Solutions:Maths HL 2007 Paper 2

From ZuluNotes - Free Leaving Cert Notes

Solution to the 2007 Higher Level Mathematics Paper 2

$\vec{r}=\frac{65t}{16}\Big(\frac{\vec{p}}{|\vec{p}|}+\frac{\vec{q}}{|\vec{q}|}\Big) =\frac{65t}{16}\Big(\frac{3\vec{i}+4\vec{j}}{\sqrt{3^2+4^2}}+\frac{5\vec{i}+12{j}}{\sqrt{5^2+12^2}}\Big) =\frac{65t}{16}\Big(\frac{3\vec{i}+4\vec{j}}{\sqrt{25}}+\frac{5\vec{i}+12{j}}{\sqrt{169}}\Big) =\frac{65t}{16}\Big(\frac{3\vec{i}+4\vec{j}}{5}+\frac{5\vec{i}+12{j}}{13}\Big)$

$=\frac{65t}{16}\Big(\frac{(3\vec{i}+4\vec{j})\times13+(5\vec{i}+12{j})\times5}{13\times5}\Big) =\frac{65t}{16}\Big(\frac{39\vec{i}+52\vec{j}+25\vec{i}+60{j}}{65}\Big) =\frac{t}{16}\Big(64\vec{i}+112\vec{j}\Big) =t(4\vec{i} + 7\vec{j})$

2(c)(i)

(ii)

$\vec{p}.\vec{r}=3\times4t+4\times7t$	$\vec{q}.\vec{r}=5\times4t+12\times7t$
$=12t+28t=40t$	$=20t+84t=104t$

2(c)(ii)

(iii)

$cos|por|=\frac{\vec{p}.\vec{r}}{|\vec{p}||\vec{r}|} =\frac{40t}{5\times\sqrt{16t^2+49t^2}} =\frac{8t}{sqrt{65t^2}} =\frac{8t}{t\sqrt{65}} =\frac{8}{\sqrt{65}}$

$cos|qor|=\frac{\vec{q}.\vec{r}}{|\vec{q}||\vec{r}|} =\frac{104t}{13\times\sqrt{16t^2+49t^2}} =\frac{8t}{sqrt{65t^2}} =\frac{8t}{t\sqrt{65}} =\frac{8}{\sqrt{65}}$

$|por|=|qor| \qquad \therefore \vec{r}$ bisects $poq$

Question 3

(a)

$(1,1)$	$(8,-5)$	$(5,-2)$
$(0,0)$	$(7,-6)$	$(4,-3)$

area $=\frac{1}{2}|x_1y_2-x_2y_1| =\frac{1}{2}|-21-(-24)| =\frac{1}{2}|-21+24| =\frac{1}{2}|3| =\frac{3}{2}$

3(a)

(b)

(i)

$x'=4x+2y$	$y'=-3x-y$
$x'=4x-6x-2y'$	$y=-3x-y'$
$x=\frac{-x'-2y'}{2}$	$y=-3\frac{-x'-2y'}{2}-y'=\frac{+3x'+4y'}{2}$

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

(b)

$p(1,-1)$		$q(3,-3)$
$x'=4-2=2$	$y'=-3+1=-2$	$x'=12-6=6$	$y'=-9+3=-6$
$f(p)=(2,-2)$		$f(q)=(6,-6)$

$|pq|=\sqrt{2^2+2^2}=\sqrt{8}=2\sqrt{2}$

$|f(p)f(q)|=\sqrt{4^2+4^2}=\sqrt{32}=4\sqrt{2}$

$|pq|:|f(p)f(q)|=2\sqrt{2}:4\sqrt{2}=1:2$

3(b)(ii)

(c)

(i)

at any point on the first line, $3x-5y+6=0$

at any point on the second line, $5x-7y+4=0$

$\therefore$ at the point of intersection, both $3x-5y+6$ and $5x-7y+4 \qquad =0$

$k(0)+l(0)=0$ The point of intersection is on that line.

(ii)

slope of the line, $m=-\frac{a}{b}=-\frac{3k+5l}{-5k-7l}=2$

$\frac{3k+5l}{5k+7l}=2$

$3k+5l=2(5k+7l)$

$3k+5l=10k+14l$

$7k+9l=0$

3(c)(ii)

(iii)

logically, no matter how large l becomes, short of reaching infinity it can never make the line $5x-7y+4=0$

to prove, slope of $5x-7y+4=0 \qquad =-\frac{5}{-7}=\frac{5}{7}$

if we put that slope $=\frac{3k+5l}{5k+7l}$ we get a contradiction:

$\frac{5}{7}=\frac{3k(1)+5l}{5(1)+7l}$

$5(5+7l)=7(3+5l)$

$25+35l=21+35l$

$25=21$ which cannot be true.

3(c)(iii)

Question 4

(a)

$(cosA+sinA)^2=cos^2A+sin^2A+2cosAsinA=1+sin2A$

(b)

$6cos^2x+sinx-5=0$

$6(1-sin^2x)+sinx-5=0$

$6-6sin^2x+sinx-5=0$

$6sin^2x-sinx-1=0$

$(3sinx+1)(2sinx-1)=0$

$sinx=\frac{-1}{3} \qquad or \qquad \frac{1}{2}$

$x=(180+19)^o \qquad or \qquad (360-19)^o \qquad or \qquad 45^o \qquad or \qquad (90+45)^o$

$=30^o \qquad , \qquad 150^o \qquad, \qquad 199^o \qquad, \qquad 341^o$

4(b)

(c)

(i)

draw the triangle abc(note:acb is a right angle, JC proof)	draw the triangle aoc
$cos\alpha=\frac{\|ac\|}{\|ab\|}$	draw a line from o to ac, splitting it perpendicularly.
$\|ac\|=\|ab\|cos\alpha$	right angled triangle of side r and $\frac{1}{2}\|ac\|$ and angle $\alpha$
$=2rcos\alpha$	$\frac{1}{2}\|ac\|=rcos\alpha$
	$\|ac\|=2rcos\alpha$

4(c)(i)

(ii)

draw a line from o to c

the sector, ocb has angle $2\alpha$ and area $\frac{2\alpha r^2}{2} = \alpha r^2$

the triangle aoc has base $|ac| \qquad=2rcos\alpha$ and height $rsin\alpha$

$\therefore$ area $=\frac{1}{2} 2rcos\alpha rsin\alpha = \frac{1}{2} r^2 sin2\alpha$

these two areas together make half the area of the semi circle $=\frac{\pi r^2}{2}$

$\alpha r^2 + \frac{1}{2}r^2sin2\alpha = \frac{\pi r^2}{4}$ [ $\qquad r^2$ cancle across the equation.]

$2\alpha + sin2\alpha = \frac{\pi}{2}$

Question 5

(a)

$\lim_{x\right0}\frac{sin2x}{sin3x} =\lim_{x\right0}\frac{sin2x}{2x}\frac{3x}{sin3x}\frac{2x}{3x} =\frac{2}{3}$

because $\lim_{x\right0}\frac{sinx}{x}=1$

5(a)

(b)

$cos(A+B)=cosAcosB-sinAsinB$

$cos(A-B)=cosAcos-B-sinAsin-B$ [remember $cos-A=cosA$ and $sin-A=-sinA$ ]

$cos(A-B)=cosAcosB+sinAsinB$

now take $A=90-A$ and remember $cos(90-A)=sinA$ and $sin(90-A)=cosA$

$sin(A+B)=cos(90-A-B)=cos(90-A)cosB+sin(90-A)sinB$

$=sinAcosB+cosAsinB$

(c)

$tan60=\frac{h}{\|pr\|}$	$tan30=\frac{h}{\|qr\|}$
$\sqrt{3}=\frac{h}{\|pr\|}$	$\frac{1}{\sqrt{3}}=\frac{h}{\|qr\|}$
$\|pr\|=\frac{h}{\sqrt{3}}$	$\|qr\|=h\sqrt{3}$

triangle of side $\frac{h}{\sqrt{3}}$ , $h\sqrt{3}$ and c.

$c^2=\big(\frac{h}{\sqrt{3}}\big)^2+\big(h\sqrt{3}\big)^2 -2\frac{h}{\sqrt{3}}h\sqrt{3}cos|prq|$

$cos|prq|=\frac{\big(\frac{h}{\sqrt{3}}\big)^2+\big(h\sqrt{3}\big)^2-c^2}{2\frac{h}{\sqrt{3}}h\sqrt{3}}$

but $c^2=\frac{13h^2}{3}$

$cos|prq|=\frac{\big(\frac{h}{\sqrt{3}}\big)^2+\big(h\sqrt{3}\big)^2-\frac{13h^2}{3}}{2\frac{h}{\sqrt{3}}h\sqrt{3}} =\frac{\frac{h^2}{3}+h^23-\frac{13h^2}{3}}{2h^2} =\frac{\frac{1}{3}+3-\frac{13}{3}}{2} =-\frac{1}{2}$

$|prq|=120^o$

5(c)

Question 6

(a)

(i)

6 possible people in first seat, 5 in second etc $=6!=720$

6(a)(i)

(ii)

take mary and john as a unit and give arrangements of five times arrangements of two(the john and mary unit) $5!\times2!=240$

6(a(ii)

(b)

$\alpha \qquad and \qquad \beta$ are roots of $px^2+qx+r=0$

$\therefore \qquad p\alpha^2+q\alpha+r=0$ and $p\beta^2+q\beta+r=0$

$u_n=l\alpha^n+m\beta^n$

$pu_{n+2}+qu_{n+1}+ru_n=p(l\alpha^{n+2}+m\beta^{n+2})+q(l\alpha^{n+1}+m\beta^{n+1})+r(l\alpha^n+m\beta^n)$

$=l\alpha^n(p\alpha^2+q\alpha+r)+m\beta^n(p\beta^2+q\beta+r)$

$=l\alpha^n(0)+m\beta^n(0)=0$

(c)

(i)

probability of first disk being red $=\frac{r}{w+r}$

probability of second disk being red, given that first disk was $=\frac{r-1}{w+r-1}$

probability of both being red $=\frac{r}{w+r}\times\frac{r-1}{w+r-1}=\frac{r(r-1)}{(w+r)(w+r-1)}$

6(c(i)

(ii)

$w=1 \qquad \qquad p=\frac{r(r-1)}{(1+r)(1+r-1)}=\frac{1}{2}$

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

$2r-2=1+r$

$r=3$

6(c(ii)

(iii)

if $w=2$ and $p=\frac{1}{2}$

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

$2r^2-2r=r^2+3r+2$

$r^2-5r-2=0$

$b^2-4ac=25+8=33$

the root of 33 is not a whole number and r must be an integer (must have a whole number of red disks)

if $w=4$ and $p=\frac{1}{2}$

$\frac{r(r-1)}{(4+r)(4+r-1)}=\frac{1}{2}$

$2r^2-2r=r^2+7r+12$

$r^2-9r-12=0$

$b^2-4ac=81+12=93$

the root of 93 is not a whole number and r must be an integer (must have a whole number of red disks)

if $w=6$ and $p=\frac{1}{2}$

$\frac{r(r-1)}{(6+r)(6+r-1)}=\frac{1}{2}$

$2r^2-2r=r^2+11r+30$

$r^2-13r-30=0$

$(r-15)(r+2)$

$r=15$ or $-2$ but r cannot be negative.

$\therefore \qquad w=6$ and $r=15$

6(c(iii)

Question 7

(a)

(i)

how many four letter selections from seven $=\begin{pmatrix} 7\\ 4 \end{pmatrix}=35$

7(a)(i)

(ii)

3 vowels and 4 consonants, only on possibility for no vowels.

$35-1=34$

7(a)(ii)

(b)

(i)

total possibilities after throwing two dice $=6\times6=36$

number of possibilities with identical numbers - 6

number of possibilities that add up to five = 4

number of possibilities that do both - 0

$\frac{6}{36}+\frac{4}{36}=\frac{5}{18}$

7(b)(i)

(ii)

list the possibilities- $(3,6)(6,3)(4,4)(4,5)(5,4)(4,6)(6,4)(5,6)(6,5)(5,5)(6,6)$

$=\frac{11}{36}$

7(b)(ii)

(c)

(i)

numbers: $a,a+d,a+2d,a+3d,a+4d,a+5d,a+6d$

mean: $\frac{7a+21d}{7}=a+3d$

7(c)(i) a + 3d

(ii)

deviations: $3d,2d,d,0,d,2d,3d$

squared: $9d^2,4d^2,d^2,0,d^2,4d^2,9d^2$

standard deviation: $\sqrt{\frac{9d^2+4d^2+d^2+0+d^2+4d^2+9d^2}{7}}=\sqrt{\frac{28d^2}{7}}=\sqrt{4d^2}=2d$

Section B

Question 8

(a)

$p+q=1 \qquad q=1-p$

$pq=p(1-p)=p-p^2$

differentiate: $1-2p$

put equal to zero to find max value $1-2p=0$

$p=\frac{1}{2}$

8(a)

(b)

(i)

$f(x)=(1+x)^m$	$f(0)=(1)^m=1$	$a_0=1$
$f'(x)=m(1+x)^{m-1}$	$f'(0)=m(1)^{m-1}=m$	$a_1=m$
$f''(x)=m(m-1)(1+x)^{m-2}$	$f''(0)=m(m-1)(1)^{m-2}=m(m-1)$	$a_2=\frac{m(m-1)}{2!}$
$f'''(x)=m(m-1)(m-2)(1+x)^{m-3}$	$f'''(0)=m(m-1)(m-2)(1)^{m-3}=m(m-1)(m-2)$	$a_3=\frac{m(m-1)(m-2)}{3!}$

$1+mx+\frac{m(m-1)}{2!}x^2+\frac{m(m-1)(m-2)}{3!}x^3$

(ii)

$U_{r+1}=\frac{m(m-1)(m-2)...(m-r+1)}{r!}x^r$

$U_r=\frac{m(m-1)(m-2)...(m-r+2)}{(r-1)!}x^{r-1}$

test for convergence:

$\lim_{r\right\infty}\frac{U_{r+1}}{U_r} =\lim_{r\right\infty}\frac{m(m-1)(m-2)...(m-r+1)}{r!}x^r\times\frac{(r-1)!}{m(m-1)(m-2)...(m-r+2)x^{r-1}}$

$=\lim_{r\right\infty}\frac{(m-r+1)}{r}x \qquad$ [ $x^r$ cancles out $x^{r-1}$ . $r!$ cancles out $(r-1)!$ . The pattern in $m(m-1)$ etc is getting smaller so the smaller number $m-r+1$ remains.]

$=\lim_{r\right\infty}\big(\frac{m}{r}-1+\frac{1}{r}\big)x$

$=\big(0-1+0\big)x$

$=-x$

$\therefore$ series converges for $-1<x<1$

(c)

$\int_0^1tan^{-1}xdx$

$u=tan^{-1}x$	$dv=dx$
$du=\frac{1}{1+x^2}dx$	$v=x$

$\int udv=uv-\int vdu$

$\int_0^1tan^{-1}xdx=\Bigg(tan^{-1}x\times x-\int x\frac{1}{1+x^2}dx\Bigg)_0^1$

$=\Bigg(xtan^{-1}x\Bigg)_0^1-\int_0^1 \frac{x}{1+x^2}dx =\Bigg(1tan^{-1}1-0\Bigg)-\int_0^1 \frac{x}{1+x^2}dx$

$=\Bigg(\frac{\pi}{4}\Bigg)-\int_0^1 \frac{x}{1+x^2}dx$

$u=1+x^2 \qquad du=2xdx \qquad \frac{1}{2}du=xdx$

change of limits: when $x=1 \qquad, \qquad u=1+1=2$ and when $x=0 \qquad, \qquad u=1+0=1$

$=\frac{\pi}{4}-\frac{1}{2}\int_1^2 \frac{1}{u}du$

$=\frac{\pi}{4}-\frac{1}{2}(ln2-ln1)$

$=\frac{\pi}{4}-\frac{1}{2}(ln2-0)$

$=\frac{\pi}{4}-\frac{ln2}{2}$

8(c)

Question 10

(a)

(i)

The set of natural numbers under subtraction is not closed. For example, $1-2=-1$ , and $-1$ is not a natural number. Subtraction is also not associative.

(ii)

$0$ has no inverse in the set of real numbers.

(b)

(i)

First, let's draw the Cayley table:

o	$I_{\pi}$	$R_{180}$	$S_{x}$	$S_{y}$
$I_{\pi}$	$I_{\pi}$	$R_{180}$	$S_{x}$	$S_{y}$
$R_{180}$	$R_{180}$	$I_{\pi}$	$S_{y}$	$S_{x}$
$S_{x}$	$S_{x}$	$S_{y}$	$I_{\pi}$	$R_{180}$
$S_{y}$	$S_{y}$	$S_{x}$	$R_{180}$	$I_{\pi}$

From this, we can clearly see that this set is closed under composition.

We can also see that $I_{\pi}$ o $a=a$ o $I_{pi}$ for all $a$ , this $I_{\pi}$ is the identity.

As $I_{pi}$ appears exactly once in every row and column, we have inverses.

Finally, we may assume associativity.

Thus, G forms a group under compisition.

(ii)

Note from the Cayley table that G is an Abelian/commutitive group - thus $Z(G)=G=\{I_{\pi},R_{180},S_{x},S_{y}\}$ .

(c)

(i)

By Lagrange's theorem, if a group has prime order, then it has no proper subgroups.

Let $n$ be the order of $a$ , $a \not= e$ . Then the group { $e$ , $a$ , $a^{2}$ ,..., $a^{n-1}$ } is a subgroup of G. But G has no proper subgroups, so this must be the whole group. Thus the group is generated by $a$ and so is cyclic.

(ii)

Let $n$ be the order of $a$ , $a \not= e$ . Then the group { $e$ , $a$ , $a^{2}$ ,..., $a^{n-1}$ } is a subgroup of G, with $n$ elements.

Therefore, by Lagrange, $n$ is a factor of the order of $G$ . Thus the order of any element is a factor of the order of the group.

$x=5+7cos\theta$	$y=7sin\theta$
$cos\theta=\frac{x-5}{7}$	$sin\theta=\frac{y}{7}$
$cos^2\theta=\frac{x^2-10x+25}{49}$	$sin^2\theta=\frac{y^2}{49}$

$x^2+y^2-4x-6y+5=0$	$x^2+y^2-6x-8y+23=0$
centre $=(2,3) \qquad r=\sqrt{4+9-5}=\sqrt{8}=2\sqrt{2}$	centre $=(3,4) \qquad r=\sqrt{9+16-23}=\sqrt{2}$

draw the triangle abc(note:acb is a right angle, JC proof)	draw the triangle aoc
$cos\alpha=\frac{\|ac\|}{\|ab\|}$	draw a line from o to ac, splitting it perpendicularly.
$\|ac\|=\|ab\|cos\alpha$	right angled triangle of side r and $\frac{1}{2}\|ac\|$ and angle $\alpha$
$=2rcos\alpha$	$\frac{1}{2}\|ac\|=rcos\alpha$
	$\|ac\|=2rcos\alpha$

$tan60=\frac{h}{\|pr\|}$	$tan30=\frac{h}{\|qr\|}$
$\sqrt{3}=\frac{h}{\|pr\|}$	$\frac{1}{\sqrt{3}}=\frac{h}{\|qr\|}$
$\|pr\|=\frac{h}{\sqrt{3}}$	$\|qr\|=h\sqrt{3}$

Solutions:Maths HL 2007 Paper 2

From ZuluNotes - Free Leaving Cert Notes

Contents

Section A

Question 1

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Section B

Question 8

Question 10

Views

Personal tools

Navigation

SEARCH

Subjects

Search

Toolbox