a}{1-r} = 2+\frac{1}{4}$\\
- $\frac{a}{1-r}=\frac{9}{4}$\\
- $(\frac{1-r}{r^{2}})x(\frac{1}{1-r}) = \frac{9}{4}$\\
- $\frac{1}{4r^{2}}= \frac{9}{4}$\\
- $4 = 36r^{2}$\\
- $9r^{2}=1$\\
- $r = \frac{1}{3}$\\
- $a = \frac{3}{2}$
- {}{}
- \\\\
- \section*{Question 4}
- 4.1
- {Answer}{}
- $a = \frac{1}{2}$ and $b = 2$\\
- {}{}
- {Memo}{}
- $f(x) = -ax^{2} + bx + 6$\\
- $f'(x) = -2ax+b$\\
- $-2ax+b=3$\\
- at $x=-1$\\
- $2a+b=3$ [1]\\
- $f(-1)=\frac{7}{2}$\\
- $-a-b+6=\frac{7}{2}$\\
- $-2a-2b+12=7$\\
- $2a+2b=5$ [2]\\
- $[2]-[1]$\\
- $b=2$\\
- $2a + 2 = 3$\\
- $a = \frac{1}{2}$\\
- OR\\
- $f'(x) = -2ax+b$\\
- $3=2a+b$\\
- $b=3-2a$\\
- $\frac{7}{2}=-a(-1)^{2}+(3-2a)(-1)+6$\\
- $a+3=\frac{7}{2}$\\
- $a=\frac{1}{2}$\\
- $b=2$
- {}{}
- \\\\
- 4.2
- {Answer}{}
- $(-2;0)$ $(6;0)$\\
- {}{}
- {Memo}{}
- $f(x) = -\frac{1}{2}x^{2} + 2x+6$\\
- x-intercepts:\\
- $-\frac{1}{2}x^{2} + 2x+6 = 0$\\
- $-x^{2} + 4x+12=0$\\
- $x^{2}-4x-12=0$\\
- $(x-6)(x+2)=0$\\
- $(-2;0)$ $(6;0)$
- {}{}
- \\\\
- 4.3
- {Answer}{}
- TP$(2;8)$\\
- {}{}
- {Memo}{}
- $f(x)=-\frac{1}{2}x^{2} + 2x+6$\\
- $f'(x) = 0$ or $x = -\frac{b}{2a}$ or $x= \frac{-2+6}{2}$\\
- $-x+2=0$ and $x=-\frac{2}{2(-\frac{1}{2})}$ and $x = 2$\\
- $x = 2$ and $x = 2$\\
- $y = -\frac{1}{2}(2)^{2} + 2(2)+6$\\
- $= -2+4+6$\\
- $= 8$\\
- TP$(2;8)$\\
- OR\\
- $y = -\frac{1}{2}(x^{2}-4x-12)$\\
- $= -\frac{1}{2}[(x-2)^{2}-4-12]$\\
- $= -\frac{1}{2}(x-2)^{2} + 8$\\
- TP$(2;8)$
- {}{}
- \\\\
- 4.4
- {Answer}{}
- look at memo\\
- {}{}
- {Memo}{}
- look at graph
- {}{}
- \includegraphics{graph3.png}\\
- \\\\\
- 4.5
- {Answer}{}
- $ 0<x<4$ or (0 ; 4)\\
- {}{}
- {Memo}{}
- $ 0<x<4$ or (0 ; 4)
- {}{}
- \\\\
- 4.6
- {Answer}{}
- look at memo\\
- {}{}
- {Memo}{}
- look at graph
- {}{}
- \includegraphics{graph3.png}\\
- \\\\
- 4.7
- {Answer}{}
- $x \leq -2 $ or $-1\leq x \leq 6$\\
- OR\\
- $(-\infty; 2] $ OR $ [-1 ; 6]$\\
- {}{}
- {Memo}{}
- $x \leq -2 $ or $-1\leq x \leq 6$\\
- OR\\
- $(-\infty; 2] OR [-1 ; 6]$
- {}{}
- \\\\
- \section*{Question 5}
- 5.1
- {Answer}{}
- $y \in R$ ; $y \neq 1$\\
- {}{}
- {Memo}{}
- $y < -1 $ or $ y>-1$\\
- OR\\
- $y \in (-\infty ; -1)$ or $ y \in (-1 ; \infty)$\\
- OR\\
- $R-\{-1\}$
- {}{}
- \\\\
- 5.2
- {Answer}{}
- D(2; -1)\\
- {}{}
- {Memo}{}
- $g(x) = \frac{2}{x-2}-1 $
- {}{}
- \\\\
- 5.3
- {Answer}{}
- t = 3\\
- {}{}
- {Memo}{}
- $f(x) = log_{3}x$\\
- $log_{3}t = 1$\\
- $t = 3$\\
- OR\\
- $g(x) = \frac{2}{x-2}-1 $\\
- $1 = \frac{2}{t-2}-1 $\\
- $2 = \frac{2}{t-2}$\\
- $2t -4 = 2$\\
- $t = 3$
- {}{}
- \\\\
- 5.4
- {Answer}{}
- $y=3^x$\\
- {}{}
- {Memo}{}
- $x = log_{3}y$\\
- $y=3^x$
- {}{}
- \\\\
- 5.5
- {Answer}{}
- $x < 1$\\
- {}{}
- {Memo}{}
- $3^x < 3^1$\\
- $x < 1$\\
- OR\\
- $3^x < 3^1$\\
- $ x \in (-\infty; 1)$\\
- {}{}
- \\\\
- 5.6
- {Answer}{}
- Point of intersection: B (1 ; 0)\\
- {}{}
- {Memo}{}
- Equation of the axis of symmetry: $y = -x+1 $\\
- $x$-intercept of the axis of symmetry is at $x$ = 1\\
- has an $x$-intercept at B(1 ; 0) which is the same as the\\
- $x$-intercept of the axis of symmetry\\
- OR\\
- Since BE = ED = 1 and D lies on the axis of symmetry\\
- and the gradient of the axis of symmetry is –1, B will\\
- also lie on the axis of symmetry. But B also lies on $f$.\\
- Therefore B(1 ; 0) is the point of intersection between $f$\\
- and the axis of symmetry with a negative gradient.
- {}{}
- \\\\
- \section*{Question 6}
- 6.1
- {Answer}{}
- $r$ = 6.5\% \\
- {}{}
- {Memo}{}
- $A = P(1 + i)^n $\\
- $12 146,72 = 10 000(1+\frac{r}{12})^{36}$\\
- $(1+\frac{r}{12})^{36} = 1,214672$\\
- $1+\frac{r}{12} = \sqrt[36]{1,214672}$\\
- = 1,005416\\
- $\frac{r}{12} = 0.005416$\\
- $r$ = 0,06500\\
- $r$ = 6.5\%
- {}{}
- \\\\
- 6.2.1
- {Answer}{}
- His monthly instalment is R 5 536,95\\
- {}{}
- {Memo}{}
- P = $\frac{x[1-(1+i)^{-n}]}{i}$\\
- 235000 = $\frac{x[1-(1+\frac{0,11}{12})^{-54}]}{\frac{0,11}{12}}$\\
- $x = \frac{235000\cdot\frac{0,11}{0,12}}{[1-(1+\frac{0,11}{12})^{-54}]}$\\
- = R 5 536,95
- {}{}
- \\\\
- 6.2.2
- {Answer}{}
- Amount of interest = R23 739.60\\
- {}{}
- {Memo}{}
- Amount paid for the year : $(5 536,95 \cdot 12) = $R66 443,40\\
- Balance=$235000(1+\frac{0,11}{12})^{12}-\frac{5536,95[(1+\frac{0,11}{12})^{12}-1]}{\frac{0,11}{0,12}}$\\
- = 192296,17\\
- Interest = $(5536,95 \cdot 12)-(235000-192296,17$\\)
- =66443,40 - 42703,83\\
- = 23 739,57\\
- OR\\
- Total amount paid in first year = $R 5 536.95 \cdot 12$ = R66 443,40\\
- Balance on loan after 1 year = P of remaining installments\\
- P = $\frac{x[1-(1+i)^{-n}]}{i}$\\
- $= \frac{5536,95[1-(1+\frac{0,11}{12})^{-42}]}{\frac{0,11}{12}}$\\
- = R192296,20\\
- Amount paid off in the first year:\\
- R235 000 – R192 296,20 = R42 703,80 \\
- Amount of interest = R66 443,40 – R42 703,80\\
- = R23 739,60\\
- OR\\
- P=$\frac{5536,95[1-(1+\frac{0.11}{12})^{-12}]}{\frac{0.11}{12}}$\\
- = R62648.18\\
- 235 000 – 62 648,18 = R172 351,82\\
- After 12 months, money owed on house is\\
- $172351.82(1+\frac{0.11}{12})^{12}$\\
- = 192296.17\\
- Amount paid after 12 months is\\
- $5 536,95 \cdot 12 =$ R 66 443, 40\\
- Amount of interest paid:\\
- R 66 443, 40 – (235 000 – 192 296,17)\\
- = R 23 739, 57\\
- {}{}
- \\\\
- \section*{Question 7}
- 7.1
- {Answer}{}
- $4x-1$\\
- {}{}
- {Memo}{}
- $f(x+h) = 2(x+h)^2 -(x+h)$\\
- $= 2(x^2+2xh+h^2)-x-h$\\
- $= 2x^2 +4xh+2h^2-x-h$\\
- $f(x+h)-f(x) = 2x^2 +4xh+2h^2-x-h-2x^2+x$\\
- $= 4xh+2h^2-h$\\
- $$f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}$$
- $$= \lim_{h\to0}\frac{4xh+2h^2-h}{h}$$
- $$= \lim_{h\to0}\frac{h(4x+2h-1)}{h}$$
- $$= \lim_{h\to0} (4x+2h-1)$$
- $$= 4x-1$$
- $$\mathbf{OR}$$\\
- $$f'(x)= \lim_{h\to0} \frac{f(x+h)-f(x)}{h}$$
- $$f'(x)= \lim_{h\to0} \frac{2(x+h)^2 -(x+h)-(2x^2-x)}{h}$$
- $$ = \lim_{h\to0}\frac{2x^2 +4xh+2h^2-x-h -2x^2+x}{h}$$
- $$ = \lim_{h\to0}\frac{4xh+2h^2-h}{h}$$
- $$ = \lim_{h\to0}\frac{h(4x+2h-1)}{h}$$
- $$ = \lim_{h\to0}(4x+2h-1)$$
- $$ = 4x-1$$
- {}{}
- \\\\
- 7.2.1
- {Answer}{}
- $6x-4$\\
- {}{}
- {Memo}{}
- $D_{x}[(x+1)(3x-7)]$\\
- $=D_{x}(3x^2-4x-7)$\\
- $= 6x-4$
- {}{}
- \\\\
- 7.2.2
- {Answer}{}
- $\frac{dy}{dx} = \frac{3}{2}x^{\frac{1}{2}}+5x^{-2}$\\
- {}{}
- {Memo}{}
- $y = \sqrt{x^3}-\frac{5}{x}+\frac{1}{2}\pi$\\
- $y = x^{\frac{3}{2}}-5x^{-1}+\frac{1}{2}\pi$\\
- $\frac{dy}{dx} = \frac{3}{2}x^{\frac{1}{2}}+5x^{-2}$
- {}{}
- \\\\
- \section*{Question 8}
- 8.1
- {Answer}{}
- Point of inflection at $x$ = 2\\
- {}{}
- {Memo}{}
- $f(x) = x^3-6x^2+9x$\\
- $f'(x) = 3x^2-12x+9$\\
- $f''(x)= 6x-12=0$\\
- $x=2$
- $f''(0) = 6(0)-12$\\
- =-12\\
- $f''(3) = 6(3)-12$\\
- $=6$\\
- Point of inflection at $x$=2
- {}{}
- \includegraphics{graph7.png}
- \\\\
- 8.2
- {Answer}{}
- look at memo\\
- {}{}
- {Memo}{}
- look at graph
- {}{}
- \includegraphics{graph4.png}
- \\\\
- 8.3
- {Answer}{}
- $y = -f(x) $ will be concave down for $ x > 2$\\
- {}{}
- {Memo}{}
- $f$ concave up for $x > 2$\\
- $y -f(x) $ will be concave down for $x > 2$\\
- {}{}
- \\\\
- 8.4.1
- {Answer}{}
- (3;7)\\
- {}{}
- {Memo}{}
- (3;7)
- {}{}
- \\\\
- 8.4.2
- {Answer}{}
- Do not agree with Claire as her statement is incorrect.\\
- {}{}
- {Memo}{}
- Between $x$ = 1 and $x$ = 3 the graph of$f$ is decreasing.
- Therefore at $x$ = 2 the gradient will have a negative value.
- OR\\
- $f'(2)=3(2)^2-12(2)+9$\\
- =-3\\
- $\neq$1\\
- {}{}
- \\\\
- \section*{Question 9}
- 9.1
- {Answer}{}
- PB = $\frac{\sqrt{3}}{2} = 0,87$\\
- {}{}
- {Memo}{}
- $y=x^2+2$\\
- $P(x;x^2+2)$\\
- B(0 ; 3)\\
- $PB^2 = (x-0)^2+(x^2+2-3)^2\\$
- $= x^2+x^4-2x^2+1$\\
- $= x^4-x^2+1$\\
- PB will be a minimum if $PB^2$ is a minimum\\
- $\frac{d(PB^2)}{dx} = 4x^3-2x$\\
- $4x^3-2x=0$\\
- $x(2x^2-1)=0$\\
- $x=0 $ or $ x^2 = \frac{1}{2}$\\
- $x = \frac{1}{\sqrt{2}}$\\
- $PB^2 = (\frac{1}{\sqrt{2}})^4-(\frac{1}{\sqrt{2}})^2+1$\\
- $= \frac{1}{4}-\frac{1}{2}+1$\\
- =$\frac{3}{4}$\\
- PB = $\frac{\sqrt{3}}{2} = 0.87$\\
- OR\\
- Gradient of tangent to curve = $2x$\\
- Gradient of line joining B and the curve = $\frac{x^2+2-3}{x-0}$\\
- $= \frac{x^2-1}{x}$\\
- Shortest distance will be where tangent to curve is perpendicular to the line joining P and the curve\\
- $ \frac{x^2-1}{x} = \frac{1}{2x}$\\
- $2x(x^2-1)=-x$\\
- $2x^3-2x=0$\\
- $x(2x^2-1)=0$\\
- $x=0$ or $x^2 = \frac{1}{2}$\\
- $x = \frac{1}{\sqrt{2}}$\\
- $PB^2 = (\frac{1}{\sqrt{2}})^4-(\frac{1}{\sqrt{2}})^2+1$\\
- $= \frac{1}{4}-\frac{1}{2}+1$\\
- =$\frac{3}{4}$
- PB = $\frac{\sqrt{3}}{2} = 0.87$\\
- OR\\
- $P(k;k^2+2)$ and B(0;3)\\
- BP $\perp$ tangent passing through $y = x^2 + 2$ at P.\\
- $m_{tangent at P} = 2k$\\
- $M_{BP} = -\frac{1}{2k}$\\
- Equation of BP: $y = (-\frac{1}{2k})x+3$\\
- $y_p = (-\frac{1}{2k})(k)+3 = 2,5$\\
- $\Rightarrow k^2+2=2,5$ and so $ k=\sqrt{0.5}$ and P$(\sqrt{0,5};2,5)$\\
- BP = $\sqrt{(\sqrt{0,5}-0)^2+(2,5-3)^2} = \sqrt{\frac{3}{4}}=\frac{\sqrt{3}}{2}=0,87$
- {}{}
- \\\\
- \section*{Question 10}
- 10.1
- {Answer}{}
- look at memo\\
- {}{}
- {Memo}{}
- memo
- {}{}
- \includegraphics{graph5.png}
- \\\\
- 10.2
- {Answer}{}
- $x=42$\\
- {}{}
- {Memo}{}
- $ (49-x) +x+ 8 + 4 + 5+2 + (60-x) + 14 = 100$\\
- $ -x + 142 = 100$\\
- $x=42$
- {}{}
- \\\\
- 10.3
- {Answer}{}
- P(use only one application) = 27\% \\
- {}{}
- {Memo}{}
- P(use only one application) $= \frac{7+2+18}{100}$\\
- $= \frac{27}{100}$ or 27\%
- {}{}
- \\\\
- \section*{Question 11}
- 11.1
- {Answer}{}
- 2250\\
- {}{}
- {Memo}{}
- $5\cdot 5\cdot 10\cdot 9$\\
- =2250
- {}{}
- \\\\
- 11.2
- {Answer}{}
- Codes of two letters and five digits will ensure unique numbers for 700 000 clients.\\
- {}{}
- {Memo}{}
- Codes of two letters and five digits will ensure unique
- numbers for 700 000 clients.\\
- {}{}
- \includegraphics{graph6.png}
- \\\\
- \end{document}
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