2011 HSC Course

Half Yearly Examination

MATHEMATICS EXTENSION 1

Name.

Maths Teacher: Maths Line:

Set by: S. Warda and C. Schuster

General Instructions • Writing time — 2 hours • Write using blue or black pen • Calculators may be used.

Total marks (77) • Attempt questions 1— 7

The following outcomes are examined in this paper HE2: uses inductive reasoning in the construction of proofs. HE3: uses a variety of strategies to investigate mathematical models of situations

involving binomial probability, projectiles, simple harmonic motion or exponential growth and decay.

HE6: determines integrals by reduction to a standard form through a given substitution. HE7: evaluates mathematical solutions to problems and communicates them in an

annrooriate form. Q1 Q2 Q3 Q4 Q5 Q6 Q7 Total

HE2 /12

HE3 /10

HE3 /11

HE7 /13

HE7 /9

HE7 /14

HE7 /8

/77

HE2 /7

HE3 /13

HE6 /23

HE7 /31

TOTAL /74

QUESTION 1 ( 12 marks) Marks Outcome

HE2 a) Write the sum 5 + 9 + 13 +....+ 4n — 3 in Sigma Notation 1

b) Use the method of mathematical induction to prove that

5 (1 + 0+ (2 + 3) + (3 + 5) + ... + (n + (2n — 0) = 2 n(3n + 1)

where n is a positive integer.

c) (i) By considering the cases where a positive integer k is even 2 (k = 2x) and odd (k = 2x + 1), show that k 2 + k is always

even. i.e. k 2 + k = 2m , where m is also an integer.

(ii) Prove, by Mathematical induction, that for all positive integral 4 values of n, n 3 + 5n is divisible by 6.

QUESTION 2 Start a new page ( 10 marks ) Marks Outcome a) When Lleyton and Bec play tennis, Bec has a 0.1 chance of winning

any particular game. On the weekend they intend to play 6 games. Calculate the probability ( to 2 significant figures) that :

(i) Bec wins no games. 2

(ii) Bec wins at least one game. 2

(iii) Bec wins 4 or more games. 2

b) The probability of George being late to class is 0.2. He is supposed to attend 20 classes in a week. The Year Coordinator has said that if George is late to three or more classes in a week he will receive a detention.

(i) Show that the probability that George receives a detention for being 2 late is 0.79 correct to two decimal places.

(ii) The principal said that if George received n or more detentions for 2 being late in a ten week term, he would be suspended. George's mates worked out that the probability of George being suspended was 0.65 (correct to two decimal places). Using your answer to part (i), find the value of n.

HE3

QUESTION 4 Start a new page ( 13 marks ) Marks Outcome

a) What is the coefficient of x 3 in the expansion of (2x — 5) 7 ? 3 HE7

b) Find the term independent of x in the expansion of fx

2 --

29

.x, 3

c) If (3+2x) '8 is expressed in ascending powers of x, find the greatest 4 coefficient. Leave your answer in index form.

d) Find the coefficient of x 2 in the expansion of (2x +4) 5 +(x2 —3)6

3

QUESTION 5 Start a new page ( 9 marks ) Marks Outcome

a) In the diagram below, 0 is the centre of the circle, AC is a tangent at 3 HE7 B, and D and E are points on the circumference. If L ABD = 80 ° and Z DBE = 40 ° , find the size of L BEO, giving reasons.

b) In the circle centred at 0, the chords AB and CD intersect at E. The length

2 of AB is x cm and of CD is y cm. AE = 4 cm and CE = 3 cm.

Show that 4x = 3y + 7

c)

The circle ABCD has centre 0. Tangents are drawn from an external point E to contact the circle at C and D.

LCBD = x° and LBAD = y° .

(i) Show that ZCED = (180 —2x) ° . 2

(ii) Show that ZBDC = (y— x) ° . 2

QUESTION 6 Start a new page ( 14 marks ) Marks Outcome

— a) Find the horizontal asymptote of the function y = 2x2 1 HE7 2 , 4x +3

X - D

b) Suppose the cubic f(x) = x 3 + axe + bx + c has a relative maximum at x = a and a relative minimum at x = (3.

(I) Prove that a + f3= -- 2 a

3

(ii) Deduce that the point of inflexion occurs at x = a + fi

2

c) Let f(x)= x2 x - 1

(I)

For what values of x is f(x) undefined

(ii) Show that y = f(x) is an odd function

(iii) Show that f/(x) <0 for all values of x for which the function is defined

(iv) What happens to the function y =f(x) as x --> -±co

(v) Hence sketch y = f(x)

3

2

1

2

2

1

il.

QUESTION 7 Start a new page ( 8 marks ) Marks Outcome

a) Given f(x) = log e R9 — x2 ) , state the domain of f(x). 2 HE7

Consider y = e kx where k is a constant.

d2 y Find —1-/-cl and

dx dx2 Determine the values of k for which y = ekx satisfies the equation

dy +7—dy +12y = 0 dx2 dx

Let n be a positive integer. By considering the graph of y = —1

below, x

explain why 1 1

n+1 dx 1 < — – n+1 . < n x n

n n+1

b)

c)

4

2

•V 0. t+ Vane

•• --7-)■/?n

(1-1-1)3 (14-70 •-0pcvd

-z u -0 iv_ Si .f -f• c v iop afar. c sro

TI u .•••

(0-F---ir)-+ 14

('-")(+-f-lc) T- (-÷ -71

r-7 -719+ E [(74--ior t (1-) ->f

CZ" (1-t :) -1

((i-1-?)41) (Ef-C) (i41) = S441

-,17-÷"10Q-"9c (-e-lz)-t 1+-1) f ((I—az)4 -)0 (E•7) t (1+1) -a% 4 -1 'a AA/. n aid

~1- -17) -)1 + • • - + r) (1-4- 1) •a1

-)1 os m.(4_ -2 wnssk/

c44 =

(11-IwT)(,)1-- = S H-y 1 4 1

•

{-1 1

~((1 - v-c) 4 v) + (.5 4 I) f ( -r) + (1+I) (9

93. sti•-•-Yr0 A

-"cr■-•-•flso.:1 rro r--sa 7-z LI oS --a

+ u‘.11 (,) 9 2 (.4 9

-+ 170 Z' (-1-c z -7) --A

-P -18-4 TIE S 4 "IS 4. I 4-1 £ 4;11

I- )

..,-,,f)sod -0 s-1 w w 121 •4I .59)

c) 419 -Or )31P‘rvr " ( 9s I 1 9 "49 —111'svni"

•

cu (")

•viaf cf2-)-0 C"yro S I 71 4 71

V3/v"7 S ) "\-rov`l r' '"g

(1.4 4 .2-X-C ) -C'

4* I 4 -X 7-797 (1"-P-n) --P 7(1 --t -x-C)

-r = -0! re° s 1 -)1

treA"-d Si

(--x

ire 4. -;x t7

%-/Diy-0 S, vaNc-i C,

frfl 1 vaisi x3 ri

"1°`-Esne

•-11.1)

•-e YI = x r-f- X. = (0

E's. 0 5(6'0) e,(ro) -; 14,0j 0)7 .= (0 )961 (9

02' 1 GO b'o 11, o

s -8 1-1 = 1 -1(

S- x s -17 -f- s--xr # 1 Sfi'))

8 s-z-o ft 60•0

0 z. -2.

=•'07: -1

b

C81 - ELY - o

E -c÷T T o T

'e Y-1 65 -x_ =

v _ ?yr

~~

0

"X -19 •i° - 6 c -)6 cf

..c.(--C-4

-D n n 1 -c

7 f[ c r --r

"1° z vi

— 7 V/

C) TOT -0) 8(10L'a ) g = u

-SZ (I7•0) 6(E:“. -0) b 6L•cr) ol)

(0'1 'I +v U v x)

L,L•0 8

// bL•0 = j‘cg Vt-a) -De_za bi o)(-to)'JOZ e,(8.

cr ( 1 -6 .% d — ,..

oZ rio 4 cl

3 - =

6'0 Z'O J 0 (9 - -v = X)d

E-100

p•0) 4 t(b.0)30.10 5-) 4 (2'0) (CO) (`,1 s •7: x ) d 610

// L 0 =

Eis • o

x) d - I -- (-) -ct )e) (.D

-f-

— I — S 3

IT - k. k_ k • ?C, . 3 .

3 (% - k... a 1e.--1 Ic- 1

= (4 +L3)-) ( 3 + 1) = I + 2_32,-. 3

Sucs-11,;r1

C2x -

- = k. 1

k- - C,_ E s-)

3 'Co c 7c.

C-0. 411- 1 C 4_ (2) 3 (-01- I75'°°°

1+. • _5 3

• 1 4- 3 .1 ( 1- 2-

# 3 2-PC -5 c.L% .+1 = 19.c, (3)

11-k (20

k.

k C.,. 3 2 7,-

i?C 3 . z x

18 - ~ 1 9 -rte) (1q-lc t 3

L' (tg-k.) ). t el 3 1- -k_

(I -10 . z . . 3

4. re,2-4-e_s w. :

I P- > 3

3 8 31e__ F

9 II '7

C . 3 .2

= Ex_ + (z.x. -s-)3

2- 2L- )

C -2;T:

"I 'l c 7C 16-2-14. • E-4 1` q C eVe... X 18 -3k

ti

u-aa.fe.101 o f 4errn I? -34_ “)

b) 24,Z

.4 4-I

I / as-17 I - SD,

'

S 7" .71Z

Cir-D • -1--z --op( 71—),

CF---) • ( -)c) =

-17 --s 0-0 .71D, 04 -x I „a_

_s

7.011 = 8S — 0.°S

‘,/ s- Lf1 f-v—r-r-, fka c-7 -c

= 7I —.5 -z

(c -2"lc )

7- ( 047.)F I

ler2A-1? -J1--ep sa

— x z

frr ° d

/ I -7 -)C

X — - = (x — ) j- -

j- ct.+ "col! t z z 09, d-

--)+ .o ,+ z (A)

Ps ro

-S I OCI <"*-- _S '1( x- ""rr -A 17

,,t R.55,0 rr S_ 7-x

r. 2 -xt 6,

"arlri c-, rr 0 ">.

a <— 0

7 (-;-0 z

-)4

(

f --

T( 'l 7-X ) o c- -X .4- I S

— —

--X - 1 -

00- <- 5- so

00 c oC/N)

(4-70(9-f?,)

•0--c _ T t-><,

erre -i- %7 z- 4-

dwr Oc 0), j .14" i Q. co- -r jvc 0-t ('`), a-f s'Nrn°9 r -731,1S

9 1C,Or + 7

0 (-9) "gra., 0 •-fon_f cr9. 0

E- "fT_

4 -1") -?1 0, ) j_ md