Band 3 – Page 53

Mechanics, EXT2* M1 2015 HSC 14a

A projectile is fired from the origin `O` with initial velocity `V` m s`\ ^(−1)` at an angle `theta` to the horizontal. The equations of motion are given by

`x = Vt\ cos\ theta, \ y = Vt\ sin\ theta − 1/2 g t^2`. (Do NOT prove this)

Show that the horizontal range of the projectile is

`(V^2\ sin\ 2theta)/g`. (2 marks)

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A particular projectile is fired so that `theta = pi/3`.

Find the angle that this projectile makes with the horizontal when

`t = (2V)/(sqrt3\ g)`. (2 marks)

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State whether this projectile is travelling upwards or downwards when

`t = (2V)/(sqrt3\ g)`. Justify your answer. (1 mark)

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Show Answers Only

`text(See Worked Solutions.)`
`30^@`
`text(Downwards)`

Show Worked Solution

i. `text(Show range is)\ \ (V^2\ sin\ 2theta)/g`

`x`	`= Vt\ cos\ theta`
`y`	`= Vt\ sin\ theta − 1/2 g t^2`

`text(Horizontal range occurs when)\ \ y = 0`

`Vt\ sin\ theta − 1/2 g t^2`	`= 0`
`t(V\ sin\ theta − 1/2 g t)`	`= 0`
`:.V\ sin\ theta − 1/2 g t`	`= 0`
`1/2 g t`	`= V\ sin\ theta`
`t`	`= (2V\ sin\ theta)/g`

`text(Find)\ \ x\ \ text(when)\ \ t = (2V\ sin\ theta)/g`

`x`	`= V · (2V\ sin\ theta)/g · cos\ theta`
	`= (V^2\ sin\ 2theta)/g\ \ …\ text(as required)`

♦♦ Mean mark part (ii) 28%.

ii.	`x`	`= Vt\ cos\ theta`
	`dot x`	`= V\ cos\ theta`
	`y`	`= Vt\ sin\ theta − 1/2 g t^2`
	`dot y`	`= V\ sin\ theta − g t`

`text(When)\ \ t = (2V)/(sqrt3\ g)\ \ text(and)\ \ theta = pi/3`

`dot x`	`= V\ cos\ pi/3 = V/2`
`dot y`	`= V\ sin\ pi/3 − g\ (2V)/(sqrt3\ g)`
	`= (sqrt3V)/2 − (2V)/sqrt3`
	`= (sqrt3(sqrt3V) − 2 xx 2V)/(2sqrt3)`
	`= -V/(2sqrt3)`

`text(Let)\ alpha =\ text(Angle of projectile with the horizontal)`

`tan\ α`	`=(\|\ doty\ \|) / dotx`
	`= (V/(2sqrt3))/(V/2)`
	`= V/(2sqrt3) xx 2/V`
	`= 1/sqrt3`
`:.α`	`= 30^@`

`:.\ text(When)\ \ t = (2V)/(sqrt3\ g),\ text(the projectile makes)`

`text(a)\ \ 30^@\ \ text(angle with the horizontal.)`

iii. `text(When)\ t = (2V)/(sqrt3\ g)`

♦♦ Mean mark part (iii) 31%.

`dot y = −V/(2sqrt3)`

`text(The negative value of)\ \ dot y\ \ text(indicates that)`

`text(the particle is travelling downwards.)`

Trigonometry, 2ADV T2 2007 HSC 4a

Solve `sqrt 2\ sin\ x = 1` for `0 <= x <= 2 pi`. (2 marks)

Show Answers Only

`pi/4 , (3 pi)/4`

Show Worked Solution

`sqrt 2\ sin\ x`	`= 1`	`\ \ \ \ \ \ \ 0 <= x <= 2 pi`
`sin\ x`	`= 1/sqrt 2`

`=>sin\ pi/4 = 1/sqrt 2\ \ text(and sin is positive in)`

`text(1st/2nd quadrants,)`

`:. x`	`= pi/4\ ,\ pi – pi/4`
	`= pi/4\ ,\ (3 pi)/4`

Financial Maths, 2ADV M1 2007 HSC 3b

Heather decides to swim every day to improve her fitness level.

On the first day she swims 750 metres, and on each day after that she swims `100` metres more than the previous day. That is, she swims 850 metres on the second day, 950 metres on the third day and so on.

Write down a formula for the distance she swims on the `n`th day. (1 mark)

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How far does she swim on the 10th day? (1 mark)

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What is the total distance she swims in the first 10 days? (1 mark)

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After how many days does the total distance she has swum equal the width of the English Channel, a distance of 34 kilometres? (2 marks)

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`T_n = 650 + 100n`
`1650\ text(metres)`
`12\ text(km)`
`34\ text(km)`

Show Worked Solution

i. `T_1`	`= 750`
`T_2`	`= 850`

`=> text(AP)\ text(where)\ a = 750\ ,\ d = 100`

`:. T_n`	`= a + (n-1) d`
	`= 750 + (n – 1) 100`
	`= 750 + 100n – 100`
	`= 650 + 100n`

ii. `T_10`	`= 650 + 100 xx 10`
	`= 1650\ text(metres)`

`:.\ text(She swims 1650 metres on the 10th day.)`

iii. `S_n = n/2 [2a + (n – 1) d]`

`:. S_10`	`= 10/2 [2 xx 750 + (10 -1) 100]`
	`= 5 [1500 + 900]`
	`= 12\ 000`

`:.\ text(She swims 12 km in the first 10 days.)`

iv. `text(Find)\ n\ text(such that)\ S_n = 34\ text(km)`

`n/2 [2 xx 750 + (n – 1) 100]`	`= 34\ 000`
`n/2 [1500 + 100n – 100]`	`= 34\ 000`
`n/2 [1400 + 100n]`	`= 34\ 000`
`700n + 50n^2`	`= 34\ 000`
`50 n^2 + 700n – 34\ 000`	`= 0`
`50 (n^2 + 14 n – 680)`	`= 0`

`text(Using the quadratic formula)`

`n`	`= {-b +- sqrt(b^2 – 4ac)}/(2a)`
	`= {-14 +- sqrt(14^2 – 4 xx 1 xx (-680))}/(2 xx 1)`
	`= (-14 +- sqrt 2916)/2`
	`= (-14 +- 54)/2`
	`= 20 or -34`
	`= 20\ ,\ n > 0`

`:.\ text(Her total distance equals 34 km after 20 days.)`

Linear Functions, 2UA 2007 HSC 3a

In the diagram, `A`, `B` and `C` are the points `(10, 5)`, `(12, 16)` and `(2, 11)` respectively.

Copy or trace this diagram into your writing booklet.

Find the distance `AC`. (1 mark)
Find the midpoint of `AC`. (1 mark)
Show that `OB_|_AC`. (2 marks)
Find the midpoint of `OB` and hence explain why `OABC` is a rhombus. (2 marks)
Hence, or otherwise, find the area of `OABC`. (1 mark)

Show Answers Only

`10\ text(units)`
`(6, 8)`
`text(Proof)\ \ text{(See Worked Solutions)}`
`(6, 8)\ \ text(See Worked Solutions)`
`100\ text(u²)`

Show Worked Solution

(i) `A (10, 5)\ \ \ C (2, 11)`

`text(dist)\ AC`	`= sqrt {(x_2 – x_1)^2 + (y_2 – y_1)^2}`
	`= sqrt {(2 – 10)^2 + (11 – 5)^2}`
	`= sqrt (64 + 36)`
	`= sqrt 100`
	`= 10\ text(units)`

(ii) `text(Midpoint)\ AC`	`= ((x_1 + x _2)/2 , (y_1 + y_2)/2)`
	`= ((10+2)/2 , (5 + 11)/2)`
	`= (6, 8)`

(iii) `B (12, 16)`

`M_(OB)`	`= (y_2 – y_1)/(x_2 – x_1)`
	`= (16 – 0)/(12 – 0)`
	`= 4/3`

`M_(AC)`	`= (11 – 5)/(2 – 10)`
	`= 6/-8`
	`= -3/4`

`M_(OB) xx M_(AC) = 4/3 xx -3/4 = -1`

`:. OB_|_ AC`

(iv) `text(Midpoint)\ OB`	`= ((12 + 0)/2 , (16 + 0)/2)`
	`= (6, 8)`

`text(S)text(ince midpoint)\ OB = text(midpoint)\ AC`

`text(and)\ OB_|_AC`

`=> text(Diagonals of)\ OABC\ text(are perpendicular)`

`text(bisectors)`

`:. OABC\ text(is a rhombus)`

(v) `text(dist)\ OB`

`= sqrt ((12 – 0)^2 + (16 – 0)^2)`

`= sqrt (144 + 256)`

`= sqrt 400`

`= 20\ text(units)`

`text(Area of)\ OABC`

`= 1/2 xx AC xx OB`

`= 1/2 xx 10 xx 20`

`= 100\ text(u²)`

Calculus, 2ADV C3 2007 HSC 2c

The point `P (pi, 0)` lies on the curve `y = x sinx`. Find the equation of the tangent to the curve at `P`. (3 marks)

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`y = – pi x + pi^2`

Show Worked Solution

`y = x sin x`

`(dy)/(dx)`	`= x xx d/(dx) (sin x) + d/(dx) x xx sin x`
	`= x cos x + sin x`

`text(When)\ \ x = pi`

`(dy)/(dx)`	`= pi xx cos pi + sin pi`
	`= pi (-1) + 0`
	`= – pi`

`text(Equation of line,)\ \ m = – pi,\ text(through)\ P(pi, 0):`

`y – y_1`	`= m(x – x_1)`
`y – 0`	`= – pi(x – pi)`
`:. y`	`= – pi x + pi^2`

Calculus, 2ADV C4 2007 HSC 2bii

Evaluate `int_1^4 8/x^2\ dx`. (3 marks)

Show Answers Only

`6`

Show Worked Solution

`int_1^4 8/x^2\ dx`

`= 8 int_1^4 x^-2\ dx`

`= 8[-1/x]_1^4`

`= 8[(-1/4) – (-1/1)]`

`= 8(3/4)`

`= 6`

Calculus, 2ADV C4 2007 HSC 2bi

Find `int (1 + cos 3x)\ dx`. (2 marks)

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`x + 1/3 sin 3x + C`

Show Worked Solution

`int (1 + cos 3x)\ dx`

`= x + 1/3 sin 3x + C`

Calculus, 2ADV C2 2007 HSC 2aii

Differentiate with respect to `x`:

`(1 + tan x)^10`. (2 marks)

Show Answers Only

`10 sec^2 x \ (1 + tan x)^9`

Show Worked Solution

`y = (1 + tan x)^10`

`(dy)/(dx)`	`= 10 (1 + tan x)^9 xx d/(dx) (tan x)`
	`= 10 sec^2 x \ (1 + tan x)^9`

Functions, 2ADV F1 2007 HSC 1f

Find the equation of the line that passes through the point `(1, 3)` and is perpendicular to `2x + y + 4 = 0`. (2 marks)

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`x-2y + 7 = 0`

Show Worked Solution

`2x + y + 4`	`= 0`
`y`	`= -2x-4`

`=>\ text(Gradient) = -2`

`:. text(⊥ gradient) = 1/2\ \ \ (m_1 m_2=-1)`

`text(Equation of line)\ \ m = 1/2, \ text(through)\ (1, 3):`

`y-y_1`	`= m (x-x_1)`
`y-3`	`= 1/2 (x-1)`
`y`	`= 1/2 x + 5/2`
`2y`	`= x + 5`
`:. x-2y + 5`	`= 0`

Financial Maths, 2ADV M1 2007 HSC 1d

Find the limiting sum of the geometric series

`3/4 + 3/16 + 3/64 + …`. (2 marks)

Show Answers Only

`1`

Show Worked Solution

`3/4 + 3/16 + 3/64 + …`

`=> text(GP where)\ \ a = 3/4,\ \ \ r = T_2/T_1 = 1/4`

`:. S_oo`	`= a/(1 – r)`
	`= (3/4)/(1 – 1/4)`
	`= 1`

Proof, EXT1 P1 2015 HSC 13c

Prove by mathematical induction that for all integers `n ≥ 1`,

`1/(2!) + 2/(3!) + 3/(4!) + … + n/((n + 1)!) = 1 − 1/((n + 1)!)`. (3 marks)

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`text{Proof (See Worked Solutions)}`

Show Worked Solution

`text(Prove for)\ n ≥ 1`

`1/(2!) + 2/(3!) + 3/(4!) + … + n/((n + 1)!) = 1 − 1/((n + 1)!)`

`text(If)\ n = 1`

`text(LHS)`	`= 1/(2!) = 1/2`
`text(RHS)`	`= 1 − 1/(2!) = 1 − 1/2 = 1/2`

`:.\ text(True for)\ n = 1`

`text(Assume true for)\ n = k`

`text(i.e.)\ \ 1/(2!) + 2/(3!) + … + k/((k + 1)!) = 1 − 1/((k + 1)!)`

`text(Prove true for)\ n = k + 1`

`text(i.e.)\ \ 1/(2!) + 2/(3!) + … + k/((k + 1)!) + (k + 1)/((k + 2)!) = 1 − 1/((k + 2)!)`

`text(LHS)`	`= 1 − 1/((k + 1)!) + (k + 1)/((k + 2)!)`
	`= 1 − (((k + 2) − (k + 1))/((k + 1)!(k + 2)))`
	`= 1 − 1/((k + 2)!)\ \ …\ text(as required)`

`=>\ text(True for)\ n = k + 1`

`:.\ text(S)text(ince true for)\ n = 1,\ text(by PMI, true for integral)\ n ≥ 1.`

Combinatorics, EXT1 A1 2015 HSC 13b

Consider the binomial expansion

`(2x + 1/(3x))^18 = a_0x^(18) + a_1x^(16) + a_2x^(14) + …`

where `a_0, a_1, a_2`, . . . are constants.

Find an expression for `a_2`. (2 marks)

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

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`(\ ^18C_2 · 2^(16))/(3^2)`
`(\ ^(18)C_9 · 2^9)/(3^9)`

Show Worked Solution

i. `text(Need co-efficient of)\ x^(14)`

`text(General term of)\ (2x + 1/(3x))^(18)`

`T_k`	`= \ ^(18)C_k(2x)^(18-k) · (1/(3x))^k`
	`= \ ^(18)C_k · 2^(18-k) · x^(18-k) · 3^(-k) · x^(-k)`
	`= \ ^(18)C_k · 2^(18-k) · 3^(-k) · x^(18-2k)`

`a_2\ text(occurs when:)`

`18-2k= 14\ \ =>\ \ k=2`

`:.a_2= \ ^(18)C_2 · 2^(18-2) · 3^(-2)= (\ ^(18)C_2 · 2^(16))/(3^2)`

ii. `text(Independent term occurs when:)`

`18-2k= 0\ \ =>\ \ k=9`

`:.\ text(Independent term)= \ ^(18)C_9 · 2^(18-9) · 3^(-9) = (\ ^(18)C_9 · 2^9)/(3^9)`

Mechanics, EXT2* M1 2015 HSC 13a

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 `v^2` as a function of `x`.

For what value(s) of `x` is the particle at rest? (1 mark)

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What is the maximum speed of the particle? (1 mark)

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The velocity `v` of the particle is given by the equation

`v^2 = n^2(a^2 − (x −c)^2)` where `a`, `c` and `n` are positive constants.

What are the values of `a`, `c` and `n`? (3 marks)

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Show Answers Only

`3\ text(or)\ 7`
`V = sqrt11\ text(m/s)`
`a = 2, c = 5,`
`n = (sqrt11)/2`

Show Worked Solution

i. `text(Particle is at rest when)\ v^2 = 0`

`:. x = 3\ \ text(or)\ \ 7`

ii. `text(Maximum speed occurs when)`

`v^2`	`= 11`
`v`	`= sqrt11\ text(m/s)`

iii. `v^2 = n^2(a^2 − (x − c)^2)`

♦ Mean mark 41%.

`text(Amplitude) = 2`

`:.a = 2`

`text(Centre of motion when)\ x = 5`

`:.c = 5`

`text(S)text(ince)\ \ v^2 = 11\ \ text(when)\ \ x = 5`

`11`	`= n^2(2^2 − (5 − 5)^2)`
	`= 4n^2`
`n^2`	`= 11/4`
`:.n`	`= sqrt11/2`

Trig Ratios, EXT1 2015 HSC 12d

A kitchen bench is in the shape of a segment of a circle. The segment is bounded by an arc of length 200 cm and a chord of length 160 cm. The radius of the circle is `r` cm and the chord subtends an angle `theta` at the centre `O` of the circle.

Show that `160^2 = 2r^2 (1 - cos\ theta)`. (1 mark)
Hence, or otherwise, show that `8 theta^2 + 25 cos\ theta - 25 = 0`. (2 marks)
Taking `theta_1 = pi` as a first approximation to the value of `theta`, use one application of Newton’s method to find a second approximation to the value of `theta`. Give your answer correct to two decimal places. (2 marks)

Show Answers Only

`text(See Worked Solutions)`
`text(See Worked Solutions)`
`2.57\ \ text{(to 2 d.p.)}`

Show Worked Solution

(i) `text(Using the cosine rule)`

`a^2`	`= b^2 + c^2 − 2bc\ cos\ A`
`160^2`	`= r^2 + r^2 −2 xx r xx r xx cos\ theta`
	`= 2r^2 − 2r^2\ cos\ theta`
	`= 2r^2(1 − cos\ theta)\ \ …\ text(as required)`

(ii) `text(Show)\ \ 8theta^2 + 25\ cos\ theta − 25 = 0`

`text(Arc length)`	`= 200`
`theta/(2pi) xx 2pir`	`= 200`
`rtheta`	`= 200`
`r`	`= 200/theta`

`text(Substitute)\ \ r = 200/theta\ text{into part (i) equation}`

`2 xx (200/theta)^2(1 – cos\ theta)`	`= 160^2`
`80\ 000(1 – cos\ theta)`	`= 25\ 600\ theta^2`
`25(1 – cos\ theta)`	`= 8theta^2`
`8theta^2 + 25\ cos\ theta – 25`	`= 0\ \ …\ text(as required.)`

(iii)	`f(theta)`	`= 8theta^2 + 25\ cos\ theta – 25`
	`f′(theta)`	`= 16theta – 25\ sin\ theta`
	`f(pi)`	`= 8pi^2 + 25\ cos\ pi – 25`
		`= 28.9568…`
	`f′(pi)`	`= 16pi – 25\ sin\ pi`
		`= 50.2654…`

`theta_2`	`= pi – (f(pi))/(f′(pi))`
	`= pi – (28.9568…)/(50.2654…)`
	`= 2.5655…`
	`= 2.57\ \ text{(to 2 d.p.)}`

Plane Geometry, EXT1 2015 HSC 12a

In the diagram, the points `A`, `B`, `C` and `D` are on the circumference of a circle, whose centre `O` lies on `BD`. The chord `AC` intersects the diameter `BD` at `Y`. The tangent at `D` passes through the point `X`.

It is given that `∠CYB = 100^@` and `∠DCY = 30^@`.

Copy or trace the diagram into your writing booklet.

What is the size of `∠ACB`? (1 mark)
What is the size of `∠ADX`? (1 mark)
Find, giving reasons, the size of `∠CAB`. (2 marks)

Show Answers Only

`60°`
`30°`
`70°`

Show Worked Solution

(i)

`∠DCB`	`= 90^@\ \ text{(angle in semi-circle)}`
`:.∠ACB`	`= 90^@ − 30^@`
	`= 60^@`

(ii)	`∠ADX`	`= ∠ACD\ \ text{(angle in alternate segment)}`
		`= 30^@`

(iii)	`∠CBY`	`= 180 − (100 + 60)\ \ text{(angle sum of}\ Delta CBY text{)}`
		`= 20^@`
	`∠CAD`	`= 20^@\ \ text{(angles in the same segment on arc}\ CD text{)}`
	`∠DAB`	`= 90^@\ \ text{(angle in semi-circle)}`
	`:.∠CAB`	`= 90^@ − 20^@`
		`= 70^@`

Functions, EXT1 F2 2015 HSC 11f

Consider the polynomials `P(x) = x^3-kx^2 + 5x + 12` and `A(x) = x - 3`.

Given that `P(x)` is divisible by `A(x)`, show that `k = 6`. (1 mark)

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Find all the zeros of `P(x)` when `k = 6`. (2 marks)

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Show Answers Only

`text(See Worked Solutions)`
`3, 4, −1`

Show Worked Solution

i.	`P(x)`	`= x^3-kx^2 + 5x + 12`
	`A(x)`	`= x-3`

`text(If)\ P(x)\ text(is divisible by)\ A(x)\ \ =>\ \ P(3) = 0`

`3^3-k(3^2) + 5 xx 3 + 12`	`= 0`
`27-9k + 15 + 12`	`= 0`
`9k`	`= 54`
`:.k`	`= 6\ \ …\ text(as required)`

ii. `text(Find all roots of)\ P(x)`

`P(x)=(x-3)*Q(x)`

`text{Using long division to find}\ Q(x):`

`:.P(x)`	`= x^3-6x^2 + 5x + 12`
	`= (x-3)(x^2-3x-4)`
	`= (x-3)(x-4)(x + 1)`

`:.\ text(Zeros at)\ \ \ x = -1, 3, 4`

Calculus, EXT1 C2 2015 HSC 11e

Use the substitution `u = 2x - 1` to evaluate `int_1^2 x/((2x - 1)^2)\ dx`. (3 marks)

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`1/4(ln 3 + 2/3)`

Show Worked Solution

`u = 2x − 1`

`⇒ 2x`	`= u + 1`
`x`	`= 1/2(u + 1)`

`(du)/(dx)`	`= 2`
`dx`	`= (du)/2`

`text(When)`	`\ \ x = 2,\ `	`u = 3`
	`\ \ x = 1,\ `	`u = 1`

`:. int_1^2 x/((2x − 1)^2) \ dx`

`= int_1^3 1/2(u + 1) · 1/(u^2) · (du)/2`

`= 1/4int_1^3 ((u + 1)/(u^2)) du`

`= 1/4 int_1^3 1/u + u^(−2) du`

`= 1/4 [ln u − u^(−1)]_1^3`

`= 1/4 [(ln 3 − 1/3) − (ln 1 − 1)]`

`= 1/4 (ln 3 − 1/3 + 1)`

`= 1/4(ln 3 + 2/3)`

Trigonometry, EXT1 T3 2015 HSC 11d

Express `5 cos x-12 sin x` in the form `A cos (x + \alpha)`, where `0 ≤ \alpha ≤ pi/2`. (2 marks)

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`13\ cos\ (x + 1.176…)`

Show Worked Solution

`5\ cos\ x-12\ sin\ x`	`= A\ cos\ (x + \alpha)`
	`= A\ cos\ x\ cos\ \alpha-A\ sin\ x\ sin\ \alpha`

`A\ cos\ \alpha = 5,\ \ A\ sin\ \alpha = 12`

`A^2= 5^2 + 12^2 = 169\ \ =>\ \ A=13`

`13\ cos\ \alpha`	`= 5`
`cos\ \alpha`	`= 5/13`
`\alpha`	`= cos^(-1)\ (5/13) ≈ 1.176…\ text(radians)`

`:. 5cos x-12sin x = 13 cos (x + 1.176…)`

Functions, EXT1 F1 2015 HSC 11c

Solve the inequality `4/(x + 3) ≥ 1`. (3 marks)

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`−3 < x ≤ 1, \ \ x ≠ −3`

Show Worked Solution

`text(Solution 1)`

`4/(x + 3) ≥ 1`

`4(x + 3)`	`≥ (x + 3)^2`
`4x + 12`	`≥ x^2 + 6x + 9`
`x^2 + 2x-3`	`≤ 0`
`(x + 3)(x-1)`	`≤ 0`

`:.−3 < x ≤ 1, \ \ x ≠ −3`

`text(Solution 2)`

`text(If)\ (x + 3)`	`> 0:`
`x`	`> -3`

`4/(x + 3)`	`≥ 1`
`4`	`≥ x + 3`
`x`	`≤ 1`

`-3 < x ≤ 1`

`text(If)\ (x + 3)`	`< 0:`
`x`	`< -3`

`4/(x + 3)`	`≥ 1`
`4`	`≤ x + 3`
`x`	`≥ 1`

`text(No solution.)`

`:. −3 < x ≤ 1`

Linear Functions, EXT1 2015 HSC 11b

Calculate the size of the acute angle between the lines `y = 2x + 5` and `y = 4 − 3x`. (2 marks)

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`45^@`

Show Worked Solution

`y = 2x + 5,`	`m_1 = 2`
`y = 4 − 3x,`	`m_2 = −3`

`tan\ theta`	`= \|(m_1 − m_2)/(1 + m_1m_2)\|`
	`= \|(2 − (−3))/(1 + 2(−3))\|`
	`= \|5/(−5)\|`
	`= 1`

`:.theta = 45^@`

Trig Calculus, EXT1 2015 HSC 11a

Find `int sin^2\ x\ dx`. (2 marks)

Show Answers Only

`x/2 − 1/4\ sin\ 2x + c`

Show Worked Solution

`int sin^2\ x\ dx`

`= 1/2 int 1 − cos\ 2x\ dx`

`= 1/2 (x − 1/2\ sin\ 2x) + c`

`= x/2 − 1/4\ sin\ 2x + c`

Calculus, EXT1 C2 2015 HSC 7 MC

What is the value of `k` such that `int_0^k 1/sqrt(4 − x^2) \ dx= pi/3 ?`

`1`
`sqrt3`
`2`
`2sqrt3`

Show Answers Only

`B`

Show Worked Solution

`int_0^1 1/sqrt(4 − x^2)dx`	`= pi/3`
`[sin^(−1)\ x/2]_0^k`	`= pi/3`
`sin^(−1)\ k/2 − sin^(−1)\ 0`	`= pi/3`
`sin^(−1)\ k/2`	`= pi/3`
`k/2`	`= sin\ pi/3`
	`= sqrt3/2`
`:.k`	`= sqrt3`

`⇒ B`

Trigonometry, EXT1 T1 2015 HSC 6 MC

What is the domain of the function `f(x) = sin^(-1)\ (2x)`?

`-pi ≤ x ≤ pi`
`-2 ≤ x ≤ 2`
`-pi/4 ≤ x ≤ pi/4`
`-1/2 ≤ x ≤ 1/2`

Show Answers Only

`D`

Show Worked Solution

`f(x)= sin^(-1)\ (2x)`

`text(Domain of)\ f(x) = sin^(-1) x\ \ text(is)`

`-1 ≤ x ≤ 1`

`:.\ text(Domain of)\ \ f(x) = sin^(-1)\ (2x)\ \ text(is)`

`-1 ≤ 2x ≤ 1`

`-1/2 ≤ x ≤ 1/2`

`=> D`

Combinatorics, EXT1 A1 2015 HSC 4 MC

A rowing team consists of 8 rowers and a coxswain.

The rowers are selected from 12 students in Year 10.

The coxswain is selected from 4 students in Year 9.

In how many ways could the team be selected?

`\ ^(12)C_8 +\ ^4C_1`
`\ ^(12)P_8 +\ ^4P_1`
`\ ^(12)C_8 ×\ ^4C_1`
`\ ^(12)P_8 ×\ ^4P_1`

Show Answers Only

`C`

Show Worked Solution

`\ ^(12)C_8`	`=\ text(Combinations of rowers)`
`\ ^4C_1`	`=\ text(Combinations of coxswains)`

`:.\ text(Number of ways to select team)`

`=\ ^12C_8 xx\ ^4C_1`

`=> C`

Functions, EXT1 F2 2015 HSC 1 MC

What is the remainder when `x^3-6x` is divided by `x + 3`?

`-9`
`9`
`x^2-2x`
`x^2-3x + 3`

Show Answers Only

`A`

Show Worked Solution

`text(Remainder)`	`= P(-3)`
	`= (-3)^3-6(-3)`
	`= -27 + 18`
	`= -9`

`=> A`

Measurement, STD2 M7 2015 HSC 27b

A patient requires 2400 mL of fluid to be delivered at a constant rate by means of a drip over 12 hours. Each mL of fluid is equivalent to 15 drops.

How many drops per minute need to be delivered? (2 marks)

--- 4 WORK AREA LINES (style=lined) ---

Show Answers Only

`50\ text(per minute)`

Show Worked Solution

`text(Fluid rate of delivery)`

`= 2400/12`

`= 200\ text(mL per hour)`

`= 200/60`

`= 3 1/3\ text(mL per minute)`

`text(S)text(ince each mL has 15 drops)`

`#\ text(Drops)`	`= 15 xx 3 1/3`
	`= 50\ text(per minute)`

Probability, STD2 S2 2015 HSC 26e

The table shows the relative frequency of selecting each of the different coloured jelly beans from packets containing green, yellow, black, red and white jelly beans.

\begin{array} {|c|c|}
\hline
\rule{0pt}{2.5ex} \textit{Colour} \rule[-1ex]{0pt}{0pt} & \textit{Relative frequency} \\
\hline
\rule{0pt}{2.5ex} \text{Green} \rule[-1ex]{0pt}{0pt} & 0.32 \\
\hline
\rule{0pt}{2.5ex} \text{Yellow} \rule[-1ex]{0pt}{0pt} & 0.13 \\
\hline
\rule{0pt}{2.5ex} \text{Black} \rule[-1ex]{0pt}{0pt} & 0.14 \\
\hline
\rule{0pt}{2.5ex} \text{Red} \rule[-1ex]{0pt}{0pt} & \\
\hline
\rule{0pt}{2.5ex} \text{White} \rule[-1ex]{0pt}{0pt} & 0.24 \\
\hline
\end{array}

What is the relative frequency of selecting a red jelly bean? (1 mark)

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Based on this table of relative frequencies, what is the probability of NOT selecting a black jelly bean? (1 mark)

--- 2 WORK AREA LINES (style=lined) ---

Show Answers Only

$0.17$
$0.86$

Show Worked Solution

i. $\text{Relative frequency of red}$

$= 1-(0.32 + 0.13 + 0.14 + 0.24)$

$= 1-0.83$

$= 0.17$

ii. $P\text{(not selecting black)}$

$= 1-P\text{(selecting black)}$

$= 1-0.14$

$= 0.86$

Algebra, STD2 A1 2015 HSC 26b

Clark’s formula is used to determine the dosage of medicine for children.

`text(Dosage) = text(weight in kg × adult dosage)/70`

The adult daily dosage of a medicine contains 3150 mg of a particular drug.

A child who weighs 35 kg is to be given tablets each containing 525 mg of this drug.

How many tablets should this child be given daily? (2 marks)

--- 4 WORK AREA LINES (style=lined) ---

Show Answers Only

`3`

Show Worked Solution

`text(Dosage)= (35 xx 3150)/70= 1575\ text(mg)`

`text(# Tablets per day)= text(Dosage)/text(mg per tablet)= 1575/525= 3`

`:.\ text(The child should be given 3 tablets per day.)`

Calculus, EXT1* C1 2015 HSC 14a

In a theme park ride, a chair is released from a height of `110` metres and falls vertically. Magnetic brakes are applied when the velocity of the chair reaches `text(−37)` metres per second.

The height of the chair at time `t` seconds is `x` metres. The acceleration of the chair is given by `ddot x = −10`. At the release point, `t = 0, x = 110 and dot x = 0`.

Using calculus, show that `x = -5t^2 + 110`. (2 marks)

--- 5 WORK AREA LINES (style=lined) ---
How far has the chair fallen when the magnetic brakes are applied? (2 marks)

--- 4 WORK AREA LINES (style=lined) ---

Show Answers Only

`text(Proof)\ \ text{(See Worked Solutions)}`
`68.45\ text(m)`

Show Worked Solution

i. `text(Show)\ \ x = -5t^2 + 110`

`ddot x`	`= -10`
`dot x`	`= int ddot x\ dt`
	`= int -10\ dt`
	`= -10t + c`

`text(When)\ \ t = 0,\ dot x = 0`

`:.\ 0`	`= -10 (0) + c`
`c`	`= 0`

`dot x`	`= -10t`
`x`	`= int dot x\ dt`
	`= int -10t\ dt`
	`= -5t^2 + c`

`text(When)\ \ t = 0,\ x = 110`

`:.\ 110`	`= -5 (0^2) + c`
`c`	`= 110`

`:.\ x = -5t^2 + 110\ \ text(… as required.)`

ii. `text(Find)\ \t\ \text(when)\ \ dot x = -37`

`-37`	`= -10t`
`t`	`= 3.7\ \ text(seconds)`

`text(When)\ \ t = 3.7`

`x`	`= -5 (3.7^2) + 110`
	`= -68.45 + 110`
	`= 41.55`

`:.\ text(Distance the chair has fallen)`

`= 110 – 41.55`

`= 68.45\ text(m)`

Algebra, STD2 A1 2015 HSC 2 MC

Which of the following is `4x+3y-x-5y` in its simplest form?

`3x-2y`
`3x+8y`
`5x-2y`
`5x+8y`

Show Answers Only

`A`

Show Worked Solution

`4x+3y-x-5y= 3x-2y`

`⇒ A`

Calculus, 2ADV C4 2015 HSC 16a

The diagram shows the curve with equation `y = x^2-7x + 10`. The curve intersects the `x`-axis at points `A and B`. The point `C` on the curve has the same `y`-coordinate as the `y`-intercept of the curve.

Find the `x`-coordinates of points `A and B.` (1 mark)

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Write down the coordinates of `C.` (1 mark)

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Evaluate `int _0^2 (x^2-7x + 10)\ dx.` (1 mark)

--- 1 WORK AREA LINES (style=lined) ---
Hence, or otherwise, find the area of the shaded region. (2 marks)

--- 4 WORK AREA LINES (style=lined) ---

Show Answers Only

`A = 2,\ \ B = 5`
`(7, 10)`
`8 2/3`
`16 1/3\ text(u²)`

Show Worked Solution

i. `y`	`= x^2-7x + 10`
	`= (x-2) (x-5)`

`:.x = 2 or 5`

`:.\ \ x text(-coordinate of)\ \ A = 2`

`x text(-coordinate of)\ \ B = 5`

ii. `y\ text(intercept occurs when)\ \ x = 0`

`=>y text(-intercept) = 10`

`C\ text(occurs at intercept:)`

`y`	`= x^2-7x + 10`	`\ \ \ \ \ text{… (1)}`
`y`	`= 10`	`\ \ \ \ \ text{… (2)}`

`(1) = (2)`

`x^2-7x + 10`	`= 10`
`x^2-7x`	`= 10`
`x (x-7)`	`= 10`

`x = 0 or 7`

`:.\ C\ \ text(is)\ \ (7, 10)`

iii. `int_0^2 (x^2 – 7x + 10)\ dx`

`= [1/3 x^3 – 7/2 x^2 + 10x]_0^2`

`= [(1/3 xx 2^3 – 7/2 xx 2^2 + 10 xx 2) – 0]`

`= 8/3 – 14 + 20`

`= 8 2/3`

iv.

`A_1 = A_2`

♦ Mean mark 49%.

`A_2 = 8 2/3\ text(u²)\ \ \ \ text{(from part (iii))}`

`text(Let)\ \ D\ \ text(be)\ \ (7, 0)`

`text(Shaded Area)`

`= text(Area of)\ \ Delta ACD – A_2`

`= 1/2 bh – 8 2/3`

`= 1/2 xx 5 xx 10 – 8 2/3`

`= 16 1/3\ text(u²)`

Calculus, EXT1* C1 2015 HSC 15a

The amount of caffeine, `C`, in the human body decreases according to the equation

`(dC)/(dt) = -0.14C,`

where `C` is measured in mg and `t` is the time in hours.

Show that `C = Ae^(-0.14t)` is a solution to `(dC)/(dt) = -0.14C,` where ` A` is a constant.

When `t = 0`, there are 130 mg of caffeine in Lee’s body. (1 mark)

--- 3 WORK AREA LINES (style=lined) ---
Find the value of `A.` (1 mark)

--- 1 WORK AREA LINES (style=lined) ---
What is the amount of caffeine in Lee’s body after 7 hours? (1 mark)

--- 4 WORK AREA LINES (style=lined) ---
What is the time taken for the amount of caffeine in Lee’s body to halve? (2 marks)

--- 8 WORK AREA LINES (style=lined) ---

Show Answers Only

`text(Proof)\ \ text{(See Worked Solutions)}`
`130`
`48.8\ text(mg)`
`4.95\ text(hours)`

Show Worked Solution

i.	`C`	`= Ae^(-0.14t)`
	`(dC)/(dt)`	`= d/(dt) (Ae^(-0.14t))`
		`= -0.14 xx Ae^(-0.14t)`
		`= -0.14\ C`

`:.\ C = Ae^(-0.14t)\ \ text(is a solution)`

ii. `text(When)\ \ t = 0,\ C = 130`

`130`	`= Ae^(-0.14 xx 0)`
`:.\ A`	`= 130`

iii. `text(Find)\ \ C\ \ text(when)\ \ t = 7`

`C`	`= 130\ e^(-0.14 xx 7)`
	`= 130\ e^(-0.98)`
	`= 48.79…`
	`= 48.8\ text{mg (to 1 d.p.)}`

`:.\ text(After 7 hours, Lee will have 48.8 mg)`

`text(of caffeine left in her body.)`

iv. `text(Find)\ \ t\ \ text(when caffeine has halved.)`

`text(When)\ \ t = 0,\ \ C = 130`

`:.\ text(Find)\ \ t\ \ text(when)\ \ C = 65`

`65`	`= 130 e^(-0.14 xx t)`
`e^(-0.14t)`	`= 65/130`
`ln e^(-0.14t)`	`= ln\ 65/130`
`-0.14t xx ln e`	`= ln\ 65/130`
`t`	`= (ln\ 65/130)/-0.14`
	`= 4.951…`
	`= 4.95\ text{hours (to 2 d.p.)}`

`:.\ text(It will take 4.95 hours for Lee’s)`

`text(caffeine to halve.)`

Financial Maths, 2ADV M1 2015 HSC 14c

Sam borrows $100 000 to be repaid at a reducible interest rate of 0.6% per month. Let `$A_n` be the amount owing at the end of `n` months and `$M` be the monthly repayment.

Show that `A_2 = 100\ 000 (1.006)^2 - M (1 + 1.006).` (1 mark)

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Show that `A_n = 100\ 000 (1.006)^n - M (((1.006)^n - 1)/0.006).` (2 marks)

--- 5 WORK AREA LINES (style=lined) ---
Sam makes monthly repayments of $780. Show that after making 120 monthly repayments the amount owing is $68 500 to the nearest $100. (1 mark)

--- 2 WORK AREA LINES (style=lined) ---

Immediately after making the 120th repayment, Sam makes a one-off payment, reducing the amount owing to $48 500. The interest rate and monthly repayment remain unchanged.

After how many more months will the amount owing be completely repaid? (3 marks)

--- 6 WORK AREA LINES (style=lined) ---

Show Answers Only

`text(Proof)\ \ text{(Show Worked Solutions)}`
`text(Proof)\ \ text{(Show Worked Solutions)}`
`text(Proof)\ \ text{(Show Worked Solutions)}`
`79\ text(months)`

Show Worked Solution

i. `A_1`	`= 100\ 000 (1.006) – M`
`A_2`	`= A_1 (1.006) – M`
	`= [100\ 000 (1.006) – M] (1.006) – M`
	`= 100\ 000 (1.006)^2 – M (1.006) – M`
	`= 100\ 000 (1.006)^2 – M (1 + 1.006)\ \ text(… as required)`

ii. `A_3 = 100\ 000 (1.006)^3 – M (1 + 1.006 + 1.006^2)`

`vdots`

`A_n = 100\ 000 (1.006)^n – M (1 + 1.006 + … + 1.006^(n-1))`

`=> text(S)text(ince)\ \ (1 + 1.006 + … + 1.006^(n-1))\ text(is a)`

`text(GP with)\ \ a = 1,\ r = 1.006`

`:.\ A_n`	`= 100\ 000 (1.006)^n – M ((a (r^n – 1))/(r – 1))`
	`= 100\ 000 (1.006)^n – M ((1 (1.006^n – 1))/(1.006 – 1))`
	`= 100\ 000 (1.006)^n – M (((1.006)^n – 1)/0.006)`

`text(… as required.)`

iii. `text(If)\ \ M = 780 and n = 120`

`A_120`	`= 100\ 000 (1.006)^120 – 780 ((1.006^120 – 1)/0.006)`
	`= 205\ 001.80… – 780 (175.0030…)`
	`= 205\ 001.80… – 136\ 502.34…`
	`= 68\ 499.45…`
	`= $68\ 500\ \ text{(to nearest $100) … as required}`

iv. `text(After the one-off payment, amount owing)=$48\ 500`

`:. A_n = 48\ 500 (1.006)^n – 780 ((1.006^n – 1)/0.006)`

`text(where)\ \ n\ \ text(is the number of months after)`

`text(the one-off payment.)`

`text(Find)\ \ n\ \ text(when)\ \ A_n = 0`

`48\ 500 (1.006)^n – 780 ((1.006^n – 1)/0.006) = 0`

`48\ 500 (1.006)^n`	`= 780 ((1.006^n – 1)/0.006)`
`48\ 500 (1.006)^n`	`= 130\ 000 (1.006^n – 1)`
	`= 130\ 000 (1.006)^n – 130\ 000`
`81\ 500 (1.006)^n`	`= 130\ 000`
`1.006^n`	`= (130\ 000)/(81\ 500)`

`n xx ln 1.006`	`= ln\ 1300/815`
`n`	`= (ln\ 1300/815)/(ln\ 1.006)`
	`= 78.055…`

`:.\ text(The amount owing will be completely repaid after)`

`text(another 79 months.)`

Probability, 2ADV S1 2015 HSC 14b

Weather records for a town suggest that:

if a particular day is wet `(W)`, the probability of the next day being dry is `5/6`
if a particular day is dry `(D)`, the probability of the next day being dry is `1/2`.

In a specific week Thursday is dry. The tree diagram shows the possible outcomes for the next three days: Friday, Saturday and Sunday.

Show that the probability of Saturday being dry is `2/3`. (1 mark)

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What is the probability of both Saturday and Sunday being wet? (2 marks)

--- 4 WORK AREA LINES (style=lined) ---
What is the probability of at least one of Saturday and Sunday being dry? (1 mark)

--- 2 WORK AREA LINES (style=lined) ---

Show Answers Only

`text(Proof)\ \ text{(See Worked Solutions)}`
`1/(18)`
`(17)/(18)`

Show Worked Solution

i. `text{Show}\ \ P text{(Sat dry)} = 2/3`

`P text{(Sat dry)}`

`= P (W,D) + P (D, D)`

`=(1/2 xx 5/6) + (1/2 xx 1/2)`

`= 5/(12) + 1/4`

`= 2/3\ \ text(… as required)`

ii. `Ptext{(Sat and Sun wet)}`

`= P (WWW) + P (DWW)`

`= (1/2 xx 1/6 xx 1/6) + (1/2 xx 1/2 xx 1/6)`

`= 1/(72) + 1/(24)`

`= 1/(18)`

iii. `Ptext{(At least Sat or Sun dry)}`

`= 1-Ptext{(Sat and Sun both wet)}`

`= 1-1/(18)`

`= (17)/(18)`

Calculus, 2ADV C3 2015 HSC 13c

Consider the curve `y = x^3 − x^2 − x + 3`.

Find the stationary points and determine their nature. (4 marks)

--- 8 WORK AREA LINES (style=lined) ---
Given that the point `P (1/3, 70/27)` lies on the curve, prove that there is a point of inflection at `P`. (2 marks)

--- 5 WORK AREA LINES (style=lined) ---
Sketch the curve, labelling the stationary points, point of inflection and `y`-intercept. (2 marks)

--- 8 WORK AREA LINES (style=lined) ---

Show Answers Only

`text(MAX at)\ (-1/3, 86/27); \ text(MIN at)\ (1, 2)`
`text(Proof)\ \ text{(See Worked Solutions)}`

Show Worked Solution

i.	`y`	`= x^3 – x^2 – x + 3`
	`(dy)/(dx)`	`= 3x^2 – 2x – 1`
	`(d^2y)/(dx^2)`	`= 6x – 2`

`text(S.P.’s when)\ (dy)/(dx) = 0`

`3x^2 – 2x – 1`	`= 0`
`(3x + 1) (x – 1)`	`= 0`

`x = -1/3 or 1`

`text(When)\ \ x = -1/3`

`f(-1/3)`	`= (-1/3)^3 – (-1/3)^2 – (-1/3) + 3`
	`= -1/27 – 1/9 + 1/3 + 3`
	`= 86/27`
`f″(-1/3)`	`= (6 xx -1/3) – 2 = -4 < 0`

`:.\ text(MAX at)\ \ (-1/3, 86/27)`

`text(When)\ \ x = 1`

`f(1)`	`= 1^3 – 1^2 – 1 + 3 =2`
`f″(1)`	`= (6 xx 1) – 2 = 4 > 0`

`:.\ text(MIN at)\ \ (1, 2)`

ii. `(d^2y)/(dx^2) = 0\ \ text(when)`

`6x-2`	`=0`
`x`	`=1/3`

`text(Checking change of concavity)`

`text(Concavity changes either side of)\ x = 1/3`

`:.\ (1/3, 70/27)\ \ text(is a P.I.)`

iii. `text(When)\ \ x`	`= 0`
`y`	`= 3`

Trigonometry, 2ADV T1 2015 HSC 13a

The diagram shows `Delta ABC` with sides `AB = 6` cm, `BC = 4` cm and `AC = 8` cm.

Show that `cos A = 7/8`. (1 mark)

--- 2 WORK AREA LINES (style=lined) ---
By finding the exact value of `sin A`, determine the exact value of the area of `Delta ABC`. (2 marks)

--- 5 WORK AREA LINES (style=lined) ---

Show Answers Only

`text(Proof)\ \ text{(See Worked Solutions)}`
`3 sqrt 15\ \ text(cm²)`

Show Worked Solution

i. `text(Show)\ cos A = 7/8`

`text(Using the cosine rule:)`

`cos A`	`= (b^2 + c^2-a^2)/(2bc)`
	`= (8^2 + 6^2-4^2)/(2 xx 8 xx 6)`
	`= (64 + 36-16)/96`
	`= 84/96`
	`= 7/8\ \ text(… as required)`

♦ Mean mark 40%.

ii.

`a^2 + 7^2`	`= 8^2`
`a^2 + 49`	`= 64`
`a^2`	`= 15`
`a`	`= sqrt 15`
`:.\ sin A`	`= (sqrt 15)/8`

`:.\ text(Area)\ Delta ABC`	`= 1/2 bc\ sin A`
	`= 1/2 xx 8 xx 6 xx (sqrt 15)/8`
	`= 3 sqrt 15\ \ text(cm²)`

Quadratic, 2UA 2015 HSC 12e

The diagram shows the parabola `y = x^2/2` with focus `S (0, 1/2).` A tangent to the parabola is drawn at `P (1, 1/2).`

Find the equation of the tangent at the point `P`. (2 marks)
What is the equation of the directrix of the parabola? (1 mark)
The tangent and directrix intersect at `Q`.
Show that `Q` lies on the `y`-axis. (1 mark)
Show that `Delta PQS` is isosceles. (1 mark)

Show Answers Only

`y = x – 1/2`
`y = -1/2`
`text(Proof)\ \ text{(See Worked Solutions)}`
`text(Proof)\ \ text{(See Worked Solutions)}`

Show Worked Solution

(i)

`y = 1/2 x^2`

`(dy)/(dx) = x`

`text(When)\ \ x = 1,\ \ (dy)/(dx) = 1`

`text(Equation of tangent,)\ m = 1,\ text(through)\ (1, 1/2):`

`y – y_1`	`= m (x – x_1)`
`y – 1/2`	`= 1 (x – 1)`
`y – 1/2`	`= x – 1`
`y`	`= x – 1/2`

(ii) `text(Directrix is)\ \ y = -1/2`

(iii) `Q\ text(is at the intersection of)`

`y = x – 1/2\ \ …\ \ text{(1)}`

`y = -1/2\ \ \ \ …\ \ text{(2)}`

`text{(1) = (2)}`

`x – 1/2`	`= -1/2`
`x`	`= 0`

`:.\ Q\ text(lies on the)\ y text(-axis)\ \ …\ \ text(as required)`

(iv) `text(Show)\ Delta PQS\ text(is isosceles.)`

`text(Distance)\ PS = 1 – 0 = 1`

`Q\ text(has coordinates)\ (0, -1/2)`

`text(Distance)\ SQ = 1/2 + 1/2 = 1`

`:. PS = SQ = 1`

`:.\ Delta PQS\ text(is isosceles)`

Calculus, 2ADV C1 2015 HSC 12c

Find `f^{′}(x)`, where `f(x) = (x^2 + 3)/(x-1).` (2 marks)

--- 5 WORK AREA LINES (style=lined) ---

Show Answers Only

`((x-3) (x + 1))/(x-1)^2`

Show Worked Solution

`f(x) = (x^2 + 3)/(x-1)`

`text(Using the quotient rule:)`

`u`	`= x^2 + 3`	`\ \ \ \ \ \ v`	`= x-1`
`u^{′}`	`= 2x`	`\ \ \ \ \ \ v^{′}`	`= 1`

`f^{′}(x)`	`= (u^{′} v-uv^{′})/v^2`
	`= (2x (x-1)-(x^2 + 3) xx 1)/(x-1)^2`
	`= (2x^2-2x-x^2-3)/(x-1)^2`
	`= (x^2-2x-3)/(x-1)^2`
	`= ((x-3) (x + 1))/(x-1)^2`

Functions, 2ADV F1 2015 HSC 12b

The diagram shows the rhombus `OABC`.

The diagonal from the point `A (7, 11)` to the point `C` lies on the line `l_1`.

The other diagonal, from the origin `O` to the point `B`, lies on the line `l_2` which has equation `y = -x/3`.

Show that the equation of the line `l_1` is `y = 3x - 10`. (2 marks)
The lines `l_1` and `l_2` intersect at the point `D`.
Find the coordinates of `D`. (2 marks)

Show Answers Only

`text(Proof)\ \ text{(See Worked Solutions)}`
`(3, –1)`

Show Worked Solution

(i) `text(Show)\ \ l_1\ \ text(is)\ \ y = 3x – 10`

`l_2\ \ text(is)\ \ y = -x/3`

`m_(l_2)`	`= -1/3`
`:.\ m_(l_1)`	`= 3\ \ \ `	`text{(diagonals of rhombus}\ OABC\ text(are)`
		`text{perpendicular bisectors.)}`

`l_1\ text(has)\ m = 3,\ text(through)\ \ A(7, 11)`

`y – y_1`	`= m (x – x_1)`
`y – 11`	`= 3 (x – 7)`
`y-11`	`= 3x – 21`
`y`	`= 3x – 10\ \ text(… as required)`

(ii) `D\ text(occurs at the intersection of)\ \ l_1 and l_2`

`y`	`= -1/3\ x`	`\ text{… (1)}`
`y`	`= 3x – 10`	`\ text{… (2)}`

`text{(1)}`	`=\ text{(2)}`
`-1/3\ x`	`= 3x – 10`
`10/3\ x`	`= 10`
`:.\ x`	`= 3`

`text(Substitute)\ \ x = 3\ \ text{into (1)}`

`y`	`= -1/3 xx 3`
	`= -1`
`:.\ D\ text{has coordinates (3, –1)}`

Trigonometry, 2ADV T2 2015 HSC 12a

Find the solutions of `2 sin theta = 1` for `0 <= theta <= 2 pi`. (2 marks)

--- 4 WORK AREA LINES (style=lined) ---

Show Answers Only

`pi/6, (5 pi)/6`

Show Worked Solution

`2 sin theta`	`= 1,\ \ \ \ 0 <= theta <= 2 pi`
`sin theta`	`= 1/2`

`=> sin (pi/6) = 1/2 \ \ text(and sin is positive)`

`text(in the 1st/2nd quadrants)`

`:. theta`	`= pi/6, pi-pi/6`
	`= pi/6, (5 pi)/6`

Calculus, 2ADV C4 2015 HSC 11h

Find `int x/(x^2 - 3)\ dx`. (2 marks)

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`1/2 ln (x^2 – 3) + C`

Show Worked Solution

`int x/(x^2 – 3)\ dx`

`= 1/2 int (2x)/(x^2 – 3)\ dx`

`= 1/2 ln (x^2 – 3) + C`

Trig Calculus, 2UA 2015 HSC 11g

Evaluate `int_0^(pi/4) cos 2x\ dx`. (2 marks)

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`1/2`

Show Worked Solution

`int_0^(pi/4) cos 2x`

`= [1/2 sin\ 2x]_0^(pi/4)`

`= [1/2 sin\ pi/2 – 1/2 sin\ 0]`

`= 1/2 – 0`

`= 1/2`

Calculus, 2ADV C2 2015 HSC 11f

Differentiate `y = (x + 4) ln\ x`. (2 marks)

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`ln\x + 4/x +1`

Show Worked Solution

`y = (x + 4) ln\ x`

`text(Using the product rule)`

`(dy)/(dx)`	`= d/(dx) (x + 4) * ln x + (x + 4) d/(dx) ln\ x`
	`= ln x + (x + 4) 1/x`
	`= ln x + 4/x + 1`

Calculus, 2ADV C2 2015 HSC 11e

Differentiate `(e^x + x)^5`. (2 marks)

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`5 (e^x + 1) (e^x + x)^4`

Show Worked Solution

`y`	`= (e^x + x)^5`
`(dy)/(dx)`	`= 5 (e^x + x)^4 xx d/(dx) (e^x + x)`
	`= 5 (e^x + x)^4 xx (e^x + 1)`
	`= 5 (e^x + 1) (e^x + x)^4`

Financial Maths, 2ADV M1 2015 HSC 11d

Find the limiting sum of the geometric series `1 - 1/4 + 1/16 - 1/64 + …`. (2 marks)

--- 5 WORK AREA LINES (style=lined) ---

Show Answers Only

`4/5`

Show Worked Solution

`1 – 1/4 + 1/16 – 1/64 + …`

`r = -1/4,\ \ a=1`

`text(S)text(ince)\ |\ r\ | = 1/4 < 1`

`S_oo`	`= a/(1 – r)`
	`= 1/(1 – (-1/4))`
	`= 1/(5/4)`
	`= 4/5`

Functions, 2ADV F1 2015 HSC 11c

Express `8/(2 + sqrt 7)` with a rational denominator. (2 marks)

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`(-8 (2-sqrt 7))/3`

Show Worked Solution

`8/(2 + sqrt 7) xx (2-sqrt 7)/(2-sqrt 7)`

`= (8(2-sqrt 7))/(2^2 -(sqrt 7)^2)`

`= (8 (2-sqrt 7))/(4-7)`

`= (-8(2-sqrt 7))/3`

Functions, 2ADV F1 2015 HSC 11b

Factorise fully `3x^2-27`. (2 marks)

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`3 (x + 3) (x-3)`

Show Worked Solution

`3x^2-27`	`= 3 (x^2-9)`
	`= 3 (x + 3) (x-3)`

Functions, 2ADV F1 2015 HSC 11a

Simplify `4x − (8 − 6x)`. (1 mark)

Show Answers Only

`2 (5x – 4)`

Show Worked Solution

`4x – (8 – 6x)`

`= 4x – 8 + 6x`

`= 10x – 8`

`= 2 (5x – 4)`

Probability, 2ADV S1 2015 HSC 4 MC

The probability that Mel’s soccer team wins this weekend is `5/7`.

The probability that Mel’s rugby league team wins this weekend is `2/3`.

What is the probability that neither team wins this weekend?

`2/21`
`10/21`
`13/21`
`19/21`

Show Answers Only

`A`

Show Worked Solution

`Ptext{(win at soccer)} = 5/7\ \ =>\ \ Ptext{(not win at soccer)} = 1-5/7 = 2/7`

`Ptext{(win at league)} = 2/3\ \ =>\ \ Ptext{(not win at league)} = 1/3`

`Ptext{(not win at both)}= 2/7 xx 1/3= 2/21`

`=> A`

Functions, 2ADV F1 2015 HSC 2 MC

What is the slope of the line with equation `2x-4y + 3 = 0`?

`-2`
`-1/2`
`1/2`
`2`

Show Answers Only

`C`

Show Worked Solution

`2x-4y + 3`	`= 0`
`4y`	`= 2x + 3`
`y`	`= 1/2 x + 3/4`

`=> C`

Functions, 2ADV F1 2015 HSC 1 MC

What is `0.005\ 233\ 59` written in scientific notation, correct to 4 significant figures?

`5.2336 xx 10^-2`
`5.234 xx 10^-2`
`5.2336 xx 10^-3`
`5.234 xx 10^-3`

Show Answers Only

`D`

Show Worked Solution

`0.005\ 233\ 59 = 5.234 xx 10^-3`

`=> D`

Functions, 2ADV 2006 HSC 7c

Write down the discriminant of `2x^2 + (k-2)x + 8` where `k` is a constant. (1 mark)
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Hence, or otherwise, find the values of `k` for which the parabola `y = 2x^2 + kx + 9` does not intersect the line `y = 2x + 1`. (2 marks)

--- 6 WORK AREA LINES (style=lined) ---

Show Answers Only

i. `\Delta= k^2-4k-60`

ii. `-6 < k < 10`

Show Worked Solution

i. `2x^2 + (k-2)x + 8`

`Delta`	`= b^2-4ac`
	`= (k-2)^2-4 xx 2 xx 8`
	`= k^2-4k + 4-64`
	`= k^2-4k-60`

ii. `y`	`= 2x^2 + kx + 9`	`\ \ text{… (1)}`
`y`	`= 2x + 1`	`\ \ text{… (2)}`

`text(Substitute)\ y = 2x + 1\ text{into (1)}`

`2x + 1 = 2x^2 + kx + 9`

`2x^2 + kx-2x + 8 = 0`

`2x^2 + (k-2)x + 8 = 0\ …\ text{(∗)}`

`text{The graphs will not intercept if (∗) has no roots, i.e.)\ \ Delta <0`

`k^2-4k-60`	`< 0`
`(k-10) (k + 6)`	`< 0`

`text(From the graph, no intersection when)`

`-6 < k < 10`

Calculus, EXT1* C1 2006 HSC 6b

A rare species of bird lives only on a remote island. A mathematical model predicts that the bird population, `P`, is given by

`P = 150 + 300 e^(-0.05t)`

where `t` is the number of years after observations began.

According to the model, how many birds were there when observations began? (1 mark)

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According to the model, what will be the rate of change in the bird population ten years after observations began? (2 marks)

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What does the model predict will be the limiting value of the bird population? (1 mark)

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The species will become eligible for inclusion in the endangered species list when the population falls below `200`. When does the model predict that this will occur? (2 marks)

--- 6 WORK AREA LINES (style=lined) ---

Show Answers Only

`450\ text(birds)`
`-9.1\ text{(to 1 d.p.)}`
`150\ text(birds)`
`text(After 35.83… years)`

Show Worked Solution

i. `P = 150 + 300 e^(-0.05t)`

`text(When)\ \ t = 0`

`P`	`= 150 + 300 e^0`
	`= 450`

`:.\ text(There were 450 birds when observations began.)`

ii. `(dP)/dt`	`= 300 xx (-0.05) xx e^(-0.05t)`
	`= -15 e^(-0.05t)`

`text(When)\ \ t = 10`

`(dP)/dt`	`= -15 e^(-0.05 xx 10)`
	`= -15 e^(-0.5)`
	`= -9.097…`
	`= -9.1\ text{(to 1 d.p.)}`

`:.\ text(After 10 years, the bird population will be)`

`text(decreasing at a rate of 9.1 birds per year.)`

iii. `text(As)\ t rarr oo`

`300 e^(-0.05t) rarr 0`

`:. P = 150 + 300 e^(-0.05t) rarr 150`

`:.\ text(The model predicts a limiting population)`

`text(of 150 birds.)`

iv. `text(Find)\ t\ text(when)\ P < 200`

`150 + 300 e^(-0.05t)`	`< 200`
`300 e^(-0.05t)`	`< 50`
`e^(-0.05t)`	`< 50/300`

`ln e^(-0.05t)`	`< ln 1/6`
`-0.05t`	`< ln 1/6`
`t`	`> (ln 1/6)/(-0.05t)`
`t`	`> 35.83…`

`:.\ text(The model predicts the population)`

`text(will fall below 200 after 35.83… years.)`

Plane Geometry, 2UA 2006 HSC 6a

In the diagram, `AD` is parallel to `BC`, `AC` bisects `/_BAD` and `BD` bisects `/_ABC`. The lines `AC` and `BD` intersect at `P`.

Copy or trace the diagram into your writing booklet.

Prove that `/_BAC = /_BCA`. (1 mark)
Prove that `Delta ABP ≡ Delta CBP`. (2 marks)
Prove that `ABCD` is a rhombus. (3 marks)

Show Answers Only

`text(Proof)\ \ text{(See Worked Solutions)}`
`text(Proof)\ \ text{(See Worked Solutions)}`
`text(Proof)\ \ text{(See Worked Solutions)}`

Show Worked Solution

(i)

`text(Prove)\ /_BAC = /_BCA`

`/_BCA`	`= /_CAD\ \ \ text{(alternate angles,}\ BC\ text(\|\|)\ AD text{)}`
`/_CAD`	`= /_BAC\ \ \ text{(}AC\ text(bisects)\ /_BAD text{)}`
`:. /_BAC`	`= /_BCA\ …\ text(as required)`

(ii) `text(Prove)\ Delta ABP ≡ Delta CBP`

`/_BAC`	`= /_BCA\ \ \ text{(from part (i))}`
`/_ABP`	`= /_CBP\ text{(}BD\ text(bisects)\ /_ABC text{)}`
`BP\ text(is common)`

`:. Delta ABP ≡ Delta CBP\ \ text{(AAS)}`

(iii) `text(Using)\ Delta ABP ≡ Delta CBP`

`AP = PC\ \ text{(corresponding sides of congruent triangles)}`

`/_BPA = /_BPC\ \ text{(corresponding angles of congruent triangles)}`

`text(Also,)\ /_BPA = /_BPC = 90^@`

`text{(}/_APC\ text{is a straight angle)}`

`text(Considering)\ Delta APD and Delta APB`

`/_DAP`	`= /_BAP\ text{(}AC\ text(bisects)\ /_BAD text{)}`
`/_DPA`	`= 90^@\ text{(vertically opposite angles)}`
`:. /_DPA`	`= /_BPA = 90^@`

`PA\ text(is common)`

`:. Delta APD ≡ Delta APB\ \ text{(AAS)}`

`BP = PD\ \ text{(corresponding sides of congruent triangles)}`

`:. ABCD\ text(is a rhombus as its diagonals are)`

`text(perpendicular bisectors.)`

Calculus, 2ADV C4 2006 HSC 5b

Show that `d/dx log_e (cos x) = -tan x.` (1 mark)

--- 2 WORK AREA LINES (style=lined) ---
The shaded region in the diagram is bounded by the curve `y =tan x` and the lines `y =x` and `x = pi/4.`

Using the result of part (i), or otherwise, find the area of the shaded region. (3 marks)

--- 6 WORK AREA LINES (style=lined) ---

Show Answers Only

`text(Proof)\ \ text{(See Worked Solutions)}`
`(1/2 log_e 2 – pi^2/32)\ text(u²)`

Show Worked Solution

i. `d/dx log_e (cos x)`	`= (-sin x)/cos x`
	`= – tan x\ …\ text(as required)`

ii. `text(Shaded Area)`

`= int_0^(pi/4) tan x\ dx – int_0^(pi/4) x\ dx`

`= int_0^(pi/4) tan x – x\ dx`

`= [-log_e (cos x) – 1/2 x^2]_0^(pi/4)`

`=[(-log_e (cos pi/4) – 1/2 xx (pi^2)/16) – (-log_e(cos 0) – 0)]`

`= -log_e 1/sqrt 2 – pi^2/32 + log_e1`

`= -log_e 2^(-1/2) – pi^2/32`

`= (1/2 log_e 2 – pi^2/32)\ text(u²)`

Calculus, 2ADV C3 2004 HSC 4b

Consider the function `f(x) = x^3 − 3x^2`.

Find the coordinates of the stationary points of the curve `y = f(x)` and determine their nature. (3 marks)

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Sketch the curve showing where it meets the axes. (2 marks)

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Find the values of `x` for which the curve `y = f(x)` is concave up. (2 marks)

--- 5 WORK AREA LINES (style=lined) ---

Show Answers Only

`text(MAX at)\ (0,0),\ \ text(MIN at)\ (2,-4)`
`f(x)\ text(is concave up when)\ x>1`

Show Worked Solution

(i)	`f(x)`	`= x^3 – 3x^2`
	`f'(x)`	`= 3x^2 – 6x`
	`f″(x)`	`= 6x – 6`

`text(S.P.’s when)\ \ f'(x) = 0`

`3x^2 – 6x`	`= 0`
`3x (x – 2)`	`= 0`
`x`	`= 0\ \ text(or)\ \ 2`

`text(When)\ x = 0`

`f(0)`	`= 0`
`f″(0)`	`= 0 – 6 = -6 < 0`
`:.\ text(MAX at)\ (0,0)`

`text(When)\ x = 2`

`f(2)`	`= 2^3 – (3 xx 4) = -4`
`f″(2)`	`= (6 xx 2) – 6 = 6 > 0`
`:.\ text(MIN at)\ (2, -4)`

(ii)

`f(x) = x^3 – 3x^2\ text(meets the)\ x text(-axis when)\ f(x) = 0`

`x^3 – 3x^2`	`= 0`
`x^2 (x-3)`	`= 0`
`x`	`= 0\ \ text(or)\ \ 3`

(iii)

`f(x)\ text(is concave up when)`

`f″(x)`	`>0`
`6x – 6`	`>0`
`6x`	`>6`
`x`	`>1`

`:. f(x)\ text(is concave up when)\ \ x>1`

Trigonometry, 2ADV T1 2004 HSC 3c

The diagram shows a point `P` which is 30 km due west of the point `Q`.

The point `R` is 12 km from `P` and has a bearing from `P` of 070°.

Find the distance of `R` from `Q`. (2 marks)

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Find the bearing of `R` from `Q`. (2 marks)

--- 4 WORK AREA LINES (style=lined) ---

Show Answers Only

`19.2\ text(km)\ \ \ text{(1 d.p.)}`
`282^@`

Show Worked Solution

i. `text(Join)\ \ RQ\ \ text(to form)\ \ Delta RPQ`

`/_RPQ = 90-70 = 20^@`

`text(Using the cosine rule:)`

`RQ^2`	`= PR^2 + PQ^2-2 xx PR xx PQ xx cos /_RPQ`
	`= 12^2 + 30^2-2 xx 12 xx 30 xx cos 20^@`
	`= 367.421…`
`RQ`	`= 19.168…= 19.2\ text(km)\ \ text{(1 d.p.)}`

ii. `text(Using sine rule:)`

`(sin /_RQP)/12`	`= (sin 20^@)/(19.168…)`
`sin/_RQP`	`= (12 xx sin 20^@)/(19.168…)= 0.214…`
`/_RQP`	`= 12.36…^@= 12^@\ \ \ text{(nearest degree)}`

`:.\ text(Bearing of)\ R\ text(from)\ Q`

`=270+12`

`=282^@`

Calculus, 2ADV C4 2004 HSC 3bi

Evaluate `int_1^2 e^(3x)\ dx`. (2 marks)

Show Answers Only

`1/3 (e^6 – e^3)`

Show Worked Solution

	`int_1^2 e^(3x)\ dx`	`= 1/3 [e^(3x)]_1^2`
		`= 1/3 (e^6 – e^3)`

Calculus, 2ADV C2 2004 HSC 3aii

Differentiate with respect to `x`:

`(1 + sin x)^5`. (2 marks)

Show Answers Only

`5 cos x\ (1 + sinx)^4`

Show Worked Solution

`y`	`= (1 + sinx)^5`
`dy/dx`	`= 5 (1 + sinx)^4 xx d/(dx)(sinx)`
	`= 5 cos x (1 + sinx)^4`

Functions, 2ADV 2004 HSC 2c

For what values of `k` does `x^2-kx + 4 = 0` have no real roots? (2 marks)

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Show Answers Only

`-4 < k < 4`

Show Worked Solution

`x^2-kx + 4 = 0`

`text(No real roots when)\ Delta < 0`

`b^2-4ac`	`< 0`
`(text(–) k^2)-4 xx 1 xx 4`	`< 0`
`k^2-16`	`< 0`
`k^2`	`< 16`
`:.\ -4 < k`	`< 4`

`:.\ text(There are no real roots when)\ \ \ -4 < k < 4.`

`tan\ theta`	`= \|(m_1 − m_2)/(1 + m_1m_2)\|`
	`= \|(2 − (−3))/(1 + 2(−3))\|`
	`= \|5/(−5)\|`
	`= 1`