Solutions:Maths HL 2007 Paper 1

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Solution to the 2007 Higher Level Mathematics Paper 1

equal roots, $\therefore b^2 - 4ac = 0$	equal roots, $\therefore \alpha = \beta \therefore\alpha+\beta=2\alpha$ and $\alpha\beta=\alpha^2$
$(k+1)^2 - 4\times1\times(-k-2)=0$	$2\alpha=-k-1$ $\alpha=\frac{-k-1}{2}$
$k^2+2k+1+4k+8=0$	$\alpha^2=\frac{k^2+2k+1}{4}=-k-2$
$k^2+6k+9=0$	$k^2+6k+9=0$
$(k+3)(k+3)=0$	$(k+3)(k+3)=0$

1(b)(i)

(ii)

$\frac{-k-1\pm\sqrt{(k+1)^2-4\times1\times(-k-2)}}{2}$

$\frac{-k-1\pm\sqrt{k^2+2k+1+4k+8}}{2}$

$\frac{-k-1\pm\sqrt{k^2+10k+9}}{2}$

$\frac{-k-1\pm\sqrt{(k+3)(k+3)}}{2}$

$\frac{-k-1\pm\sqrt{(k+3)^2}}{2}$

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

$\frac{-k-1+k+3}{2}$	$\frac{-k-1-k-3}{2}$
$1$	$\frac{-2k-4}{2}$
$1$	$-k-2$

1(b)(ii)roots are 1 and -k-2

(iii) 1 is positive. If -k-2 is positive $-k-2>0$ and $k<-2$

1(b)(iii)k<-2

(c)(i) If $x+p$ is a factor then $f(-p)=0$

$ap^2+b=0$	$ap^2-pb-ac=0$
$p^2=\frac{-b}{a}$	$a\times\frac{-b}{a}-pb-ac=0$
	$p=\frac{-b-ac}{b}$

(ii) $p^2+p^3=p^2(1+p)$

$\frac{-b}{a}\left( 1+\frac{-b-ac}{b}\right)=\frac{-b}{a}\left(\frac{b-b-ac}{b}\right)$

$=c$

Question 2

(a) $(x+y+z=2) - (2x+y+z=3) => x=1$

$2(2+y+z=3)+(1-2y+2z=15) => z=4$

$1+y+4=2 => y=-3$

2(a) x=1, y=-3, z=4

(b) $\alpha$ and $\beta$ of $x^2-4x+6=0$

(i) $\alpha+\beta=+4 \qquad \alpha\beta=+6$

$\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha+\beta}{\alpha\beta}=\frac{+4}{6}=\frac{2}{3}$

2(b)(i)

(ii) $x^2-\frac{2}{3}x+\frac{1}{6}=6x^2-4x+1=0$

2(b)(ii) 6x^2 - 4x + 1 = 0

(c) (i) $x+\frac{9}{x+2}\ge4$

$x(x+2)+9\ge4x+8$ [we can multiply by x+2 because we know x+2>0]

$x^2+2x+9-4x-8\ge0$

$x^2-2x+1\ge0$

$(x-1)^2\ge0$ which is true because anything squared>0

(ii) $x+\frac{9}{x+a}\ge6-a$

$x(x+a)+9\ge(6-a)(x+a)$ [we can multiply by x+a because we x+a>0]

$x^2+xa+9\ge6x-ax+6a-a^2$

$x^2+x(2a-6)+9-6a+a^2\ge0$

$x^2+2x(a-3)+(a-3)^2\ge0$

$(x+a-3)^2\ge0$ which is true because anything squared>0

Question 3

(a) $A=\begin{pmatrix} \frac{1}{2}&\frac{1}{4} \\ 3&\frac{3}{2} \end{pmatrix} \qquad A^2=\begin{pmatrix} \frac{1}{2}&\frac{1}{4} \\ 3&\frac{3}{2} \end{pmatrix}\begin{pmatrix} \frac{1}{2}&\frac{1}{4} \\ 3&\frac{3}{2} \end{pmatrix}=\begin{pmatrix} \frac{1}{2}\times\frac{1}{2}+\frac{1}{4}\times3&\frac{1}{2}\times\frac{1}{4}+\frac{1}{4}\times\frac{3}{2} \\ 3\times\frac{1}{2}+\frac{3}{2}\times3&3\times\frac{1}{4}+\frac{3}{2}\times\frac{3}{2} \end{pmatrix}=\begin{pmatrix} \1&\frac{1}{2} \\ 6&3 \end{pmatrix}$

$A^2-2A=\begin{pmatrix} \1&\frac{1}{2} \\ 6&3 \end{pmatrix}-\begin{pmatrix} \1&\frac{1}{2} \\ 6&3 \end{pmatrix}=\begin{pmatrix} 0&0 \\ 0&0 \end{pmatrix}$

3(a)

(b) $z=-1+i=\sqrt{2}\left(cosine \frac{3\pi}{4}+isin\frac{3\pi}{4}\right)$

$z^5=\sqrt{2^5}\left(cosine \frac{15\pi}{4}+isin\frac{15\pi}{4}\right)=4\sqrt{2}\left(cosine \frac{-\pi}{4}+isin\frac{-\pi}{4}\right)=4\sqrt{2}\left(\frac{1}{\sqrt{2}}-\frac{i}{\sqrt{2}}\right)=4-4i$

$z^9=\sqrt{2^9}\left(cosine \frac{27\pi}{4}+isin\frac{27\pi}{4}\right)=16\sqrt{2}\left(cosine \frac{3\pi}{4}+isin\frac{3\pi}{4}\right)=16\sqrt{2}\left(-\frac{1}{\sqrt{2}}+\frac{i}{\sqrt{2}}\right)=-16+16i$

3(b)(i)

(ii) $z^5+z^9=4-4i-16+16i=-12+12i=12(-1+i)=12z$

(c)

(i) $(a+bi)^2=15+8i$

$a^2-b^2+2abi=15+8i$

$2ab=8$	$a^2-b^2=15$
$b=\frac{4}{a}$	$a^2-\frac{16}{a^2}=15$
	$a^4-16-15a^2=0=a^4-15b^2-16$
	$(a^2-16)(a^2+1)$

$a^2=16$ or -1 but cannot=-1

$a^=2 \qquad a=4 \qquad b=1$

since it was squared, $a+bi=\pm(4+i)$

3(c)(i) 4+i, -4-i

(c)(ii) $iz^2+(2-3i)z+(-5+5i)=0 \qquad a=i \qquad b=(2-3i) \qquad c=(-5+5i)$

$roots=\frac{-(2-3i)\pm\sqrt{(2-3i)^2-4\times i\times(-5+5i)}}{2i}$

$=\frac{-2+3i\pm\sqrt{4-12i-9+20i+20}}{2i}$

$=\frac{-2+3i\pm\sqrt{15+8i}}{2i}$

$=\frac{-2+3i\pm(4+i)}{2i}$ [from part i]

$=\frac{-2+3i+(4+i)}{2i}$	$=\frac{-2+3i-(4+i)}{2i}$
$=\frac{+2+4i}{2i}$	$=\frac{-6+2i}{2i}$
$=\frac{+2+4i}{2i}\times\frac{-2i}{-2i}$	$=\frac{-6+2i}{2i}\times\frac{-2i}{-2i}$
$=\frac{-4i+8}{4}$	$=\frac{+12i+4}{4}$
$=2-i$	$=1+3i$

3(c)(ii)

Question 4

(a) $\begin{pmatrix} n \\ 1 \end{pmatrix} + \begin{pmatrix} n \\ 2 \end{pmatrix}=\begin{pmatrix} n+1 \\ 2 \end{pmatrix}$

$n+\frac{n(n-1)}{2!}=\frac{n(n+1)}{2!}$

$\frac{2n+n^2-n}{2}=\frac{n^2+n}{2}$

$\frac{n^2+n}{2}=\frac{n^2+n}{2}$

(b)

(i) $u_2=\frac{1}{2}\times5=\frac{5}{2}$

$u_3=\frac{2}{3}\times\frac{5}{2}=\frac{5}{3}$

$u_4=\frac{3}{4}\times\frac{5}{3}=\frac{5}{4}$

4(b)(i)

4(b)(ii) un = 5/n

(iii) $u_k=\frac{5}{k} \qquad u_1=5=\frac{5}{1} \therefore$ true for at least one value of k

$u_{k+1}=\frac{5}{k+1}=\frac{k}{k+1}\times u_k$

$\frac{5}{k+1}=\frac{k}{k+1} \times \frac{5}{k}$

$\frac{5}{k+1}=\frac{5}{k+1}$ is true $\therefore$ true for all $k\ge1$

(c)

(i) $U_n=S_{n}-S_{n-1}=n^2log_e3-(n^2-2n+1)log_e3=(2n-1)log_e3$

to prove arithmetic show $U_{n}-U{n-1}=$ constant

$(2n-1)log_e3-(2n-3)log_e3=2log_e3=$ constant

4(c)(i)

(ii)less than $12log_e27=$ less than $12log_e3^3=$ less than $36log_e3$

when is $(2n-1)=36$ ? $2n=37 \qquad n=18.5 \qquad u_{18.5}=12log_e27 \therefore$ the first 18 terms are below that.

4(c)(ii) 18 terms

Question 5

(a) $|x+1|\le2$

$x^2+2x+1\le4$

$x^2+2x-3\le0$

$(x-1)(x+3)\le0$

it's a u-shaped graph, that's below the x-axis between -3 and 1. Mark in the integers only.

5(a) -3, -2, -1, 0, 1

(b)

5(b)(i)

(ii) $x^{9-r}\times \left(\frac{1}{x^{2r}}\right)->9-r-2r=0$

$r=3 \qquad \qquad U_4=\begin{pmatrix} 9 \\ 3 \end{pmatrix}\times(2x)^{6}\times\left(-\frac{1}{x^2}\right)^3=84\times64x^6\times\frac{-1}{x^6}=-5376$

5(b)(ii) -5376

(c)

(i) $S_n=x+2x^2+3x^3+...+nx^n$

$- xS_n=\qquad x^2+2x^3+...+(n-1)x^n +nx^{n+1}$

$S_n-xS_n=x+x^2+x^3...x^n - nx^{n+1}$

$S_n(1-x)=\frac{x(1-x^n)}{1-x} - nx^{n+1}$ [use the S_n of a geometric series taking a=x and r=x]

$S_n=\frac{x(1-x^n)}{(1-x)^2} - \frac{nx^{n+1}}{1-x}$

5(c)(i)

(ii) $\lim_{n \to \infty}\frac{x(1-x^n)}{(1-x)^2} - \frac{nx^{n+1}}{1-x}$

$=\frac{x(1-0)}{(1-x)^2} - \frac{n0}{1-x}$ [since $|x|<1 \qquad x^n=0$

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

5(c)(ii)

Question 6

(a) $\frac{d}{dx}\left(\frac{x^2-1}{x^2+1}\right)=\frac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^2}=\frac{(x^2+1)\times2x-(x^2-1)\times2x}{(x^2+1)^2}=\frac{4x}{(x^2 + 1)^2}$

6(a)

(b)

(i) $f(x)=\frac{1}{x} \qquad f(x+h)=\frac{1}{x+h}$

$f'(x)=\lim_{h \to \0}\frac{f(x+h)-f(x)}{h}=\lim_{h \to \0}\frac{\frac{1}{x+h}-\frac{1}{x}}{h}=\lim_{h \to \0}\frac{\frac{x-x-h}{x(x+h)}}{h}=\lim_{h \to \0}\frac{\frac{-h}{x(x+h)}}{h}=\lim_{h \to \0}\frac{-1}{x(x+h)}=-\frac{1}{x^2}$

(ii) $y=1/x \qquad \frac{dy}{dx}=-\frac{1}{x^2} \qquad at(2,\frac{1}{2})=-\frac{1}{4}$

$m=-\frac{1}{4} \qquad x_1=2 \qquad y_1=\frac{!}{2}$

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

$4y-2=-x+2 \qquad => \qquad x + 4y - 4 = 0$

6(b)(ii)

(c)

(i)

$f(x)=tan^{-1}\frac{x}{2}$	$g(x)tan^{-1}\frac{2}{x}$
$f'(x)=\frac{2}{4+x^2}$	$g'(x)=\frac{1}{1+\frac{4}{x^2}}\times\frac{-2}{x^2}=\frac{-2}{x^2+4}$

6(c)(i)

(ii) $f'(x)+g'(x) = 0 \qquad \therefore \qquad f(x)+g(x)=$ constant

(iii)take $x=2\sqrt{3}$ [since $f(x)+g(x)$ is constant, any value of x will give the same answer]

$f(x)+g(x)=tan^{-1}\frac{2\sqrt{3}}{2}+tan^{-1}\frac{2}{2\sqrt{3}}=tan^{-1}\sqrt{3}+tan^{-1}\frac{1}{\sqrt{3}}=\frac{\pi}{3}+\frac{\pi}{6}=\frac{\pi}{2}$

6(c)(iii)

Question 7

(a) $f(x)=x^3+2x-4=0 \qquad f'(x)=3x^2+2$

$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$

$x_2=x_1-\frac{f(x_1)}{f'(x_1)}=1-\frac{1+2-4}{3+2}=1-\frac{-1}{5}=\frac{6}{5}$

7(a)

(b) (i) $3x^2+y^2=28$

$6x+2y\frac{dy}{dx}=0$

$\frac{dy}{dx}=-\frac{3x}{y}$ at $(2,-4)=-\frac{6}{-4}=\frac{3}{2}$

$m=\frac{3}{2} \qquad x_1=2 \qquad y_1=-4$

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

$2y+8=3x-6 \qquad => \qquad 3x-2y-14=0$

7(b)(i)

(ii)

$x=e^tcost$	$y=e^tsint$
$\frac{dx}{dt}=e^tcost-e^tsint$	$\frac{dy}{dt}=e^tsint+e^tcost$
$=x-y$	$=x+y$

$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{x+y}{x-y}$

(c) (i) $f(x)=log_e3x-3x$

$f'(x)=\frac{1}{3x}\times3 - 3$

$=\frac{1}{x}-3=0 \qquad => \qquad x=\frac{1}{3} \qquad => \qquad y=0-1=-1$

$\therefore$ local maximum or minimum at $(\frac{1}{3},-1)$

$f''(x)=-\frac{1}{x^2}$ at $(\frac{1}{3},-1) = -9 <0 \therefore$ maximum point

7(c)(i) f'(x)=0 => x = 1/3, f(1/3) = -1, f(x)<0

(ii)no other solutions for $f'(x)$

$\therefore$ no other min or max points and no upturn

$\therefore \qquad (\frac{1}{3},-1)$ is the only maximum point

$\therefore$ y is never greater than -1 and never touches x-axis

7(c)(ii) max value of f(x)=-1 => doesn't cross x-axis

Question 8

(a)

(i) $\int x^3dx=\frac{x^4}{4} + c$

(i)

(ii) $\int\frac{1}{x^3}dx=\frac{-1}{2x^2} + c$

(ii)

(b)

(i) $\int\limits_0^4 x\sqrt{x^2+9}dx$

$\int\limits_0^4 x\sqrt{x^2+9}dx$	$u=x^2+9$	$du=2x$
$\frac{1}{2}\int\limits_9^{25}\sqrt{u}du$	$x=4, u=25$	$x=0, u=9$
$\frac{1}{2}\left(\frac{2}{3}\sqrt{u^3}\right)_9^{25}$	$=\frac{125}{3}-\frac{27}{3}$	$=\frac{98}{3}$

8(b)(i)

(ii) $f'(x)=6-sinx$

$f(x)=6x+cosx+c$

$f\left(\frac{\pi}{3}\right)=2\pi=6\frac{\pi}{3}+cos\frac{\pi}{3}+c$

$c=-cos\frac{\pi}{3}=-\frac{1}{2}$

$=>f(x) = 6x + cosx - \frac{1}{2}$

8(b)(ii)

(c)

point of intersection $=2x-(x^2-9)-10=0$	=>point where curve meets x-axis $x^2-9=0$	=>point where line meets x-axis $2x-10=0$
$x^2-2x+1=0$	$(x-3)(x+3)=0 \qquad x=\pm3$	$2x=10 \qquad x=5$
$(x-1)^2=0 \qquad x=1 \qquad y=1-9=-8$