Band 3 – Page 35

Functions, EXT1 F1 2004 HSC 1b

Solve `4/(x + 1) < 3.` (3 marks)

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`x < −1\ \ text(or)\ \ x > 1/3`

Show Worked Solution

`4/(x + 1) < 3`

`text(Multiply both sides by)\ (x + 1)^2`

`4(x + 1)`	`< 3(x + 1)^2`
`4x + 4`	`< 3(x^2 + 2x + 1)`
`4x + 4`	`< 3x^2 + 6x + 3`

`3x^2 + 2x − 1`	`> 0`
`(3x − 1)(x + 1)`	`> 0`

`text(LHS) = 0\ \ text(when)\ \ x = 1/3\ \ text(or)\ \ −1`

`text(From the graph,)`

`x < −1\ text(or)\ x > 1/3`

Functions, EXT1 F1 2004 HSC 1a

Indicate the region on the number plane satisfied by `y ≥ |\ x + 1\ |.` (2 marks)

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`text(See Worked Solution)`

Show Worked Solution

`y ≥ |\ x + 1\ |`

`text(Test)\ (0, 0)`

`0 ≥ |\ 0 + 1\ |`

`0 ≥ 1\ \ \ \ \ \ ⇒\ text(False)`

`:.\ text(Shaded area represents)`

`y ≥ |\ x + 1\ |`

Calculus, EXT1 C1 2006 HSC 5c

A hemispherical bowl of radius `r\ text(cm)` is initially empty. Water is poured into it at a constant rate of `k\ text(cm³)` per minute. When the depth of water in the bowl is `x\ text(cm)`, the volume, `V\ text(cm³)`, of water in the bowl is given by

`V = pi/3 x^2 (3r - x).` (Do NOT prove this)

Show that `(dx)/(dt) = k/(pi x (2r - x).` (2 marks)

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Hence, or otherwise, show that it takes 3.5 times as long to fill the bowl to the point where `x = 2/3r` as it does to fill the bowl to the point where `x = 1/3r.` (2 marks)

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

Show Worked Solution

i. `text(Show)\ \ (dx)/(dt) = k/(pi x (2r – x))`

`(dV)/(dt)`	`= k`
`V`	`= pi/3 x^2 (3r – x)`
	`= r pi x^2 – pi/3 x^3`
`(dV)/(dx)`	`= 2 pi r x – pi x^2`
	`= pi x (2r – x)`

`(dV)/(dt)`	`= (dV)/(dx) * (dx)/(dt)`
`k`	`= pi x (2r – x) * (dx)/(dt)`
`:. (dx)/(dt)`	`= k/(pi x (2r – x))\ \ text(… as required)`

ii. `(dx)/(dt)`	`= k/(pi x (2r – x))`
`(dt)/(dx)`	`= 1/k pi x (2r – x)`
`t`	`= 1/k int 2 pi r x – pi x^2\ dx`
	`= 1/k [pi r x^2 – 1/3 pix^3] + c`

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

`:.\ c = 0`

`:.t = 1/k [pi r x^2 – 1/3 pi x^3]`

`text(Find)\ \ t_1,\ \ text(when)\ \ x = 1/3r`

`t_1`	`= 1/k [pi r (r/3)^2 – 1/3 pi (r/3)^3]`
	`= 1/k [(pi r^3)/9 – (pi r^3)/81]`
	`= 1/k ((9 pi r^3)/81 – (pi r^3)/81)`
	`= (8 pi r^3)/(81k)`

`text(Find)\ \ t_2\ \ text(when)\ \ x = 2/3r`

`t_2`	`= 1/k [pi r((2r)/3)^2 – 1/3 pi ((2r)/3)^3]`
	`= 1/k [(4 pi r^3)/9 – (8 pi r^3)/81]`
	`= 1/k ((36 pi r^3)/81 – (8 pi r^3)/81)`
	`= (28 pi r^3)/(81k)`
	`= 3.5 xx (8 pi r^3)/(81k)`
	`= 3.5 xx t_1`

`:.\ text(It takes 3.5 times longer to fill the bowl.)`

L&E, EXT1 2006 HSC 5a

Show that `y = 10e^(-0.7t) + 3` is a solution of

`(dy)/(dt) = -0.7(y - 3).` (2 marks)

Show Answers Only

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

Show Worked Solution

`y = 10e^(-0.7t) + 3`

`(dy)/(dt)`	`= -0.7 xx 10e^(-0.7t)`
	`= -0.7 (10e^(-0.7t) + 3 – 3)`
	`= -0.7 (y – 3)`

Polynomials, EXT1 2006 HSC 3b

By considering `f(x) = 3log_e x - x`, show that the curve `y = 3 log_e x` and the line `y = x` meet at a point `P` whose `x`-coordinate is between `1.5` and `2`. (1 mark)
Use one application of Newton’s method, starting at `x = 1.5`, to find an approximation to the `x`-coordinate of `P`. Give your answer correct to two decimal places. (2 marks)

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

Show Worked Solution

(i)	`f(x)`	`= 3 log_e x – x`
	`f(1.5)`	`= 3 log_e 1.5 – 1.5`
		`= -0.283… <0`
	`f(2)`	`= 3 log_e 2 – 2`
		`= 0.079… >0`

`:.\ text(S)text(ince the sign changes, a zero must)`

`text(exist between 1.5 and 2.)`

(ii)	`f(x)`	`= 3 log_e x – x`
	`f prime(x)`	`= 3/x – 1`
	`f(1.5)`	`= -0.283…`
	`f prime (1.5)`	`= 3/1.5 – 1 = 1`

`x_1`	`= x_0 – (f(x_0))/(f prime (x_0))`
	`= 1.5 – {(-0.283…)}/1`
	`= 1.783…`
	`= 1.78\ \ text{(to 2 d.p.)}`

Quadratic, EXT1 2006 HSC 2c

The points `P(2ap, ap^2), Q(2aq, aq^2)` and `R(2ar, ar^2)` lie on the parabola `x^2 = 4ay`. The chord `QR` is perpendicular to the axis of the parabola. The chord `PR` meets the axis of the parabola at `U`.

The equation of the chord `PR` is `y = 1/2(p + r)x - apr.` (Do NOT prove this.)

The equation of the tangent at `P` is `y = px - ap^2.` (Do NOT prove this.)

Find the coordinates of `U.` (1 mark)
The tangents at `P` and `Q` meet at the point `T`. Show that the coordinates of `T` are `(a(p + q), apq).` (2 marks)
Show that `TU` is perpendicular to the axis of the parabola. (1 mark)

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`(0, –apr)`
`text(Proof)\ \ text{(See Worked Solutions)}`
`text(Proof)\ \ text{(See Worked Solutions)}`

Show Worked Solution

(i) `U\ \ text(is the)\ \ y\ \ text(intercept of)\ \ PR`

`y = 1/2 (p + r)x – apr`

`text(When)\ \ x = 0`

`y = -apr`

`:.\ U\ \ text(has coordinates)\ \ (0, –apr)`

(ii) `text(Show)\ \ T\ \ text(is)\ \ (a(p + q), apq)`

`text(T)text(angents at)\ \ P and Q\ \ text(are)`

`y`	`= px – ap^2\ \ \ \ text{… (1)}`
`y`	`= qx – aq^2\ \ \ \ text{… (2)}`

`T\ \ text(occurs when)\ \ \ (1) = (2)`

`px – ap^2`	`= qx – aq^2`
`px – qx`	`= ap^2 – aq^2`
`x(p – q)`	`= a (p^2 – q^2)`
	`= a (p – q) (p + q)`
`:.\ x`	`= a (p + q)`

`text(Substitute)\ \ x = a (p + q)\ \ text{into (1)}`

`y`	`= p * a (p + q) – ap^2`
	`= ap^2 + apq – ap^2`
	`= apq`

`:.\ T (a (p + q), apq)\ \ text(… as required.)`

(iii) `text(Axis of parabola)\ \ x^2 = 4ay\ \ text(is)\ \ y text(-axis)`

`=>TU\ \ text(will be perpendicular to the)\ \ y text(-axis if it has a)`

`text(gradient of)\ \ 0\ \ text{(i.e.}\ \ T and U\ text{have same} y text{-values)}`

`:.\ text(Need to show)\ \ \ -apr = apq.`

`text(Consider)\ \ QR`

`text(S)text(ince it is perpendicular to)\ \ y text(-axis)`

`text(and the parabola is symmetrical)`

`=>\ \ 2aq`	`= -2ar`
`q`	`= -r`

`:.\ apq`	`= ap (-r)`
`apq`	`= -apr`

`:.\ TU\ \ text(is perpendicular.)`

Binomial, EXT1 2006 HSC 2b

By applying the binomial theorem to `(1 + x)^n` and differentiating, show that
`n(1 + x)^(n - 1) = ((n), (1)) + 2((n), (2)) x + … + r((n), (r)) x^(r - 1) + … + n((n), (n)) x^(n - 1).` (1 mark)
Hence deduce that
`n3^(n - 1) = ((n), (1)) + … + r((n), (r)) 2^(r - 1) + … + n((n), (n)) 2^(n - 1).` (1 mark)

Show Answers Only

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

Show Worked Solution

(i) `text(Show)`

`n (1 + x)^(n – 1) = ((n), (1)) + 2((n), (2)) x + … + r((n), (r)) x^(r – 1)`

`+ … + n((n), (n)) x^(n – 1)`

`text(Using binomial expansion)`

`(1 + x)^n = ((n), (0)) + ((n), (1)) x + ((n), (2)) x^2 + … + ((n), (r)) x^r`

`+ … + ((n), (n)) x^n`

`text(Differentiate both sides)`

`n (1 + x)^(n – 1) = ((n), (1)) + 2((n), (2)) x + … + r((n), (r)) x^(r – 1)`

`+ … + n((n), (n)) x^(n – 1)\ \ \ …\ text(as required)`

(ii) `text(Substitute)\ \ x = 2\ \ text{into above equation}`

`n (1 + 2)^(n – 1) = ((n), (1)) + 2((n), (2)) 2 + … + r((n), (r)) 2^(r – 1)`

`+ … + n((n), (n)) 2^(n – 1)`

`n3^(n – 1) = ((n), (1)) + … + r((n), (r)) 2^(r – 1) + … + n((n), (n)) 2^(n – 1)`

`text(… as required.)`

Calculus, EXT1 C2 2006 HSC 2a

Let `f(x) = sin^-1 (x + 5).`

State the domain and range of the function `f(x).` (2 marks)

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Find the gradient of the graph of `y = f(x)` at the point where `x = -5.` (2 marks)

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Sketch the graph of `y = f(x).` (2 marks)

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`text(Domain):\ -6 <= x <= -4, \ \ \ text(Range):\ -pi/2 <= y <= pi/2`
`1`

Show Worked Solution

i. `f(x) = sin^-1 (x + 5)`

`text(Domain)`

`-1 <= x + 5 <= 1`

`-6 <= x <= -4`

`text(Range)`

`-pi/2 <= y <= pi/2`

ii. `y = sin^-1 (x + 5)`

`(dy)/(dx) = 1/sqrt(1-(x + 5)^2)`

`text(When)\ \ x = -5`

`(dy)/(dx)`	`= 1/sqrt(1-(-5 + 5)^2)`
	`= 1/sqrt(1-0)`
	`= 1`

`:.\ text(Gradient of)\ \ y = f(x)\ \ text(at)\ \ x = -5\ \ text(is)\ \ 1.`

iii.

Trigonometry, EXT1 T2 2006 HSC 1d

Simplify `(sin theta + cos theta) (sin^2 theta - sin theta cos theta + cos^2 theta)` (1 mark)

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Hence, or otherwise, express `(sin^3 theta + cos^3 theta)/(sin theta + cos theta) - 1`, in its simplest form for `0 < theta < pi/2.` (2 marks)

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`sin^3 theta + cos^3 theta`
`-sin theta cos theta`

Show Worked Solution

i. `(sin theta + cos theta) (sin^2 theta – sin theta cos theta + cos^2 theta)`

`=sin^3 theta -sin^2thetacos theta + sin theta cos^2 theta + cos theta sin^2 theta – sin theta cos^2 theta + cos^3 theta`

`=sin^3 theta + cos^3 theta`

ii. `(sin^3 theta + cos^3 theta)/(sin theta + cos theta) – 1`

`= {(sin theta + cos theta) (sin^2 theta – sin theta cos theta + cos^2 theta)}/(sin theta + cos theta) – 1`

`= sin^2 theta + cos^2 theta – sin theta cos theta – 1`

`= 1 – sin theta cos theta – 1`

`= -sin theta cos theta`

Calculus, EXT1 C2 2006 HSC 1b

Using the substitution `u =x^4 + 8`, or otherwise, find

`int x^3 sqrt (x^4 + 8)\ dx.` (3 marks)

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`1/6 (x^4 + 8)^(3/2) + c`

Show Worked Solution

`u = x^4 + 8`

`(du)/(dx)`	`= 4x^3`
`1/4 du`	`= x^3 dx`

`:. int x^3 sqrt (x^4 + 8)\ dx`

`= int u^(1/2) *1/4 * du`

`= 1/4 int u^(1/2) du`

`= 1/4 * 2/3 * u^(3/2) + c`

`= 1/6 u^(3/2) + c`

`= 1/6 (x^4 + 8)^(3/2) + c`

Plane Geometry, EXT1 2005 HSC 5b

Two chords of a circle, `AB` and `CD`, intersect at `E`. The perpendiculars to `AB` at `A` and `CD` at `D` intersect at `P`. The line `PE` meets `BC` at `Q`, as shown in the diagram.

Explain why `DPAE` is a cyclic quadrilateral. (1 mark)
Prove that `/_ APE = /_ ABC.` (2 marks)
Deduce that `PQ` is perpendicular to `BC.` (1 mark)

Show Answers Only

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

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

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

Show Worked Solution

(i)

`/_ PAE = /_ PDE = 90°\ \ text{(given)}`

`:. DPAE\ \ text(is a cyclic quadrilateral)`

`text{(opposite angles are supplementary)}`

(ii) `text(Prove)\ \ /_ APE = /_ ABC`

`/_ ABC = /_ ADE`

`text{(angles in the same segment on arc}\ AC text{)}`

`text(S) text(ince)\ \ DPAE\ \ text(is a cyclic quad)\ \ text{(from (i))}`

`/_ APE = /_ ADE`

`text{(angles in the same segment on arc}\ AE text{)}`

`:.\ /_ APE = /_ ABC\ \ text(… as required.)`

(iii) `/_ APE = /_ ABC\ \ \ text{(from part (ii))}`

`/_ AEP = /_ BEQ\ \ \ text{(vertically opposite angles)}`

`:.\ Delta APE\ \ text(|||)\ \ Delta QBE\ \ text{(equiangular)}`

`:.\ /_ EQB = /_ EAP = 90°`

`text{(corresponding angles of similar triangles)}`

`:.\ PQ _|_ BC\ \ text(… as required.)`

Differentiation, EXT1 2005 HSC 3c

Use the definition of the derivative,

`f prime (x) = lim_(h -> 0) (f (x + h) - f(x))/h,`

to find `f prime (x)` when

`f(x) = x^2 + 5x.` (2 marks)

Show Answers Only

`2x + 5`

Show Worked Solution

`f(x) = x^2 + 5x`

`f(x + h)`	`= (x + h)^2 + 5(x + h)`
	`= x^2 + 2xh + h^2 + 5x + 5h`

`f prime (x)`	`= lim_(h -> 0) (f (x + h) – f(x))/h`
	`= lim_(h -> 0) (x^2 + 2xh + 5x + h^2 + 5h – x^2 – 5x)/h`
	`= lim_(h -> 0) (2xh + h^2 + 5h)/h`
	`= lim_(h -> 0) 2x + h + 5`
	`= 2x + 5`

Calculus, EXT1 C2 2005 HSC 3b

By expanding the left-hand side, show that
`qquad sin(5x + 4x) + sin(5x-4x) = 2 sin (5x) cos(4x)` (1 mark)

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Hence find `int sin(5x) cos (4x)\ dx.` (2 marks)

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`text(Proof)\ \ text{(See Worked Solutions)}`
`-1/18 cos(9x)-1/2 cos(x) + c`

Show Worked Solution

i. `sin (5x + 4x) + sin (5x-4x) = 2 sin(5x) cos(4x)`

`text(LHS)`	`= sin (5x) cos (4x)-sin(4x) cos (5x) + sin (5x) cos (4x)+ sin (4x) cos (5x)`
	`= 2 sin (5x) cos (4x)\ \ text(… as required)`

ii. `int sin (5x) cos (4x)\ dx`

`= 1/2 int 2 sin (5x) cos (4x)\ dx`

`= 1/2 int sin (5x + 4x) + sin (5x-4x)\ dx`

`= 1/2 int sin (9x) + sin (x)\ dx`

`= 1/2 [-1/9 cos(9x)-cos(x)] + c`

`= -1/18 cos(9x)-1/2 cos(x) + c`

Calculus, EXT1 C1 2005 HSC 2d

A salad, which is initially at a temperature of 25°C, is placed in a refrigerator that has a constant temperature of 3°C. The cooling rate of the salad is proportional to the difference between the temperature of the refrigerator and the temperature, `T`, of the salad. That is, `T` satisfies the equation

`(dT)/(dt) = -k (T-3),`

where `t` is the number of minutes after the salad is placed in the refrigerator.

Show that `T = 3 + Ae^(–kt)` satisfies this equation. (1 mark)

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The temperature of the salad is 11°C after 10 minutes. Find the temperature of the salad after 15 minutes. (3 marks)

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`text(Proof)\ \ text{(See Worked Solution)}`
`7.8°`

Show Worked Solution

i.	`T`	`= 3 + Ae^(-kt)`
	`(dT)/(dt)`	`= -k xx Ae^(-kt)`
		`= -k [3 + Ae^(-kt) – 3]`
		`= -k (T – 3)`

`:.\ T = 3 + Ae^(-kt)\ \ text(satisfies the equation)`

ii. `T = 3 + Ae^(-kt)`

`text(When)\ \ t = 0\ ,\ T = 25`

`25`	`= 3 + Ae°`
`A`	`= 22`
`:.\ T`	`= 3 + 22 e^(-kt)`

`text(When)\ \ t = 10\ ,\ T = 11`

`:.\ 11`	`= 3 + 22e^(-10k)`
`8`	`= 22e^(-10k)`
`e^(-10k)`	`= 8/22`

`log_e e^(-10k)`	`= log_e\ 4/11`
`-10k`	`= log_e\ 4/11`
`k`	`= -1/10(log_e\ 4/11)`
	`=0.1011…`

`text(Find)\ \ T\ \ text(when)\ \ t = 15:`

`T`	`= 3 + 22 e^(-15k),\ \ text(where)\ \ k = 0.1011…`
	`= 3 + 22 e^(-1.5174…)`
	`= 7.8241…`
	`= 7.8°\ \ text{(to 1 d.p.)}`

`:.\ text(After 15 minutes, the salad will have a)`

`text(temperature of 7.8°.)`

Trig Calculus, EXT1 2005 HSC 2c

Differentiate `e^(3x) (cos x - 3 sin x).` (2 marks)
Hence, or otherwise, find
`int e^(3x) sin x\ dx.` (1 mark)

Show Answers Only

`-10 e^(3x) sin x`
`-1/10 e^(3x) (cos x – 3 sin x) + c`

Show Worked Solution

(i) `y = e^(3x) (cos x – 3 sin x)`

`text(Using the product rule)`

`(dy)/(dx)`	`= e^(3x) (-sin x – 3 cos x) + 3e^(3x) (cos x – 3 sin x)`
	`= -e^(3x) sin x – 3e^(3x) cos x + 3e^(3x) cos x – 9e^(3x) sin x`
	`= -10 e^(3x) sin x`

(ii) `int e^(3x) sin x\ dx`

`= -1/10 int-10 e^(3x) sin x\ dx`

`= -1/10 e^(3x) (cos x – 3 sin x) + c`

Linear Functions, EXT1 2005 HSC 1e

The point `P (1, 4)` divides the line segment joining `A (text(–1), 8)` and `B (x, y)` internally in the ratio `2:3`. Find the coordinates of the point `B.` (2 marks)

Show Answers Only

`(4, text(–2))`

Show Worked Solution

`P (1, 4)\ \ text(divides)\ \ A (text(–1), 8)\ and\ B (x, y)`

`text(internally in ratio)\ \ 2:3.`

`P`	`-= ((nx_1 + mx_2)/(m + n)\ ,\ (ny_1 + my_2)/(m + n))`
`(1, 4)`	`-= ((3 (-1) + 2x)/(2 + 3)\ ,\ (3 (8) + 2y)/(2 + 3))`
	`-= ((2x – 3)/5\ ,\ (24 + 2y)/5)`

`:. (2x -3)/5`	`= 1`	`\ \ \ \ \ \ \ (24 + 2y)/5`	`= 4`
`2x – 3`	`= 5`	`24 + 2y`	`= 20`
`2x`	`= 8`	`2y`	`= -4`
`x`	`= 4`	`y`	`= -2`

`:.\ B\ \ text(has coordinates)\ \ (4, text(–2))`

Trigonometry, EXT1 T1 2005 HSC 1c

State the domain and range of `y = cos^-1 (x/4).` (2 marks)

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`text(Domain)\ \ -4 <= x <= 4`

`text(Range)\ \ 0 <= y <= pi`

Show Worked Solution

`y = cos^-1\ x/4`

`text(Domain of)\ \ y =cos^-1 x\ \ text(is)`

`-1 <= x <= 1`

`:.\ text(Domain of)\ \ y = cos^-1\ x/4\ \ text(is)`

`-1 <= x/4 <= 1`

`-4 <= x <= 4`

`text(Range)\ \y = cos^-1\ x\ \ text(is)`

`0 <= y <= pi`

`:.\ text(Range)\ \ y = cos^-1\ x/4\ \ text(is)`

`0 <= y <= pi`

Calculus, EXT1 C2 2005 HSC 1a

Find `int 1/(x^2 + 49)\ dx.` (1 mark)

Show Answers Only

`1/7\ tan^-1\ x/7 + c`

Show Worked Solution

`int 1/(x^2 + 49)\ dx`	`= 1/7 int 7/(x^2 + 7^2)\ dx`
	`= 1/7\ tan^-1\ x/7 + c`

Calculus, EXT1* C3 2007 HSC 9a

The shaded region in the diagram is bounded by the curve `y = x^2 + 1`, the `x`-axis, and the lines `x = 0` and `x = 1.`

Find the volume of the solid of revolution formed when the shaded region is rotated about the `x`-axis. (3 marks)

Show Answers Only

`(28 pi)/15\ \ text(u³)`

Show Worked Solution

`V`	`= pi int_0^1 y^2\ dx`
	`= pi int_0^1 (x^2 + 1)^2\ dx`
	`= pi int_0^1 x^4 + 2x^2 + 1\ dx`
	`= pi [1/5 x^5 + 2/3 x^3 + x]_0^1`
	`= pi[(1/5 + 2/3 + 1) – 0]`
	`= pi [3/15 + 10/15 + 1]`
	`= (28 pi)/15\ \ text(u³)`

Quadratic, 2UA 2007 HSC 7a

Find the coordinates of the focus, `S`, of the parabola `y = x^2 + 4`. (2 marks)
The graphs of `y = x^2 + 4` and the line `y = x + k` have only one point of intersection, `P`. Show that the `x`-coordinate of `P` satisfies.
1. `x^2 - x + 4 - k = 0`. (1 mark)
Using the discriminant, or otherwise, find the value of `k`. (1 mark)
Find the coordinates of `P`. (2 marks)
Show that `SP` is parallel to the directrix of the parabola. (1 mark)

Show Answers Only

`(0, 4 1/4)`
`text(Proof)\ \ text{(See Worked Solutions)}`
`15/4`
`(1/2, 4 1/4)`
`text(Proof)\ \ text{(See Worked Solutions)}`

Show Worked Solution

(i) `y`	`= x^2 + 4`
`x^2`	`= y – 4`

`text(Using)\ (x – x_0) = 4a (y – y_0)`

`x_0`	`= 0`
`a`	`= 1/4`
`y_0`	`= 4`

`(x – 0) = 4 xx 1/4 xx (y – 4)`

`text(Vertex)`	`= (0, 4)`
`:.\ text(Focus)`	`= (0, 4 1/4)`

(ii) `y`	`= x^2 + 4`	`\ \ …\ \ (1)`
`y`	`= x + k`	`\ \ …\ \ (2)`

`text(Solving simultaneously)`

`x^2 + 4`	`= x + k`
`x^2 – x + 4 – k`	`= 0`

(iii) `text(If only 1 point of intersection,)`

`b^2 – 4ac`	`= 0`
`(–1)^2 – 4 xx 1 xx (4 – k)`	`= 0`
`1 – 16 + 4k`	`= 0`
`4k`	`= 15`
`k`	`= 15/4`

(iv) `text(Finding)\ \ P`

`x^2 – x + 4 – 15/4 = 0`

`x^2 – x + 1/4 = 0`

`(x – 1/2)^2 = 0`

`x = 1/2`

`text(When)\ x = 1/2`

`y`	`= (1/2)^2 + 4`
	`= 4 1/4`

`:. P\ text(has coordinates)\ \ (1/2, 4 1/4)`

(v) `text(Focus)\ (S)`	`= (0, 4 1/4)`
`P`	`= (1/2, 4 1/4)`

`:.\ text(S)text(ince)\ y text(-values are the same,)\ SP\ text(is parallel)`

`text(with the)\ x text(-axis.)`

`=>text(Directrix has the equation)\ \ y = 3 3/4`

`:. SP\ text(is parallel with the directrix.)`

Calculus, 2ADV C3 2007 HSC 6b

Let `f (x) =x^4 - 4x^3`.

Find the coordinates of the points where the curve crosses the axes. (2 marks)

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

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Find the coordinates of the points of inflection. (1 mark)

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Sketch the graph of `y = f (x)`, indicating clearly the intercepts, stationary points and points of inflection. (3 marks)

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

Show Answers Only

`(0, 0)\ , \ (4, 0)`
`text(Minimum S.P. at)\ (3,\ text(-27))`
`(0, 0) and (2, 16)`

Show Worked Solution

i. `f (x) = x^4 – 4x^3`

`text(Cuts)\ x text(-axis when)\ f(x) = 0`

`x^4 – 4x`	`= 0`
`x^3 (x – 4)`	`= 0`

`x = 0 or 4`

`:. text(Cuts the)\ x text(-axis at)\ (0, 0)\ ,\ (4, 0)`

`text(Cuts the)\ y text(-axis when)\ x = 0`

`:. text(Cuts the)\ y text(-axis at)\ (0, 0)`

ii. `f(x) = x^4 – 4x^3`

`f prime (x) = 4x^3 – 12x^2`

`f″ (x) = 12x^2 – 24x`

`text(S.P.’s when)\ f prime (x) = 0`

`4x^3 – 12x^2`	`= 0`
`4x^2 (x – 3)`	`= 0`

`x = 0 or 3`

`text(When)\ x = 0`

`f(0) = 0`

`f″(0) = 0`

`text(S)text(ince concavity changes, a P.I.)`

`text(occurs at)\ (0, 0)`

`text(When)\ x = 3`

`f (3)`	`= 3^4 – 4 xx 3^3`
	`= -27`

`f″ (3)`	`= 12 xx 3^2 – 24 xx 3`
	`= 36 > 0`

`:. text(Minimum S.P. at)\ (3,\ text(–27))`

iii. `text(P.I. when)\ f″(x) = 0`

`12x^2 – 24x`	`= 0`
`12x(x – 2)`	`= 0`

`x = 0 or 2`

`text(P.I. at)\ (0, 0)\ \ \ text{(from(ii))}`

`text(When)\ x = 2`

`text(S)text(ince concavity changes, a P.I.)`

`text(occurs when)\ x = 2`

`f (2)`	`= 2^4 – 4 xx 2^3`
	`= 16`

`:. text(P.I.’s at)\ (0, 0) and (2, 16)`

iv.

Calculus, EXT1* C1 2007 HSC 5b

A particle is moving on the `x`-axis and is initially at the origin. Its velocity, `v` metres per second, at time `t` seconds is given by

`v = (2t)/(16 + t^2).`

What is the initial velocity of the particle? (1 mark)

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Find an expression for the acceleration of the particle. (2 marks)

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Find the time when the acceleration of the particle is zero. (1 mark)

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Find the position of the particle when `t = 4`. (3 marks)

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Show Answers Only

`0`
`{2(16 – t^2)}/(16 + t^2)^2`
`4\ text(seconds)`
`log_e 2\ \ text(metres)`

Show Worked Solution

i. `v = (2t)/(16 + t^2)`

`text(When)\ t`	`= 0`
`v`	`= 0`

`:.\ text(Initial velocity is 0.)`

ii. `a = d/(dt) ((2t)/(16 + t^2))`

`text(Using quotient rule)`

`u`	`= 2t`	`v`	`= 16 + t^2`
`u prime`	`= 2`	`v prime`	`= 2t`

`(dv)/(dt)`	`= (u prime v – uv prime)/v^2`
	`= {2(16 + t^2) – 2t * 2t}/(16 + t^2)^2`
	`= (32 + 2t^2 – 4t^2)/(16 + t^2)^2`
	`= {2(16 – t^2)}/(16 + t^2)^2`

iii. `text(Find)\ t\ text(when)\ (dv)/(dt) = 0`

`{2 (16 – t^2)}/(16 + t^2)^2`	`= 0`
`2 (16 – t^2)`	`= 0`
`t^2`	`= 16`
`t`	`= 4\ ,\ t >= 0`

`:.\ text(The acceleration is zero when)`

`t = 4\ text(seconds.)`

iv. `v = (2t)/(16 + t^2)`

`x`	`= int v\ dt`
	`= int (2t)/(16 + t^2)`
	`= log_e (16 + t^2) + c`

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

`0 = log_e (16 + 0) + c`

`c = -log_e 16`

`:. x = log_e(16 + t^2) – log_e 16`

`text(When)\ t = 4,`

`x`	`= log_e (16 + 4^2) – log_e 16`
	`= log_e 32 – log_e 16`
	`= log_e (32/16)`
	`= log_e 2\ \ text(metres)`

`:.\ text(When)\ t = 4\ , \ text(the position of the)`

`text(particle is)\ log_e 2 \ text(metres.)`

Plane Geometry, 2UA 2007 HSC 5a

In the diagram, `ABCDE` is a regular pentagon. The diagonals `AC` and `BD` intersect at `F`.

Copy or trace this diagram into your writing booklet.

Show that the size of `/_ABC` is `108°`. (1 mark)
Find the size of `/_BAC`. Give reasons for your answer. (2 marks)
By considering the sizes of angles, show that `Delta ABF` is isosceles. (2 marks)

Show Answers Only

`text(See Worked Solutions)`
`36°`
`text(See Worked Solutions)`

Show Worked Solution

(i)

`text(Sum of all internal angles`

`= (n – 2) xx 180°`

`= (5 – 2) xx 180°`

`= 540°`

`:. /_ABC`

`= 540/5= 108°`

(ii) `BA = BC`

`text{(equal sides of a regular pentagon)}`

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

`/_BAC`	`= 1/2 (180 – 108)\ \ \ text{(base angle of}\ Delta BAC text{)}`
	`= 36°`

(iii) `text(Consider)\ Delta BCD and Delta ABC`

`BC = CD = BA`

`text{(equal sides of a regular pentagon)}`

`/_BCD = /_ABC = 108°`

`text{(internal angles of a regular pentagon)}`

`:. Delta BCD -= Delta ABC\ \ \ text{(SAS)}`

`:. CBF`	`= 36°`	`text{(corresponding angles in}`
		`text{congruent triangles)}`

`/_FBA`	`= 108 – 36`
	`= 72°`

`/_BFA`	`= 180 – (72 + 36)\ \ \ \ text{(angle sum of}\ Delta ABF text{)}`
	`= 72°`

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

Trigonometry, 2ADV T1 2007 HSC 4c

An advertising logo is formed from two circles, which intersect as shown in the diagram.

The circles intersect at `A` and `B` and have centres at `O` and `C`.

The radius of the circle centred at `O` is 1 metre and the radius of the circle centred at `C` is `sqrt 3` metres. The length of `OC` is 2 metres.

Use Pythagoras’ theorem to show that `/_OAC = pi/2`. (1 mark)

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Find `/_ ACO` and `/_ AOC`. (2 marks)

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Find the area of the quadrilateral `AOBC`. (1 mark)

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Find the area of the major sector `ACB`. (1 mark)

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Find the total area of the logo (the sum of all the shaded areas). (2 marks)

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

Show Answers Only

`text(Proof)\ \ text{(See Worked Solutions)}`
`/_ ACO = pi/6\ ,\ /_ AOC = pi/3`
`sqrt 3\ \ text(m²)`
`(5 pi)/2\ text(m²)`
`((19 pi + 6 sqrt 3)/6)\ text(m²)`

Show Worked Solution

`text(In)\ Delta AOC`

`AO^2 + AC^2`	`= 1^2 + sqrt 3^2`
	`=1 + 3`
	`= 4`
	`= OC^2`

`:. Delta AOC\ \ text(is right-angled and)\ \ /_OAC = pi/2`

ii. `sin\ /_ACO`	`= 1/2`
`:. /_ACO`	`= pi/6`

`sin\ /_AOC`	`= sqrt 3/2`
`:. /_AOC`	`= pi/3`

iii. `text(Area)\ AOBC`

`= 2 xx text(Area)\ Delta AOC`

`= 2 xx 1/2 xx b xx h`

`= 2 xx 1/2 xx 1 xx sqrt 3`

`= sqrt 3\ \ text(m²)`

iv. `/_ACB = pi/6 + pi/6 = pi/3`

`:. /_ACB\ text{(reflex)}`	`= 2 pi – pi/3`
	`= (5 pi)/3`

`text(Area of major sector)\ ACB`

`= theta/(2 pi) xx pi r^2`

`= {(5 pi)/3}/(2 pi) xx pi(sqrt 3)^2`

`= (5 pi)/6 xx 3`

`= (5 pi)/2\ text(m²)`

v. `/_AOB = pi/3 + pi/3 = (2 pi)/3`

`:. /_AOB\ text{(reflex)}`	`= 2 pi – (2 pi)/3`
	`= (4 pi)/3`

`text(Area of major sector)\ AOB`

`= {(4 pi)/3}/(2 pi) xx pi xx 1^2`

`= (2 pi)/3\ text(m²)`

`:.\ text(Total area of the logo)`

`= (5 pi)/2 + (2 pi)/3 + text(Area)\ AOBC`

`= (15 pi + 4 pi)/6 + sqrt 3`

`= ((19 pi + 6 sqrt 3)/6)\ text(m²)`

Probability, 2ADV S1 2007 HSC 4b

Two ordinary dice are rolled. The score is the sum of the numbers on the top faces.

What is the probability that the score is 10? (2 marks)

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What is the probability that the score is not 10? (1 mark)

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

Show Answers Only

`1/12`
`11/12`

Show Worked Solution

`text{P (score = 10)}`

`= 3/36`

`= 1/12`

ii. `text{P (score is not ten)}`

`= 1 – text{P (score is ten)}`

`= 1 – 1/12`

`= 11/12`

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)

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

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)

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

Show Answers Only

`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)

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

Show Answers Only

`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)

Show Answers Only

`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)

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

Show Answers Only

`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 + 7/2`
`2y`	`= x + 7`
`:. x-2y + 7`	`= 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)

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

Show Answers Only

`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)

--- 6 WORK AREA LINES (style=lined) ---
Find an expression for the term independent of `x`. (2 marks)

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

Show Answers Only

`(\ ^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`
`2k`	`= 4`
`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`
`2k`	`= 18`
`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)

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

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)

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

Show Answers Only

`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 + α)`, where `0 ≤ α ≤ pi/2`. (2 marks)

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Show Answers Only

`13\ cos\ (x + 1.176…)`

Show Worked Solution

`5\ cos\ x – 12\ sin\ x`	`= A\ cos\ (x + α)`
	`= A\ cos\ x\ cos\ α – A\ sin\ x\ sin\ α`

`:. A\ cos\ α = 5,\ \ \ A\ sin\ α = 12`

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

`⇒ 13\ cos\ α`	`= 5`
`cos\ α`	`= 5/13`
`α`	`= 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)

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

Show Answers Only

`−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)

Show Answers Only

`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)

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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)

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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)

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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)

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What is the probability of at least one of Saturday and Sunday being dry? (1 mark)

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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)

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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)

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Sketch the curve, labelling the stationary points, point of inflection and `y`-intercept. (2 marks)

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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`