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© R J Aus SHS ME1 Harder page 1

Singleton High School Mathematics Extension 1, Harder Questions 2018 2017

6 The point P2p

,1

p2

⎛

⎝⎜

⎞

⎠⎟ , where p ≠ 0, lies on the parabola x2 = 4y.

What is the equation of the normal at P?

A. px − x = − p B. p2 y + px = −1

C. p2 y − p3x = 1− 2 p2 D. p2 y + p3x = 1+ 2 p2

7 Which diagram represents the domain of the function f (x) = sin−1 3

x⎛⎝⎜

⎞⎠⎟

?

8 A stone drops into a pond, creating a circular ripple. The radius of the ripple increases from 0 cm,

at a constant rate of 5 cm s–1.At what rate is the area enclosed within the ripple increasing when the radius is 15 cm?

A. 25π cm2s−1 B. 30π cm2s−1 C. 150π cm2s−1 D. 225π cm2s−1

9 When expanded, which expression has a non-zero constant term?

A. x + 1

x2⎛⎝⎜

⎞⎠⎟

7

B. x2 + 1

x3⎛⎝⎜

⎞⎠⎟

7

C. x3 + 1

x4⎛⎝⎜

⎞⎠⎟

7

D. x4 + 1

x5⎛⎝⎜

⎞⎠⎟

7

10 Three squares are chosen at random from the 3 × 3 grid shown, and a cross is placed in each chosen square.What is the probability that all three crosses lie in thesame row, column or diagonal?

A.128

B.221

C.13

D.89

2016

6 What is the general solution of the equation 2sin2 x − 7sin x + 3= 0?

(A) nπ − (−1)n π3

(B) nπ + (−1)n π3

(C) nπ − (−1)n π6

(D) nπ + (−1)n π6

7 The displacement x of a particle at time t is given by x = 5sin4t +12cos4t.What is the maximum velocity of the particle?(A) 13 (B) 28 (C) 52 (D) 68

8 A team of 11 students is to be formed from a group of 18 students. Among the 18 studentsare 3 students who are left-handed.What is the number of possible teams containing at least 1 student who is left-handed?(A) 19448 (B) 30459 (C) 31824 (D) 58344

9 The diagram shows the graph of y = f (x).


Which of the following is a correct statement?(A) ′′f (1) < f (1) <1< ′f (1) (B) ′′f (1) < ′f (1) < f (1) <1(C) f (1) <1< ′f (1) < ′′f (1) (D) ′f (1) < f (1) <1< ′′f (1)

10 Consider the polynomial p(x) = ax3 + bx2 + cx − 6 with a and b positive.Which graph could represent p(x)?

2015

6 What is the domain of the function f (x) = sin−1(2x)?

(A) −π ≤ x ≤ π (B) −2 ≤ x ≤ 2 (C) − π4≤ x ≤ π

4(D) − 1

2≤ x ≤ 1

2

7 What is the value of k such that 1

4− x2dx

0

k

∫ = π3

?

(A) 1 (B) 3 (C) 2 (D) 2 3

8 What is the value of limx→3

sin(x − 3)(x − 3)(x + 2

?

(A) 0 (B)15

(C) 5 (D) Undefined

9 Two particles oscillate horizontally. The displacement of the first is given by x = 3sin4t and thedisplacement of the second is given by x = asin nt. In one oscillation, the second particle coverstwice the distance of the first particle, but in half the time.What are the values of a and n?(A) a = 1.5, n = 2 (B) a = 1.5, n = 8 (C) a = 6, n = 2 (D) a = 6, n = 8

10 The graph of the function y = cos 2t − π3

⎛⎝⎜

⎞⎠⎟

is shown below.

What are the coordinates of the point P?

(A)5π12

,0⎛⎝⎜

⎞⎠⎟

(B)2π3

,0⎛⎝⎜

⎞⎠⎟

(C)11π12

,0⎛⎝⎜

⎞⎠⎟

(D)7π6

,0⎛⎝⎜

⎞⎠⎟

2017

13(a) A particle is moving along the x-axis in simple harmonic motion centred at the origin. 3When x = 2 the velocity of the particle is 4. When x = 5 the velocity of the particle is 3.Find the period of the motion.

13(b) Let n be a positive EVEN integer.

(i) Show that (1+ x)n + (1− x)n = 2n0

⎛⎝⎜

⎞⎠⎟+

n2

⎛⎝⎜

⎞⎠⎟

x2 + ...+nn

⎛⎝⎜

⎞⎠⎟

xn⎡

⎣⎢

⎤

⎦⎥. 2

(ii) Hence show that n (1+ x)n−1 − (1− x)n−1⎡⎣

⎤⎦ = 2 2

n2

⎛⎝⎜

⎞⎠⎟

x + 4n4

⎛⎝⎜

⎞⎠⎟

x3 + ...+ nnn

⎛⎝⎜

⎞⎠⎟

xn−1⎡

⎣⎢

⎤

⎦⎥. 1

(iii) Hence show that n2

⎛⎝⎜

⎞⎠⎟+ 2

n4

⎛⎝⎜

⎞⎠⎟+ 3

n6

⎛⎝⎜

⎞⎠⎟

...+ n2

nn

⎛⎝⎜

⎞⎠⎟= n2n−3. 2


13(c) A golfer hits a ball with initial speed V ms–1 at an angle θ to the horizontal.The golf ball is hit from one side of a lake and must have a horizontal rangeof 100 m or more to avoid landing in the lake.Neglecting the effects of air resistance, the equations describing the motionof the ball are x =Vt cosθ

y =Vt sinθ − 12

gt2

where t is the time in seconds after the ball is hit and g is the acceleration

due to gravity in ms−2. Do NOT prove these equations.

(i) Show that the horizontal range of the golf ball is V 2 sin2θg

metres. 2

(ii) Show that if V 2 <100g then the horizontal range of the ball is less than 100 m. 1

It is now given that V 2 = 200g and that the horizontal range of the ball is 100m or more.

(iii) Show that π12

≤θ ≤ 5π12

. 2

(iv) Find the greatest height the ball can achieve. 2

2016

13(a) The tide can be modelled using simple harmonic motion.At a particular location, the high tide is 9 metres and the low tide is 1 metre.At this location the tide completes 2 full periods every 25 hours.Let t be the time in hours after the first high tide today.

(i) Explain why the tide can be modelled by the function x = 5+ 4cos4π25

t⎛⎝⎜

⎞⎠⎟

. 2

(ii) The first high tide tomorrow is at 2 am. What is the earliest time tomorrow at which the tide is 2increasing at the fastest rate?

13(b) The trajectory of a projectile fired with speed u ms−1 at an angle θ to the horizontal is represented

by the parametric equations x = ut cosθ and y = ut sinθ −5t2, where t is the time in seconds.

(i) Prove that the greatest height reached by the projectile is u2 sin2θ

20. 2

A ball is thrown from a point 20 m above the horizontal

ground. It is thrown with speed 30 ms−1 at an angle of 30° to the horizontal. At its highest point the ball hits a wall,as shown in the diagram.

(ii) Show that the ball hits the wall at a height of 1254

m above the ground. 2

The ball then rebounds horizontally from the wall with speed 10 ms−1. You may assume that

the acceleration due to gravity is 10 ms−2.(iii) How long does it take the ball to reach then ground after it rebounds from the wall? 2(iv) How far from the wall is the ball when it hits the ground? 1


13(c) The circle centred at O has a diameter AB. From the point Moutside the circle the line segments MA and MB are drawnmeeting the circle at C and D respectively, as shown in the diagram.The chords AD and BC meet at E. The line segment ME producedmeets the diameter AB at F.Copy or trace the diagram into your writing booklet.

(i) Show that CMDE is a cyclic quadrilateral. 2(ii) Hence, or otherwise, prove that MF is perpendicular to AB. 2

2015

13(a) A particle is moving along the x-axis in simple harmonic motion.

The displacement of the particle is x metres and its velocity is v ms−1.

The parabola below shows v2 as a function of x.(i) For what value(s) of x is the particle at rest? 1(ii) What is the maximum speed of the particle? 1(iii) The velocity of the particle is given by the equation 3

v2 = n2 a2 − (x − c)2( )where a, c and n are positive constants.What are the values of a, c and n?

13(b) Consider the binomial expansion 2x + 13x

⎛⎝⎜

⎞⎠⎟

18

= a0x18 + a1x16 + a2x14 + ...

where a0,a1,a2,... are constants.

(i) Find an expression for a2. 2

(ii) Find an expression for the term independent of x. 2

13(c) Prove by mathematical induction that for all integers n ≥1, 312!+ 2

3!+ 3

4!+ ...+ n

(n+1)!= 1− 1

(n+1)!.

13(d) Let f (x) = cos−1(x)+ cos−1(−x), where −1≤ x ≤1.(i) By considering the derivative of f (x), prove that f (x) is constant. 2

(ii) Hence deduce that cos−1(−x) = π − cos−1(x). 1

2017

14(a) Prove by mathematical induction that 82n+1 + 62n−1 is divisible by 7, for any integer n ≥1. 3

14(b) Let P 2 p, p2( ) be a point on the parabola x2 = 4y.

The tangent to the parabola at P meets the parabola x2 = −4ay,a > 0, at Q and R. Let M be the midpoint of QR.

(i) Show that the x coordinates of R and Q satisfy 2

x2 + 4apx − 4ap2 = 0.

(ii) Show that the coordinates of M are −2ap,− p2(2a +1)( ). 2

(iii) Find the value of a so that the point M always lies on 2

the parabola x2 = −4y.


14(c) The concentration of a drug in a body is F(t), where t is the time in hours after the drug is taken.Initially the concentration of the drug is zero. The rate of change of concentration of the drug is

given by ′F (t) = 50e−0.5t − 0.4F(t).

(i) By differentiating the product F(t)e0.4t show that ddt

F(t)e0.4t( ) = 50e−0.1t . 2

(ii) Hence, or otherwise, show that F(t) = 500 e−0.4t − e−0.5t( ). 2

(iii) The concentration of the drug increases to a maximum. 2For what value of t does this maximum occur?

2016

14(a)(i) Show that 4n3 +18n2 + 23n+ 9 can be written as (n+1) 4n2 +14n+ 9( ). 1

(ii) Using the result in part (i), or otherwise, prove by mathematical induction that, for n ≥1, 3

1× 3+ 3×5+5× 7 + ...+ (2n−1)(2n+1) = n3

4n2 + 6n−1( ).

14(b) Consider the expansion of (1+ x)n , where n is a positive integer.

(i) Show that 2n =n0

⎛⎝⎜

⎞⎠⎟+

n1⎛⎝⎜

⎞⎠⎟+

n2

⎛⎝⎜

⎞⎠⎟+!+

nn

⎛⎝⎜

⎞⎠⎟

1

(ii) Show that n2n−1 =n1⎛⎝⎜

⎞⎠⎟+ 2

n2

⎛⎝⎜

⎞⎠⎟+ 3

n3

⎛⎝⎜

⎞⎠⎟+!+ n

nn

⎛⎝⎜

⎞⎠⎟

. 1

(iii) Hence, or otherwise, show that nr

⎛⎝⎜

⎞⎠⎟

(2r − n)

r=1

n

∑ = n. 2

14(c) The point T 2at,at2( ) lies on the parabola P1 with equation x2 = 4ay.

The tangent to the parabola P1 at T meets the directrix at D. The normal

to the parabola P1 at T meets the vertical line through D at the point R,

as shown in the diagram.

(i) Show that the point D has coordinates at − at

,−a⎛⎝⎜

⎞⎠⎟

. 1

(ii) Show that the locus of R lies on another parabola P2. 3

(iii) State the focal length of the parabola P2. 1

It can be shown that the minimum distance between R and T occurs when thenormal to P1 at T is also the normal to P2 at R. (Do NOT prove this.)

(iv) Find the values of t so that the distance between R and T is a minimum. 2

2015

14(a) A projectile is fired from then origin O with initial velocity V ms−1 at anangle θ to the horizontal. The equations of motion are given by

x =VT cosθ , y =Vt sinθ − 12

gt2. (Do NOT prove this.)

(i) Show that the horizontal range of the projectile is V 2 sin2θg

. 2

A projectile is fired so that θ = π3

.

(ii) Find the angle that this projectile makes with the horizontal when t = 2V3g

. 2


(iii) State whether this projectile is travelling upwards or downwards when t = 2V3g

. 1

Justify your answer.

14(b) A particle is moving horizontally. Initially the particle is at the origin O moving with velocity 1 ms−1.The acceleration of the particle is given by !!x = x −1, where x is its displacement at time t.

(i) Show that the velocity of the particle is given by !x = 1− x. 3(ii) Find an expression for x as a function of t. 2(iii) Find the limiting position of the particle. 1

14(c) Two players a and B play a series of games against each other to get a prize. In any game,either of the players is equally likely to win.To begin with, the first player who wins a total of 5 games gets the prize.

(i) Explain whay the probability of player A getting the prize in exactly 7 games is 64

⎛⎝⎜

⎞⎠⎟

12

⎛⎝⎜

⎞⎠⎟

7

. 1

(ii) Write an expression for the probability of player A getting the prize in at most 7 games. 1(iii) Suppose now that the prize is given to the first player tio win a total of (n+1) games, 2

where n is a positive integer.By considering the probability that A gets the prize, prove that

nn

⎛⎝⎜

⎞⎠⎟

2n +n+1

n⎛⎝⎜

⎞⎠⎟

2n−1 +n+ 2

n⎛⎝⎜

⎞⎠⎟

2n−2 + ...+2nn

⎛⎝⎜

⎞⎠⎟= 22n.

2014

14(b) Two players A and B play a game that consists of taking turns until a winner is determined.Each turn consists of spinning the arrow on a spinner once. The spinner has three sectors P, Q and R.The probabilities that the arrow stops in sectors P, Q and R are p, q and r respectively.

The rules of the game are as follows:• If the arrow stops in sector P, then the player having the turn wins.• If the arrow stops in sector Q, the the player having the turn loses and the other player wins.• If the arrow stops in sector R, then the other player takes a turn.Player A takes the first turn.

(i) Show that the probability of player A winning on the first or the second turn of the game 2is (1− r)( p + r).

(ii) Show that the probability that player Aeventually wins the game is p + r1+ r

. 3


SHS ME1 Harder Solutions 2018 2017

6 D

dydx

= x2

, mnormal = − p. Equ'n normal: y − 1

p2= − p x − 2

p⎛⎝⎜

⎞⎠⎟

, p2 y −1= − p3x + 2 p2.

p2 y + p3x = 1+ 2 p2

7 A −1≤ 3

x≤1, x > 0 : − x ≤ 3≤ x, x ≥ 3. x < 0 : − x ≥ 3≥ x, x ≤ −3

8 C drdt

= 5. A = πr2, dAdr

= 2πr. dAdt

= 10πr. r = 15, dAdt

= 150π cm2s−1

9 C x7−r 1

x2⎛⎝⎜

⎞⎠⎟

r

= x7−3r , x2( )7−r 1x3

⎛⎝⎜

⎞⎠⎟

r

= x14−5r , x3( )7−r 1x4

⎛⎝⎜

⎞⎠⎟

r

= x21−7r , x = 3

10 B

Choose 3 squares in 93

⎛⎝⎜

⎞⎠⎟

ways. 3 horizontal rows, 3 vertical columns, 2 diagonals, 8 outcomes.

89C3

= 8× 3× 29×8× 7

= 221

2016

6 D (2sin x −1)(sin x − 3) = 0, sin x = 1

2, x = π

6,5π

6

7 C x = 13

513

sin4t + 1213

cos4t⎛⎝⎜

⎞⎠⎟= 13sin(4t +α ). !x = 4×13cost(4t +α ). Max = 52

8 B

1811⎛⎝⎜

⎞⎠⎟−

1511⎛⎝⎜

⎞⎠⎟= 31824−1365 = 30 459

9 A ′′f (1) = 0

10 A a > 0 so y increases as x increases so C and D are not possible answers.

′p (x) = 3ax2 + bx + c. When ′p (x) = 0 the quadratic equation has two roots.Since a > 0,b > 0,c < 0 then these roots will have different signs so x < 0 for one stationary pointand x > 0 for the other stationary point. Hence answer is A.

2015

6 D −1≤ 2x ≤1, − 1

2≤ x ≤ 1

2

7 B

1

4− x2dx

0

k

∫ = π3

. sin−1 x2

⎡

⎣⎢

⎤

⎦⎥

0

k

= π3

. sin−1 k2− 0 = π

3. k

2= sinπ

3. k = 3.

8 B limx→3

sin(x − 3)(x − 3)(x + 2)

= 15

limx→3

sin(x − 3)(x − 3)

= 15

9 D x = 3sin4t : a = 3, T = 2π

4= π

2. x = asin nt : a = 6, T = π

4 so

2πn

= π4

, n = 8

10 C cos 2t − π

3⎛⎝⎜

⎞⎠⎟= 0, 2t − π

3= 3π

2, 2t = 3π

2+ π

3= 11π

6, t = 11π

12


2017

13(a) x = 2, !x = 4. x = 5, !x = 3.

v2 = n2 a2 − x2( )16 = n2 a2 − 4( ) (i)

9 = n2 a2 − 25( ) (ii)

(i) ÷ (ii)169

= a2 − 4

a2 − 2516a2 − 400 = 9a2 − 36

7a2 = 364

a2 = 52

16 = n2 52− 4( )n2 = 16

48= 1

3

n = 1

3

Period = 2πn

= 2π 3

13(b)(i) (1+ x)n =n0

⎛⎝⎜

⎞⎠⎟+

n1

⎛⎝⎜

⎞⎠⎟

x +n2

⎛⎝⎜

⎞⎠⎟

x2 +!+nn

⎛⎝⎜

⎞⎠⎟

xn

(1− x)n =n0

⎛⎝⎜

⎞⎠⎟−

n1

⎛⎝⎜

⎞⎠⎟

x +n2

⎛⎝⎜

⎞⎠⎟

x2 +!+ (−1)n nn

⎛⎝⎜

⎞⎠⎟

xn

(1+ x)n + (1− x)n =n0

⎛⎝⎜

⎞⎠⎟+

n1

⎛⎝⎜

⎞⎠⎟

x +n2

⎛⎝⎜

⎞⎠⎟

x2 +!+nn

⎛⎝⎜

⎞⎠⎟

xn +n0

⎛⎝⎜

⎞⎠⎟−

n1

⎛⎝⎜

⎞⎠⎟

x +n2

⎛⎝⎜

⎞⎠⎟

x2 +!+ (−1)n nn

⎛⎝⎜

⎞⎠⎟

xn

= 2n0

⎛⎝⎜

⎞⎠⎟+

n2

⎛⎝⎜

⎞⎠⎟

x2 +!+nn

⎛⎝⎜

⎞⎠⎟

xn⎛

⎝⎜⎞

⎠⎟ as negative terms cancel out.

(ii)ddx

(1+ x)n + (1− x)n( ) = n(1+ x)n−1 − n(1− x)n−1

= n (1+ x)n−1 − (1− x)n−1⎡⎣

⎤⎦

ddx

2n0

⎛⎝⎜

⎞⎠⎟+

n2

⎛⎝⎜

⎞⎠⎟

x2 +!+nn

⎛⎝⎜

⎞⎠⎟

xn⎛

⎝⎜⎞

⎠⎟⎛

⎝⎜

⎞

⎠⎟ = 2 2

n2

⎛⎝⎜

⎞⎠⎟

x + 4n4

⎛⎝⎜

⎞⎠⎟

x3 +!+ nnn

⎛⎝⎜

⎞⎠⎟

xn−1⎡

⎣⎢

⎤

⎦⎥

So n (1+ x)n−1 − (1− x)n−1⎡⎣

⎤⎦ = 2 2

n2

⎛⎝⎜

⎞⎠⎟

x + 4n4

⎛⎝⎜

⎞⎠⎟

x3 +!+ nnn

⎛⎝⎜

⎞⎠⎟

xn−1⎡

⎣⎢

⎤

⎦⎥

(iii) 2 2n2

⎛⎝⎜

⎞⎠⎟

x + 4n4

⎛⎝⎜

⎞⎠⎟

x3 +!+ nnn

⎛⎝⎜

⎞⎠⎟

xn−1⎡

⎣⎢

⎤

⎦⎥ = n (1+ x)n−1 − (1− x)n−1⎡

⎣⎤⎦

x = 1: 2 2n2

⎛⎝⎜

⎞⎠⎟+ 4

n4

⎛⎝⎜

⎞⎠⎟+!+ n

nn

⎛⎝⎜

⎞⎠⎟

⎡

⎣⎢

⎤

⎦⎥ = n (1+1)n−1 − (1−1)n−1⎡

⎣⎤⎦

4n2

⎛⎝⎜

⎞⎠⎟+ 2

n4

⎛⎝⎜

⎞⎠⎟+ 3

n6

⎛⎝⎜

⎞⎠⎟+!+ n

2nn

⎛⎝⎜

⎞⎠⎟

⎡

⎣⎢

⎤

⎦⎥ = n2n−1

n2

⎛⎝⎜

⎞⎠⎟+ 2

n4

⎛⎝⎜

⎞⎠⎟+ 3

n6

⎛⎝⎜

⎞⎠⎟+!+ n

2nn

⎛⎝⎜

⎞⎠⎟= n2n−3


13(c) V ms−1, R ≥100 m. x =Vt cosθ , y =Vt sinθ − 12

gt2

(i) For range, y = 0 : Vt sinθ − 12

gt2 = 0

t V sinθ − gt2

⎛⎝⎜

⎞⎠⎟= 0

t = 0 initially or V sinθ = gt2

t = 2V sinθg

t = 2V sinθg

: x =V cosθ × 2V sinθg

x = V 2 sin2θg

is the horizontal range.

(ii) Maximum range when θ = π4

Maximum range = V 2

g×1= V 2

g

Less than 100m when V 2

g<100 or V 2 <100g

(iii) V 2 = 200g

V 2 sin2θg

≥100

200sin2θ ≥100sin2θ ≥ 0.5π6≤ 2θ ≤ π − π

6π6≤ 2θ ≤ 5π

6π12

≤θ ≤ 5π12

(iv) y =Vt sinθ − 12

gt2 ⇒ dydt

=V sinθ − gt

dydt

= 0 : V sinθ − gt = 0

t = V sinθg

t = V sinθg

: y =V sinθ × V sinθg

− 12

g × V sinθg

⎛⎝⎜

⎞⎠⎟

2

y = V 2 sin2θg

− V 2 sin2θg

= V 2 sin2θ2g

V 2 = 200g : y = 100sin2θ

Largest value of θ is 5π12

.

Hence greatest height of ball is 100sin2 5π12

⎛⎝⎜

⎞⎠⎟= 93.3 m

2016

13(a)(i) H = 9 m, L = 1 m. a = 9−12

= 4. Tide measured from x = 5

T = 12.5 hours = 2πn

, n = 4π25

Assume : x = 5+ 4cos 4π25

t +α⎛⎝⎜

⎞⎠⎟

t = 0, x = 9 : 9−5 = 4cosαcosα = 1 ⇒ α = 0

Hence: x = 5+ 4cos 4π25

t⎛⎝⎜

⎞⎠⎟

models the tide.

13(a)(ii) The tide is inrcreasing at its greatest rate half-way between low tide and high tide when !!x = 0.The first Low tide will be at 2 am + 6.25 hours, at 8:15 amThe tide is inrcreasing at its greatest rate at 8:15 am + 3.125 hours, or 11:22.5 am.

OR For stationary points of !x you require !!x = 0.


13(a)(ii) !!x = −n2(x −5)

!!x = − 4π25

⎛⎝⎜

⎞⎠⎟

2× 4cos 4π

25t

⎛⎝⎜

⎞⎠⎟

!!x = 0 : 4π25

t = π2

, 3π2

t = 258

, 758

t = 258

is between High and Low tides when the rate is decreasing (<0)

t = 1258

is between Low and High tides when the rate of increase is greatest.

Time is 2 am + 9 38

hours or 11:22.5 am.

13(b) x = ut cosθ , y = ut sinθ −5t2

(i) Greatest height when dydt

= 0. (ii) t = 0, y = 20, u = 30 ms−1,θ = 30°.

dydt

= usinθ −10t Greatest height = 20+ 302 × sin2 30°20

dydt

= 0 : usinθ −10t = 0 = 20+ 3×15× 12

⎛⎝⎜

⎞⎠⎟

2

t = usinθ10

= 1254

m above the ground.

Greatest height: y = usinθ × usinθ10

−5× usinθ10

⎛⎝⎜

⎞⎠⎟

2

= u2 sin2θ10

− u2 sin2θ20

= u2 sin2θ20

(iii) t = 0, !x = 10, y = 1254

, g = 10,θ = 0, (iv) x = 10t cos0

y = 10t sin0−5t2. x = 10t

y = 5t2 t = 2.5: x = 25

5t2 = 1254

Hits the ground 25 m from the wall.

t = 52= 2.5 s

13(c)(i) ∠ACB = 90° (angle in a semi-circle)∴ ∠MCE = 90° (MCA is a straight line)∠ADB = 90° (angle in a semi-circle)∴ ∠MDE = 90° (MDB is a straight line)∴ CMDE is a cyclic quadrilateral (pair of opposite angles supplementary)


13(c)(ii) Join CD∠CME = ∠CDE (angles in the same segment in circle CMDE)∠CDA = ∠CBA (angles in the same segment in circle CDBA)∴ ∠CME = ∠EBA (each equal to ∠CDE)In Δs CME and FBE∠CME = ∠EBF (shown above)∠CEM = ∠FEB (vertically opposite angles)∴ ΔCME |||ΔFBE∴ ∠BFE = ∠MCE = 90° (corresponding angles in similar triangles)∴ MF ⊥ AB

2015

13(a)(i) v = 0 when v2 = 0. x = 3, 7

(ii) Maximum speed = 11ms−1

(iii) c = 3+ 72

= 5 as it is the centre.

v2 = n2 a2 − (x − c)2( )v2 = n2 a2 − (x −5)2( )a = 7 −5 = 2Maximum speed when x = 5 (passing through the centre of motion)

11= n2 22 − (5−5)2( )11= n222

n2 = 114

a = 2, c = 5, n = 112

13(b) 2x + 13x

⎛⎝⎜

⎞⎠⎟

18

= a0x18 + a1x16 + a2x14 + ...

(i) Term 3=182

⎛⎝⎜

⎞⎠⎟

(2x)16 13x

⎛⎝⎜

⎞⎠⎟

2

= 18×172

× 216 x16 × 1

9x2= 17 × 216 x14. a2 = 17 × 216 18

2⎛⎝⎜

⎞⎠⎟× 216

9

⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪

(ii) Tr+1 =18r

⎛⎝⎜

⎞⎠⎟

(2x)18−r 13x

⎛⎝⎜

⎞⎠⎟

r

=18r

⎛⎝⎜

⎞⎠⎟× 218−r x18−r × 1

3r xr=

18r

⎛⎝⎜

⎞⎠⎟× 218−r

3r× x18−2r

Term is independent of x when r = 9. T10 =189

⎛⎝⎜

⎞⎠⎟× 218−9

39=

189

⎛⎝⎜

⎞⎠⎟× 29

39


13(c) 12!+ 2

3!+ 3

4!+ ...+ n

(n+1)!= 1− 1

(n+1)!, n ≥1.

n = 1: LHS = 12!

= 12

RHS = 1− 12!

= 1− 12= 1

2= LHS

Result is true for n = 1.

Assume result is true for n = k, i.e. assume that 12!+ 2

3!+ 3

4!+ ...+ k

(k +1)!= 1− 1

(k +1)!.

Prove true for n = k +1, i.e. prove that 12!+ 2

3!+ 3

4!+ ...+ k

(k +1)!+ k +1

(k + 2)!= 1− 1

(k + 2)!

LHS = 12!+ 2

3!+ 3

4!+ ...+ k

(k +1)!+ k +1

(k + 2)!

= 1− 1(k +1)!

+ k +1(k + 2)!

= 1− k + 2− (k +1)(k + 2)!

⎛⎝⎜

⎞⎠⎟= 1− 1

(k + 2)!= RHS

Hence result is true for n = k +1 if it is true for n = k.Result is true for n = 1 so it is true for n = 1+1= 2 and hence by the Principle of MathematicalInduction is true for all n ≥1.

13(d) f (x) = cos−1(x)+ cos−1(−x), −1≤ x ≤1. (ii) f (0) = cos−1(0)+ cos−1(0)

(i) ′f (x) = −1

1− x2+ −(−1)

1− x2= 0 = π

2+ π

2

Since ′f (x) = 0 then f (x) is a constant. = π

Hence cos−1(x)+ cos−1(−x) = π

So cos−1(−x) = π − cos−1(x)

2017

14(a) 82n+1 + 62n−1 is divisible by 7.

n = 1: Exp = 83 + 6 = 512+ 6 = 518 = 7 × 74Hence result is true for n = 1.

Assume result is true for n = k, i.e. assume that 82k+1 + 62k−1 = 7 M

Prove result is true for n = k +1, i.e. prove that 82k+3 + 62k+1 is divisible by 7.

Exp = 82k+3 + 62k+1

= 64×82k+1 + 36× 62k−1

= 36 82k+1 + 62k−1( ) + 28×82k+1

= 36× 7 M + 28×82k+1

= 7 36M + 4×82k+1( )Hence result is true for n = k +1 if it is true for n = k. The result is true for n = 1 so by thePrinciple of Mathematical Induction it is true for all n ≥1.


14(b) P 2 p, p2( ). x2 = 4y. a > 0. MQ = MR. tangent at P meets x2 = 4ay at Q, R.

(i) Equation of tangent: y = px − p2

Parabola x2 = −4ay

Intersection: x2 = −4a px − p2( )x2 + 4apx − 4ap2 = 0

Roots of this equation are the x-coordinates of Q and R.(ii) Let the roots of the quadratic be x1 and x2.

Sum of roots = −4ap

Hence x-coordinate of M is x1 + x2

2= −4ap

2= −2ap

Hence Q x1, px1 − p2( ) and R x2, px2 − p2( )So M is

x1 + x22

,px1 − p2 + px2 − p2

2

⎛

⎝⎜⎜

⎞

⎠⎟⎟

⇒ −2ap,p x1 + x2( )− 2 p2

2

⎛

⎝⎜⎜

⎞

⎠⎟⎟

−2ap,−2ap2 − p2( ) ⇒ −2ap,− p2(2a +1)( )Or Substitute x = −2ap into y = px − p2 : y = −2ap2 − p2 = − p2(2a +1)

M −2ap,− p2(2a +1)( )(iii) M lies on x2 = −4y : 4a2 p2 = −4× − p2(2a +1)( )

a2 p2 = p2(2a +1)

a2 = 2a +1

a2 − 2a −1= 0

a = 2 ± 4+ 42

= 1± 2

a > 0 so a = 1+ 2

14(c) F(0) = 0. ′F (t) = 50e−0.5t − 0.4F(t)

(i) ddt

F(t)e0.4t⎡⎣

⎤⎦ = ′F (t)× e0.4t + F(t)× 0.4e0.4t = 50e−0.5t − 0.4F(t)( )× e0.4t + F(t)× 0.4e0.4t

= 50e−0.5t × e0.4t − 0.4F(t)× e0.4t + F(t)× 0.4e0.4t = 50e−0.1t

(ii) ddt

F(t)e0.4t⎡⎣

⎤⎦dt

0

t

∫ = 50e−0.1t dt0

t

∫F(t)e0.4t⎡⎣

⎤⎦0

t= −500e−0.1t⎡⎣

⎤⎦0

t

F(t)e0.4t − F(0)e0 = −500 e−0.1t −1( )F(t)e0.4t = −500 e−0.1t −1( )F(t) = −500 e−0.5t − e−0.4t( ) = 500 e−0.4t − e−0.5t( )


(iii) Maximum when ′F (t) = 0 : 50e−0.5t − 0.4F(t) = 0

50e−0.5t − 0.4 500 e−0.4t − e−0.5t( )( ) = 0

50e−0.5t − 200e−0.4t + 200e−0.5t = 0

250e−0.5t − 200e−0.4t = 0

50e−0.5t 5− 4e0.1t( ) = 0

e0.1t = 1.250.1t = ln1.25t = 10ln1.25 = 2.231 hours ≈ 2 hours 14 minutes

2016

14(a)(i) Exp = (n+1) 4n2 +14n+ 9( ) = 4n3 +14n2 + 9n+ 4n2 +14n+ 9 = 4n3 +18n2 + 23n+ 9

(ii) n = 1:LHS = 1× 3= 3. RHS = 1×13

(4+ 6−1) = 13× 9 = 3= LHS

Hence result is true for n = 1.Assume result is true for n = k, i.e. assume that

1× 3+ 3×5+5× 7 + ...+ (2k −1)(2k +1) = k3

4k2 + 6k −1( )Prove result is true for n = k +1, i.e. prove that

1× 3+ 3×5+5× 7 + ...+ (2k −1)(2k +1)+ (2k +1)(2k + 3) = (k +1)3

4(k +1)2 + 6(k +1)−1( )= (k +1)

34k2 +14k + 9( )

LHS = 1× 3+ 3×5+5× 7 + ...+ (2k −1)(2k +1)+ (2k +1)(2k + 3)

= k3

4k2 + 6k −1( ) + (2k +1)(2k + 3)

= 13

4k3 + 6k2 − k( ) + 4k2 +8k + 3( )= 1

34k3 + 6k2 − k +12k2 + 24k + 9( )

= 13

4k3 +18k2 + 23k + 9( ) = (k +1)3

4k2 +14k + 9( ) from (i) = RHS

Hence result is true for n = k +1 if it is true for n = k. The result is true for n = 1, hence it istrue for n = 1+1= 2 and by the principle of mathematical induction it is true for all n ≥1.

14(b)(i) (1+ x)n =n0

⎛⎝⎜

⎞⎠⎟+

n1⎛⎝⎜

⎞⎠⎟

x +n2

⎛⎝⎜

⎞⎠⎟

x2 +n3

⎛⎝⎜

⎞⎠⎟

x3 + ...+nr

⎛⎝⎜

⎞⎠⎟

xr + ...+nn

⎛⎝⎜

⎞⎠⎟

xn

x = 1: 2n =n0

⎛⎝⎜

⎞⎠⎟+

n1⎛⎝⎜

⎞⎠⎟+

n2

⎛⎝⎜

⎞⎠⎟+

n3

⎛⎝⎜

⎞⎠⎟+ ...+

nr

⎛⎝⎜

⎞⎠⎟+ ...+

nn

⎛⎝⎜

⎞⎠⎟

(ii) n(1+ x)n−1 =n1⎛⎝⎜

⎞⎠⎟+ 2

n2

⎛⎝⎜

⎞⎠⎟

x + 3n3

⎛⎝⎜

⎞⎠⎟

x2 + ...+ rnr

⎛⎝⎜

⎞⎠⎟

xr−1 + ...+ nnn

⎛⎝⎜

⎞⎠⎟

xn−1

x = 1: n2n−1 =n1⎛⎝⎜

⎞⎠⎟+ 2

n2

⎛⎝⎜

⎞⎠⎟+ 3

n3

⎛⎝⎜

⎞⎠⎟+ ...+ r

nr

⎛⎝⎜

⎞⎠⎟+ ...+ n

nn

⎛⎝⎜

⎞⎠⎟


(iii)nr

⎛⎝⎜

⎞⎠⎟

r=1

n

∑ (2r − n) = 2nr

⎛⎝⎜

⎞⎠⎟

rr=1

n

∑ − nnr

⎛⎝⎜

⎞⎠⎟

r=1

n

∑nr

⎛⎝⎜

⎞⎠⎟

rr=1

n

∑ =n1⎛⎝⎜

⎞⎠⎟+ 2

n2

⎛⎝⎜

⎞⎠⎟+ 3

n3

⎛⎝⎜

⎞⎠⎟+ ...+ n

nn

⎛⎝⎜

⎞⎠⎟= n2n−1 from (ii)

nr

⎛⎝⎜

⎞⎠⎟

r=1

n

∑ =n1⎛⎝⎜

⎞⎠⎟+

n2

⎛⎝⎜

⎞⎠⎟+

n3

⎛⎝⎜

⎞⎠⎟+ ...+

nn

⎛⎝⎜

⎞⎠⎟= 2n −1 from (i)

nr

⎛⎝⎜

⎞⎠⎟

r=1

n

∑ (2r − n) = 2n2n−1 − n 2n −1( ) = n2n − n2n + n = n

14(c) T 2at,at2( ), x2 = 4ay, directrix y = −a.

(i) Tangent at T : y = tx − at2

For D, y = −a : −a = tx − at2

tx = at2 − a

x = at − at

⇒ D at − at

,−a⎛⎝⎜

⎞⎠⎟

(ii) Normal at T : x + ty = at3 + 2at

For R, x = at − at

: at − at+ ty = at3 + 2at

ty = at3 + 2at − at + at

y = at2 + a + a

t2⇒ R at − a

t,at2 + a + a

t2⎛⎝⎜

⎞⎠⎟

Locus of R : y = at2 − 2a + a

t2+ 3a

y = a t − 1t

⎛⎝⎜

⎞⎠⎟

2

+ 3a

x = a t − 1t

⎛⎝⎜

⎞⎠⎟

: y = a × xa

⎛⎝⎜

⎞⎠⎟

2

+ 3a

y = x2

a+ 3a ⇒ x2 = a( y − 3a) which is another parabola.

(iii) Focal length of parabola 2 is a4

.

(iv) Gradient of normal to P1 at T : m1 =−1t

Equation of parabola P2 : y = xa

2+ 3a ⇒ dy

dx= 2x

a

At R at − at

,at2 + a + a

t2⎛⎝⎜

⎞⎠⎟

:dydx

= 2a

at − at

⎛⎝⎜

⎞⎠⎟= 2 t − 1

t⎛⎝⎜

⎞⎠⎟=

2 t2 −1( )t

Gradient of normal to P2 at R : m2 = t

2 t2 −1( )Since these two normals are the same: m1 = m2


−1t= −t

2 t2 −1( ) ⇒ 2 t2 −1( ) = t2

2t2 − 2+ t2 = 0 ⇒ t2 = 2 ⇒ t = ± 2OR Gradient of Tangent at R = Gradient of tangent at T

2 t2 −1( )t

= t ⇒ 2 t2 −1( ) = t2 ⇒ t = ± 2

2015

14(a) x =V t cosθ , y =V t sinθ − 12

gt2.

(i) y = 0 : 0 =V t sinθ − 12

gt2

t V sinθ − 12

gt⎛⎝⎜

⎞⎠⎟= 0

t = 0 or V sinθ − 12

gt

t = 0 or t = 2V sinθg

t = 2V sinθg

: x =V cosθ × 2V sinθg

⇒ x = V 2 sin2θg

is the range.

14(a)(ii) θ = π3

: x =V t cosπ3

, y =V t sinπ3− 1

2gt2

x = V t2

, y = 3V t2

− 12

gt2

dxdt

= V2

,dydt

= 3V2

− gt

dydx

=

3V2

− gt

V2

= 3V − 2gtV

t = 2V

3g:

dydx

=3V − 2g × 2V

3gV

=3V − 4V

3V

= 3V − 4V

3V= − 1

3

tanθ = − 1

3= 1

3⇒ θ = π

6

(iii)dydx

< 0 at t = 2V

3g so the particle is moving downwards.


14(b) t = 0, x = 0, v = 1. Moving horizontally. !!x = x −1.

(i) !!x = x −1 (ii)dxdt

= 1− x

ddx

12

v2⎛⎝⎜

⎞⎠⎟= x −1

dx1− x

= dt

12

v2 = x2

2− x +C t = − ln(1− x)+C2

x = 0,v = 1: C = 12

t = 0,x = 0 : C2 = 0

12

v2 = x2

2− x + 1

2t = − ln(1− x)

v2 = x2 − 2x +1 1− x = e−t

v2 = (x −1)2 x = 1− e−t

v > 0 so !x = 1− x as !x = x −1 gives v = −1 when x = 0.

(iii) As t →∞ then e−t → 0 and x →1.Limiting position is x = 1.

14(c)(iii) wins n+1( ) games.

Wins in n+1( ) games:nn

⎛⎝⎜

⎞⎠⎟

12

⎛⎝⎜

⎞⎠⎟

n+1

Wins in n+ 2( ) games:n+1

n⎛⎝⎜

⎞⎠⎟

12

⎛⎝⎜

⎞⎠⎟

n+2

Wins in n+ 3( ) games:n+ 2

n⎛⎝⎜

⎞⎠⎟

12

⎛⎝⎜

⎞⎠⎟

n+3 etc.

P( A wins (n+1) games) =nn

⎛⎝⎜

⎞⎠⎟

12

⎛⎝⎜

⎞⎠⎟

n+1+

n+1n

⎛⎝⎜

⎞⎠⎟

12

⎛⎝⎜

⎞⎠⎟

n+2+

n+ 2n

⎛⎝⎜

⎞⎠⎟

12

⎛⎝⎜

⎞⎠⎟

n+3+ ...+

2nn

⎛⎝⎜

⎞⎠⎟

12

⎛⎝⎜

⎞⎠⎟

2n+1

P(B wins (n+1) games) =nn

⎛⎝⎜

⎞⎠⎟

12

⎛⎝⎜

⎞⎠⎟

n+1+

n+1n

⎛⎝⎜

⎞⎠⎟

12

⎛⎝⎜

⎞⎠⎟

n+2+

n+ 2n

⎛⎝⎜

⎞⎠⎟

12

⎛⎝⎜

⎞⎠⎟

n+3+ ...+

2nn

⎛⎝⎜

⎞⎠⎟

12

⎛⎝⎜

⎞⎠⎟

2n+1

But P( A wins (n+1) games)+ P(B wins (n+1) games) = 1

2nn

⎛⎝⎜

⎞⎠⎟

12

⎛⎝⎜

⎞⎠⎟

n+1+

n+1n

⎛⎝⎜

⎞⎠⎟

12

⎛⎝⎜

⎞⎠⎟

n+2+

n+1n

⎛⎝⎜

⎞⎠⎟

12

⎛⎝⎜

⎞⎠⎟

n+3+ ...+

2nn

⎛⎝⎜

⎞⎠⎟

12

⎛⎝⎜

⎞⎠⎟

2n+1⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪= 1

222n+1

nn

⎛⎝⎜

⎞⎠⎟

2n +n+1

n⎛⎝⎜

⎞⎠⎟

2n−1 +n+ 2

n⎛⎝⎜

⎞⎠⎟

2n−2 + ...+2nn

⎛⎝⎜

⎞⎠⎟

⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪= 1

nn

⎛⎝⎜

⎞⎠⎟

2n +n+1

n⎛⎝⎜

⎞⎠⎟

2n−1 +n+ 2

n⎛⎝⎜

⎞⎠⎟

2n−2 + ...+2nn

⎛⎝⎜

⎞⎠⎟= 22n

nn

⎛⎝⎜

⎞⎠⎟

2n +n+1

n⎛⎝⎜

⎞⎠⎟

2n−1 +n+ 2

n⎛⎝⎜

⎞⎠⎟

2n−2 + ...+2nn

⎛⎝⎜

⎞⎠⎟= 22n


2014

14(b)(i) A wins on first turn: Prob(AW1) = pB loses on first turn and A wins: Prob(BL1) = rqBut q = 1− ( p + r)Prob( A wining on first or second turn) = p + r(1− p − r)

= p + r − rp − r2

= p(1− r)+ r(1− r)= (1− r)( p + r)

(ii) q = 1− p − rA wins on first turn: Prob(AW1) = pB loses on first turn and A wins: Prob(BL1) = rq

A wins on second turn: Prob(AW 2) = r2 p

B loses on second turn and A wins: Prob(BL2) = r3q

A wins on third turn: Prob(AW 3) = r4 p

B loses on third turn and A wins: Prob(BL3) = r5q

P( A wins eventually) = p + r2 p + r4 p + ....+ rq + r3q + r5q + ...

= p1− r2

+ rq1− r2

= p + r(1− p − r)(1− r)(1+ r)

= ( p + r)(1− r)(1− r)(1+ r)

= p + r1+ r

SHS ME1 Harder Notes 18web3.singleton-h.schools.nsw.edu.au/HSC Seminar Day 2016/Presen… · © R J...

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