Band 3 – Page 35

Functions, EXT1′ F1 2011 HSC 3a

Draw a sketch of the graph

`quad y = sin\ pi/2 x` for `0 < x < 4.` (1 mark)

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Find `lim_(x -> 0) x/(sin\ pi/2 x).` (1 mark)

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Draw a sketch of the graph

`quad y = x/(sin\ pi/2 x)` for `0 < x < 4.` (2 marks)
(Do NOT calculate the coordinates of any turning points.)

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ii. `2/pi`

iii.

Show Worked Solution

ii. `lim_(x->0) x/(sin\ pi/2 x)`	`= 2/pi lim_(x->0) (pi/2 x)/(sin\ pi/2 x)`
	`= 2/pi xx 1`
	`= 2/pi`

iii.

Complex Numbers, EXT2 N2 2011 HSC 2d

Use the binomial theorem to expand `(cos theta + i sin theta)^3.` (1 mark)

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Use de Moivre’s theorem and your result from part (i) to prove that

`cos^3 theta = 1/4 cos 3 theta + 3/4 cos theta.` (3 marks)

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Hence, or otherwise, find the smallest positive solution of

`4 cos^3 theta - 3 cos theta = 1.` (2 marks)

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`cos^3 theta + 3 i cos^2 theta sin theta – 3 cos theta sin^2 theta – i sin^3 theta`
`text(Proof)\ \ text{(See Worked Solutions)}`
`theta = (2 pi)/3`

Show Worked Solution

i. `(cos theta + i sin theta)^3`

`=sum_(k=0)^3 \ ^3C_k (cos theta)^(3-k) (i sin theta)^k`

`= cos^3 theta + 3 cos^2 theta (i sin theta)+ 3 cos theta (i sin theta)^2 + (i sin theta)^3`

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

ii. `text(Using De Moivre’s Theorem)`

`(cos theta + i sin theta)^3 = cos 3 theta + i sin 3 theta`

`text(Equate real parts)`

`cos 3 theta`	`= cos^3 theta – 3 cos theta sin^2 theta`
`cos 3 theta`	`= cos^3 theta – 3 cos theta (1 – cos^2 theta)`
`cos 3 theta`	`= 4 cos^3 theta – 3 cos theta`
`4 cos^3 theta`	`=cos 3 theta+3cos theta`
`:.cos^3 theta`	`= 1/4 cos 3 theta + 3/4 cos theta\ \ \ text(… as required)`

iii. `text(If)\ \ \ 4 cos^3 theta – 3 cos theta = 1`

`=>cos 3 theta = 1\ \ \ \ text{(from part (ii))}`

`3 theta`	`= 2 k pi`
`:. theta`	`= (2 k pi)/3`

`:.\ text(Smallest positive solution occurs when)`

`theta = (2 pi)/3\ \ \ \ text{(i.e. when}\ k = 1 text{)}`

Complex Numbers, EXT2 N2 2011 HSC 2c

Find, in modulus-argument form, all solutions of `z^3 = 8.` (2 marks)

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`2text(cis)\ 0,\ 2text(cis)\ (2pi)/3,\ 2text(cis)((-2pi)/3)`

Show Worked Solution

`z^3 = 8`

`z^3 = 8 [cos (2 k pi) + i sin (2 k pi)],\ \ k=0, +-1, +-2`

`z = 2[cos ((2 k pi)/3) + i sin ((2 k pi)/3)]\ \ \ text{(De Moivre)}`

`text(When)\ \ k = 0,`

`z`	`= 2 (cos 0 + i sin 0)`
	`= 2text(cis)\ 0`

`text(When)\ \ k = 1,`

`z`	`= 2 [cos ((2 pi)/3) + i sin((2 pi)/3)]`
	`=2text(cis)\ (2pi)/3`

`text(When)\ \ k = -1,`

`z`	`= 2 [cos ((-2 pi)/3) + i sin ((-2 pi)/3)]`
	`=2 text(cis)((-2 pi)/3)`

Complex Numbers, EXT2 N2 2011 HSC 2b

On the Argand diagram, the complex numbers `0, 1 + i sqrt 3 , sqrt 3 + i` and `z` form a rhombus.

Find `z` in the form `a + ib`, where `a` and `b` are real numbers. (1 mark)

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An interior angle, `theta`, of the rhombus is marked on the diagram.

Find the value of `theta.` (2 marks)

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`(1 + sqrt 3) + i(1 + sqrt 3)`
`(5 pi)/6`

Show Worked Solution

i. `z`	`= 1 + i sqrt 3 + sqrt 3 + i`
	`= (1 + sqrt 3) + i (1 + sqrt 3)`

ii. `text(arg)\ z = tan^-1 ((1 + sqrt 3)/(1 + sqrt 3)) = pi/4`

`text(arg)\ (sqrt 3 + i) = tan^-1 (1/sqrt 3) = pi/6`

`text(Difference) = pi/4 – pi/6 = pi/12`

`=>\ text(Opposite angles of a rhombus are equal)`

`=>\ text(The diagonals of a rhombus bisect the angles)`

`:.theta`	`= pi – 2 xx pi/12`
	`= (5 pi)/6\ \ text{(angle sum of triangle)`

Calculus, EXT2 C1 2011 HSC 1e

Evaluate `int_-1^1 1/(5 - 2t + t^2) \ dt.` (3 marks)

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`pi/8`

Show Worked Solution

`int_-1^1 1/{(5 – 2t + t^2)}dt`	`= int_-1^1 1/{(4 + 1 – 2t + t^2)dt}`
	`= int_-1^1 1/(4 + (t – 1)^2)dt`
	`= 1/2[tan^-1 ((t – 1)/2)]_-1^1`
	`= 1/2 [tan^-1 0 – tan^(-1)(-1)]`
	`= 1/2 [0 – (-pi/4)]`
	`= pi/8`

Calculus, EXT2 C1 2011 HSC 1d

Find `int cos^3 theta\ d theta` (3 marks)

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`sin theta-(sin^3 theta)/3 + c`

Show Worked Solution

`int cos^3 theta\ d theta`	`= int cos^2 theta cos theta\ d theta`
	`= int (1-sin^2 theta) cos theta\ d theta`
	`= int (cos theta-sin^2 theta cos theta) d theta`
	`= sin theta-(sin^3 theta)/3 + c`

Calculus, EXT2 C1 2011 HSC 1c

Find real numbers `a, b` and `c` such that

`1/(x^2 (x - 1)) = a/x + b/x^2 + c/(x - 1).` (2 marks)

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Hence, find `int 1/(x^2 (x - 1))\ dx` (2 marks)

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`a = −1,\ \ \ b = −1,\ \ \ c = 1`
`log_e\ ((x – 1)/x) + 1/x + c`

Show Worked Solution

i. `1/(x^2 (x – 1)) = a/x + b/x^2 + c/(x – 1)`

`1 = ax (x – 1) + b (x – 1) + cx^2`

`1 = ax^2 + cx^2 – ax + bx – b`

`1=(a+c)x^2+(b-a)x-b`

`text(Equating coefficients:)`

`=>0 = a + c,\ \ \ b – a = 0,\ \ \ -b = 1`

`:.a = -1,\ \ \ b = -1,\ \ \ c = 1`

ii. `int 1/(x^2 (x – 1)) \ dx`	`= int(-1/x – 1/x^2 + 1/(x – 1))\ dx`
	`= -log_e x + 1/x + log_e (x – 1) + c`
	`= log_e\ ((x – 1)/x) + 1/x + c`

Polynomials, EXT2 2012 HSC 15b

Let `P(z) = z^4 − 2kz^3 + 2k^2z^2 − 2kz + 1`, where `k` is real.

Let `α = x + iy`, where `x` and `y` are real.

Suppose that `α` and `iα` are zeros of `P(z)`, where `bar α ≠ iα`.

Explain why `bar α` and `-i bar α` are zeros of `P(z)`. (1 mark)
Show that `P(z) = z^2(z − k)^2 + (kz − 1)^2`. (1 mark)
Hence show that if `P(z)` has a real zero then
1. `P(z) = (z^2 + 1)(z+ 1)^2` or `P(z) = (z^2 + 1)(z − 1)^2.` (2 marks)
Show that all zeros of `P(z)` have modulus `1`. (2 marks)
Show that `k = x − y`. (1 mark)
Hence show that `-sqrt2 ≤ k ≤ sqrt2`. (2 marks)

Show Answers Only

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

Show Worked Solution

(i) `P(z) = z^4 − 2kz^3 + 2k^2z^2 − 2kz +1`

`P(α) = P(iα) = 0, \ \ bar α ≠ iα.`

`text(S)text(ince)\ \ P(z)\ text(has real coefficients,)`

`=>\ text(Its zeros occur in conjugate pairs.)`

`text(i.e.)\ \ P(α) = 0\ text(and)\ P(bar α) = 0`

`bar α`	`= x − iy`
`bar(i α)`	`=bar (i(x+iy))`
	`=bar (ix-y)`
	`=-y-ix`
	`=-i(x-iy)`
	`=-i barα`

`:. -i bar α\ text(is a zero of)\ P(z)\ text(as it is the conjugate of)\ iα.`

(ii)	`H(z)`	`= z^2(z − k)^2 + (kz − 1)^2`
		`= z^2 (z^2 − 2kz + k^2) + (k^2z^2 − 2kz + 1)`
		`= z^4 − 2kz^3 + k^2z^2 + k^2z^2 − 2kz +1`
		`= z^4 − 2kz^3 + 2k^2z^2 − 2kz + 1`
		`= P(z)`

(iii) `text(If)\ z\ text(is real then)\ z^2(z − k)^2\ text(and)\ (kz −1)^2\ text(are real and)`

`text(either positive or zero.)`

♦♦♦ Mean mark part (iii) 11%.

`:.text(If)\ \ P(z) = z^2(z − k)^2 + (kz − 1)^2 = 0`

`=>(z − k)^2 = 0\ text(and)\ (kz − 1)^2 = 0`

`text(When)\ \ z = k,\ \ \ k^2 − 1 = 0\ \ text(or)\ k = ±1.`

`text(If)\ k = 1`

`P(z)`	`= z^2(z − 1)^2 + (z − 1)^2`
	`= (z^2 + 1)(z − 1)^2.`

`text(If)\ k = -1`

`P(z)`	`= z^2(z + 1)^2 + (-z − 1)^2`
	`= (z^2 + 1)(z + 1)^2`

(iv) `text(Product of the roots) = e/a=1`

♦♦♦ Mean mark part (iv) 20%.

`:. α bar α iα *(-i barα)`	`=1`
`(α bar α)^2`	`=1`
`(\|α\|)^4`	`=1`
`\|α\|`	`=1`

`:.\ text(All zeros have modulus 1.)`

(v) `text(Sum of zeroes)\ = -b/a=-(-2k)/1 = 2k`

♦♦♦ Mean mark part (v) 20%.

`:. 2k`	`=α + bar α + iα + (-i bar α )`
	`=x + iy + x − iy + (-y + ix) − i(x − iy)`
	`=2x − y + ix − ix − y`
	`=2x − 2y`
`:. k`	`=x-y`

(vi) `text(S)text(ince)\ \ |α| = 1\ \ \ text{(part (iv))}`

♦♦♦ Mean mark part (vi) 2%.

`=>x^2 + y^2`	`=1`
`text(Substitute)\ \ y=x-k\ \ \ text{(part (v))}`
`x^2 + (x − k)^2`	`=1`
`2x^2 − 2kx + k^2 − 1`	`=0`

`text(For a real solution to exist), Δ ≥ 0`

`4k^2 − 8(k^2 − 1) `	`≥ 0`
`-4k^2 + 8`	`≥ 0`
`k^2`	`≥ 2`

`:. -sqrt2 ≤ k ≤ sqrt2`

Integration, EXT2 2012 HSC 14a

Find `int(3x^2 + 8)/(x(x^2 +4))\ dx`. (3 marks)

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` 2log_e x + 1/2log_e(x^2 + 4) + c`

Show Worked Solution

`(3x^2 + 8)/(x(x^2 +4))`	`= a/x + (bx + c)/(x^2 + 4)`
`3x^2 +8`	`=ax^2 + 4a+ bx^2 + cx`
	`=(a+b)x^2+cx+4a`

`a + b = 3, \ c = 0, \ 4a = 8`

`:.a = 2, b = 1, c = 0`

`int(3x^2 + 8)/(x(x^2 +4))\ dx`	`= int2/x\ dx + int x/(x^2 + 4)\ dx`
	`= 2log_e x + 1/2log_e(x^2 + 4) + c`

Harder Ext1 Topics, EXT2 2012 HSC 13b

The diagram shows `ΔS′SP`. The point `Q` is on `S′S` so that `PQ` bisects `∠S′PS`. The point `R` is on `S′P` produced so that `PQ\ text(||)\ RS`.

Show that `PS = PR`. (1 mark)
Show that `(PS)/(QS) = (PS′)/(QS′)`. (2 marks)

Show Answers Only

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

Show Worked Solution

(i)	`∠S′PQ`	`= ∠PRS = α`	`text{(corresponding angles,}\ QP\ text(\|\|)\ SR)`
	`∠QPS`	`= ∠PSR = α`	`text{(alternate angles,}\ QP\ text(\|\|)\ SR)`
	`:. ∠PSR`	`= ∠PRS = α`

`:. ΔPSR\ text{is isosceles (two angle equal)}`

`:. PS = PR\ \ \ text{(opposite equal angles in}\ Delta PSR text{)}`

(ii) `text(Draw a line through)\ S′\ text(parallel to)\ QP`

`:.(PS′)/(PR)`	`= (QS′)/(QS)\ \ \ text{(parallel lines preserve ratios)}`
`(PS′)/(QS′)`	`= (PR)/(QS)`
`text(S)text(ince)\ \ PS = PR\ \ \ text{(from part (i))}`
`(PS′)/(QS′)`	`= (PS)/(QS)\ \ \ text(… as required)`

Mechanics, EXT2 M1 2012 HSC 13a

An object on the surface of a liquid is released at time `t = 0` and immediately sinks. Let `x` be its displacement in metres in a downward direction from the surface at time `t` seconds.

The equation of motion is given by

`(dv)/(dt) = 10 − (v^2)/40`,

where `v` is the velocity of the object.

Show that `v = (20(e^t − 1))/(e^t + 1)`. (4 marks)

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Use `(dv)/(dt) = v (dv)/(dx)` to show that

`x = 20\ log_e(400/(400 − v^2))` (2 marks)

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How far does the object sink in the first 4 seconds? (2 marks)

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`text(See Worked Solutions.)`
`text(See Worked Solutions.)`
`40log_e((e^4 + 1)/(2e^2))\ text(m)`

Show Worked Solution

i.	`(dv)/(dt)`	`= 10 − (v^2)/40`
	`(dv)/(dt)`	`= (400 − v^2)/40`
	`dt`	`= 40/(400 −v^2)\ dv`
	`int dt`	`=int 40/(400 −v^2)\ dv`
	`t`	`= int (1/(20 + v) + 1/(20 − v))\ dv`
		`= log_e(20 + v) − log_e(20 − v) + c`

`text(When)\ \ t = 0, v = 0\ \ => \ c = 0`

`t=`	` log_e((20 + v)/(20 − v))`
`e^t=`	` (20 + v)/(20 − v)`
`20e^t-ve^t=`	` 20 + v`
`v+ ve^t=`	` 20e^t − 20`
`v(1+e^t)=`	`20(e^t − 1)`
`v=`	` (20(e^t − 1))/(e^t + 1)\ \ \ \ text(… as required)`

ii.	`v (dv)/(dx)`	`= 10 − (v^2)/40`
	`(40v\ dv)/(400 − v^2)`	`= dx`

`int dx`	`= int (40v)/(400 − v^2)\ dv`
`x`	`= -20log_e(400 − v^2) + c`

`text(When)\ \ x = 0, v = 0\ \ \ => c = 20log_e 400`

`:.x`	`= 20log_e400 − 20log_e(400 − v^2)`
	`= 20log_e((400)/(400 − v^2))\ \ \ \ text(… as required)`

iii. `text(When)\ \ t = 4,\ v = (20(e^4 − 1))/(e^4 + 1)`

`x`	`= 20log_e[400/(400 −((20(e^4 − 1))/(e^4 + 1))^2)]`
	`= 20log_e[((e^4 + 1)^2)/((e^4 + 1)^2 − (e^4 − 1)^2)]`
	`= 20log_e(((e^4 + 1)^2)/((e^4+1 + e^4 − 1)(e^4 + 1 − e^4 + 1)))`
	`= 20log_e(((e^4 + 1)^2)/(4e^4))`
	`= 40log_e\ (e^4 + 1)/(2e^2)\ \ text(metres)`

Calculus, EXT2 C1 2012 HSC 12c

For every integer `n ≥ 0` let `I_n = int_1^(e^2)(log_e x)^n\ dx`.

Show that for `n ≥ 1,`

`I_n = e^2 2^n − nI_(n − 1)`. (3 marks)

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`text{Proof (See Worked Solutions)}`

Show Worked Solution

`I_n = int_1^(e^2)(log_e x)^n\ dx`

`u`	`= (log_e x)^n,`	`v′`	`= 1`
`u′`	`= n (log_e x)^(n − 1) xx 1/x`	`v`	`= x`

`I_n`

`=int_1^(e^2) 1*(ln x)^n\ dx`

`= [x (log_e x)^n]_1^(e^2) − int_1^(e^2)n/x(log_e x)^(n − 1) xx x\ dx`

`= [e^2(log_e e^2)^n − 1*ln\ 1] − n int_1^(e^2)(log_e x)^(n − 1)\ dx`

`= [e^2(2 log_ee)^n-0 ]− nI_(n − 1)`

`= e^2 2^n − nI_(n − 1)`

The diagram shows the ellipse `(x^2)/(a^2) + (y^2)/(b^2) = 1` with `a > b`. The ellipse has focus `S` and eccentricity `e`. The tangent to the ellipse at `P(x_0, y_0)` meets the `x`-axis at `T`. The normal at `P` meets the `x`-axis at `N`.

Show that the tangent to the ellipse at `P` is given by the equation
1. `y − y_0 = -(b^2x_0)/(a^2y_0)(x − x_0)`. (2 marks)
Show that the `x`-coordinate of `N` is `x_0e^2`. (2 marks)
Show that `ON xx OT = OS^2` (2 marks)

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(i) `text(See Worked Solutions.)`

(ii) `text(See Worked Solutions.)`

(iii) `text(See Worked Solutions.)`

Show Worked Solution

(i) `(x^2)/(a^2) + (y^2)/(b^2) = 1`

`(2x)/(a^2) + (2y)/(b^2) *(dy)/(dx)`	`=0`
`(dy)/(dx)`	`=-(2x)/(a^2) xx (b^2)/(2y)`
	`= -(b^2x)/(a^2y)`

`text(At)\ P(x_0, y_0),\ \ m_(tan)= -(b^2x_0)/(a^2y_0)`

`:.text(Equation of tangent at)\ P\ text(is)`

`y − y_0 = -(b^2x_0)/(a^2y_0)(x − x_0)`

(ii) `text(At)\ P(x_0, y_0),\ \ m_(norm) = (a^2y_0)/(b^2x_0)`

`:.text(Equation of normal at)\ P\ text(is)`

`y − y_0 = (a^2y_0)/(b^2x_0)(x − x_0)`

`N\ \ text(occurs when)\ \ y=0`

`0-y_0=`	` (a^2y_0)/(b^2x_0)(x − x_0)`
`-b^2x_0y_0=`	` a^2y_0x − a^2x_0y_0`
`a^2y_0x=`	` a^2x_0y_0 − b^2x_0y_0`
`a^2x=`	` x_0(a^2 − b^2)`
`x=`	` (x_0(a^2 − b^2))/(a^2)\ \ \ \ \ text{(using}\ \ a^2e^2=a^2-b^2 text{)}`
`=`	`x_0e^2`

(iii) `ON = x_0e^2`

`OS=ae\ \ \ \ text{(given}\ \ S(ae,0) text{)}`

`T\ text(is the)\ x text(-axis intercept of the tangent)\ PT`

`0-y_0`	`= -(b^2x_0)/(a^2y_0)(x − x_0)`
`a^2y_0^2`	`=b^2x_0x-b^2x_0^2`
`b^2x_0x`	`=b^2x_0^2+a^2y_0^2`
`x`	`=(b^2x_0^2+a^2y_0^2)/(b^2x_0)`

`text(S)text(ince)\ P(x_0,y_0)\ text(lies on the ellipse,)`

`=> (x_0^2)/(a^2) + (y_0^2)/(b^2)`	` = 1`
`b^2 x_0^2+a^2y_0^2`	`=a^2b^2`

`:.x=OT`

`=(a^2b^2)/(b^2x_0)=a^2/x_0`

`:.ON xx OT`	`= x_0e^2 xx a^2/x_0`
	`=a^2e^2`
	`=OS^2`

Calculus, EXT2 C1 2012 HSC 12a

Using the substitution `t = tan\ theta/2`, or otherwise, find `int(d theta)/(1 − cos\ theta)`. (3 marks)

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`-cot\ theta/2 + c`

Show Worked Solution

`t = tan\ theta/2`

`dt=1/2 sec^2 (theta/2)\ d theta,\ \ \ d theta=(2\ dt)/sec^2 (theta/2)=2/(1+t^2)\ dt`

`cos\ theta = (1 − t^2)/(1 + t^2)`

`int(d theta)/(1 − cos\ theta)`	`= int1/(1 − ((1 − t^2)/(1 + t^2))) xx 2/(1 + t^2)\ dt`
	`= int 2/(1 + t^2 − (1 − t^2))`
	`= int(dt)/(t^2)`
	`= -1/t + c`
	`= -1/(tan\ theta/2) + c`
	`= -cot\ theta/2 + c`

Calculus, EXT2 C1 2012 HSC 11e

Evaluate `int_0^1 (e^(2x))/(e^(2x) + 1)\ dx`. (3 marks)

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`1/2log_e((e^2 + 1)/2)`

Show Worked Solution

`int_0^1 (e^(2x))/(e^(2x) + 1)\ dx`	`= 1/2[log_e(e^(2x) + 1)]_0^1`
	`= 1/2[log_e(e^2 + 1) − log_e2]`
	`= 1/2log_e((e^2 + 1)/2)`

Complex Numbers, EXT2 N2 2012 HSC 11b

Shade the region on the Argand diagram where the two inequalities

`|\ z + 2\ | ≥ 2` and `|\ z − i\ | ≤ 1`

both hold. (2 marks)

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Show Worked Solution

`|\ z + 2\ | ≥ 2 and |\ z − i\ | ≤ 1`

`(x + 2)^2 + y^2`	`≥ 4`
`x^2 + (y − 1)^2`	`≤ 1`

Mechanics, EXT2 M1 2013 HSC 15d

A ball of mass `m` is projected vertically into the air from the ground with initial velocity `u`. After reaching the maximum height `H` it falls back to the ground. While in the air, the ball experiences a resistive force `kv^2`, where `v` is the velocity of the ball and `k` is a constant.

The equation of motion when the ball falls can be written as

`m dot v = mg-kv^2.` (Do NOT prove this.)

Show that the terminal velocity `v_T` of the ball when it falls is
`sqrt ((mg)/k).` (1 mark)

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Show that when the ball goes up, the maximum height `H` is
`H = (v_T^2)/(2g) ln (1 + u^2/(v_T^2)).` (3 marks)

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When the ball falls from height `H` it hits the ground with velocity `w`.
Show that `1/w^2 = 1/u^2 + 1/(v_T^2).` (2 marks)

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

Show Worked Solution

i. `m dot v = mg-kv^2`

`t = 0,\ \ \ v = 0,\ \ \ x = 0\ \ \ text(Ball falling)`

`text{For terminal velocity}\ \(v_T),\ \ \ dot v = 0`

`v_T^2`	`= (mg)/k`
`:.v_T`	`= sqrt ((mg)/k)`

ii. `text(When the ball rises),\ \ m dot v = -mg-kv^2`

♦ Mean mark 43%.
 MARKER’S COMMENT: More than half of students incorrectly wrote the equation to solve as `m dot v=mg-kv^2!`

`text(Using)\ \ dot v`	`= v (dv)/(dx)`
`mv (dv)/(dx)`	`= -mg-kv^2`
`dx`	`=(-mv)/(mg + kv^2) dv`

`int_0^H dx`	`= -int_u^0 (mv)/(mg + kv^2) dv`
`[x]_0^H`	`= -m/(2k) [log_e (mg + kv^2)]_u^0`
`H`	`= -m/(2k) (log_e (mg)-log_e (mg + ku^2))`
	`= m/(2k) log_e ((mg + ku^2)/(mg))`
	`= m/(2k) log_e (1 + (ku^2)/(mg))\ \ \ \ text{(using}\ \ k = (mg)/v_T^2 text{)}`
`:.H`	`=(v_T^2)/(2g) log_e (1 + u^2/(v_T^2))`

iii. `text(When the ball falls),\ \ m dot v = mg-kv^2`

♦♦ Mean mark 14%.

`mv (dv)/(dx)`	`= mg-kv^2`
`dx`	`=(mv)/(mg-kv^2) dv`

`int_0^H dx`	`= int_0^w (mv)/(mg-kv^2)dv`
`[x]_0^H`	` =-m/(2k)[log_e (mg-kv^2)]_0^w`
`H`	`= -m/(2k) (log_e (mg-kw^2)-log_e (mg))`
	`= -m/(2k) log_e ((mg-kw^2)/(mg))`
	`= -m/(2k) log_e (1-(kw^2)/(mg))\ \ \ \ text{(using}\ \ k = (mg)/v_T^2 text{)}`
`H`	`=-(v_T^2)/(2g) log_e (1-w^2/(v_T^2))`
	`=-(v_T^2)/(2g) log_e ((v_T^2-w^2)/(v_T^2))`
	`=(v_T^2)/(2g) log_e ((v_T^2)/(v_T^2-w^2))`

`text{Using part (ii):}`

`(v_T^2)/(2g) log_e ((v_T^2)/(v_T^2-w^2))`	`=(v_T^2)/(2g) log_e (1 + u^2/(v_T^2))`
`(v_T^2)/(v_T^2-w^2)`	`=(v_T^2 + u^2)/(v_T^2)`
`(v_T^2 + u^2)(v_T^2-w^2)`	`=v_T^4`
`v_T^4-v_T^2 w^2+v_T^2 u^2-w^2 u^2`	`=v_T^4`
`v_T^2 w^2+w^2 u^2`	`=v_T^2 u^2`
`(v_T^2 w^2)/(v_T^2w^2u^2)+(w^2 u^2)/(v_T^2w^2u^2)`	`=(v_T^2 u^2)/(v_T^2w^2u^2)`
`:.1/u^2 + 1/(v_T^2)`	`=1/w^2`

Harder Ext1 Topics, EXT2 2013 HSC 15c

Eight cars participate in a competition that lasts for four days. The probability that a car completes a day is `0.7`. Cars that do not complete a day are eliminated.

Find the probability that a car completes all four days of the competition. (1 mark)
Find an expression for the probability that at least three cars complete all four days of the competition. (2 marks)

Show Answers Only

`0.2401`
`1-(0.76^8 + \ ^8C_1 * 0.76^7 * 0.24 + \ ^8C_2 * 0.76^6 * 0.24^2)`

Show Worked Solution

(i) `P text{(Completes 1 day)} = 0.7`

`:.P text{(Completes 4 days)}`	`= 0.7^4`
	`~~ 0.2401`

(ii) `P text{(Does not complete all 4 days)`	`= 1 – 0.7^4`
	`= 0.7599`
	`~~ 0.76`

♦♦ Mean mark 26%. 
STRATEGY: The use of complementary events reduced the required calculations and subsequent errors in this part.

`text(Let)\ \ C =\ text(number of cars that complete 4 days)`

`:. P text{(At least 3 cars complete all 4 days)}`

`= 1 – [P (C=0) + P (C = 1) + P (C = 2)]`

`= 1 – (\ ^8C_0 *0.76^8 + \ ^8C_1 * 0.76^7 * 0.24 + \ ^8C_2 * 0.76^6 * 0.24^2)`

`=1-(0.76^8 + \ ^8C_1 * 0.76^7 * 0.24 + \ ^8C_2 * 0.76^6 * 0.24^2)`

Functions, EXT1′ F2 2013 HSC 15b

The polynomial `P(x) = ax^4 + bx^3 + cx^2 + e` has remainder `-3` when divided by `x - 1`. The polynomial has a double root at `x = -1.`

Show that `4a + 2c = -9/2.` (2 marks)

--- 7 WORK AREA LINES (style=lined) ---
Hence, or otherwise, find the slope of the tangent to the graph `y = P(x)` when `x = 1.` (1 mark)

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

Show Answers Only

`text(Proof)\ \ text{(See Worked Solutions)}`
`-9`

Show Worked Solution

i.	`P(x)`	`= ax^4 + bx^3 + cx^2 + e`
	`P prime (x)`	`=4ax^3 + 3bx^2 + 2cx`

`P(1)=-3`
`a+b+c+e`	`=-3\ \ \ \ …\ (1)`
`P(-1)=0`
`a-b+c+e`	`=0\ \ \ \ …\ (2)`
`P′(-1)=0`
`-4a+3b-2c`	`=0\ \ \ \ …\ (3)`

`(1)-(2)`

`2b`	`=-3`
`b`	`=-3/2`

`text(Substitute into)\ \ (3)`

`:.4a+2c`	`=3b`
	`=-9/2\ \ \ \ text(… as required)`

ii. `P prime (1)`	`= 4a + 3b + 2c`
	`= -9/2 – 9/2`
	`=-9`

Functions, EXT1′ F1 2013 HSC 13b

The diagram shows the graph of a function `f(x).`

Sketch the following curves on separate half-page diagrams.

`y^2 = f(x).` (2 marks)

--- 8 WORK AREA LINES (style=lined) ---
`y = 1/(1 - f(x)).` (3 marks)

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

Show Answers Only

`text(See Worked Solutions)`
`text(See Worked Solutions)`

Show Worked Solution

i. `y^2 = f(x)`

`text(Only exist for)\ \ f(x) >= 0`

`y^2 = 1,\ \ \ y = +- 1`

ii. `y = 1/(1 – f(x))`

MARKER’S COMMENT: Correct working sketches such as `y=-f(x)` and `y=1-f(x)` meant that students could obtain some marks, even if their final sketch was wrong. Note this important advice.

`f(x) = 1,\ \ \ y\ text(undefined.)`

`f(x) > 1,\ \ \ y < 0`

`f(x) <= 0, \ \ \ y <= 1`

Conics, EXT2 2013 HSC 12d

The points `P (cp, c/p)` and `Q (cq, c/q)`, where `|\ p\ | ≠ |\ q\ |`, lie on the rectangular hyperbola with equation `xy = c^2.`

The tangent to the hyperbola at `P` intersects the `x`-axis at `A` and the `y`-axis at `B`. Similarly, the tangent to the hyperbola at `Q` intersects the `x`-axis at `C` and the `y`- axis at `D`.

Show that the equation of the tangent at `P` is `x + p^2 y = 2cp.` (2 marks)
Show that `A, B and O` are on a circle with centre `P.` (2 marks)
Prove that `BC` is parallel to `PQ.` (1 mark)

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) `P (cp, c/p),\ \ Q (cq, c/q),\ \ xy = c^2,\ \ |\ p\ | ≠ |\ q\ |`

`xy = c^2`

`y + x*(dy)/(dx) = 0,\ \ (dy)/(dx) = -y/x`

`text(At)\ \ (cp, c/p)`

`(dy)/(dx) = (-c/p)/(cp) = -1/p^2`

`text(Equation of tangent at)\ \ P`

`y – c/p`	`= -1/p^2 (x – cp)`
`p^2 y – cp`	`= -x + cp`
`:. x + p^2 y`	`= 2cp\ \ \ text(… as required)`

(ii) `text(Solution 1)`

`text(At)\ \ A, y = 0,\ \ \ x = 2cp`

`text(At)\ \ B, x = 0,\ \ \ y = (2c)/p`

`OP`	`= sqrt (c^2 p^2 + c^2/p^2) = c/p sqrt(p^4 + 1)`
`PA`	`= sqrt {(2cp – cp)^2 + c^2/p^2} = c/p sqrt (p^4 + 1)`
`PB`	`= sqrt{c^2 p^2 + ((2c)/p – c/p)^2} = c/p sqrt (p^4 + 1)`

`OP = PA = PB`

`:. A, B and O\ \ text(are on the circle centre)\ \ P.`

`text(Alternate Solution)`

`text(Mid-point of)\ \ AB`

`=((2cp+0)/2,\ (0+(2c)/p)/2)`

`=(cp,\ c/p)`

`=>P\ \ text(is the midpoint of)\ \ AB`

`text(S)text(ince)\ \ ∠AOB=90^@`

`:. A, B and O\ \ text(are on the circle centre)\ \ P.`

(iii) `text(Equation of tangent at)\ \ Q\ \ \ text{(using part (i))}`

`x + q^2y = 2cq`

`=> C\ \ text(is)\ \ (2cq, 0)`

`m_(BC)`	`= {(2c)/p – 0}/(0 – 2cq)`
	`= – 1/(pq)`
`m_(PQ)`	`= (c/p – c/q)/(cp – cq)`
	`=((cq-cp)/(pq))/(c(p-q))`
	`= {-c(p – q)}/{cpq (p – q)}`
	`= -1/(pq)`

`m_(BC) = m_(PQ)`

`:. BC\ text(||)\ PQ`

Graphs, EXT2 2013 HSC 12b

The equation `log_e y - log_e (1000 - y) = x/50 - log_e 3` implicitly defines `y` as a function of `x`.

Show that `y` satisfies the differential equation

`(dy)/(dx) = y/50 (1 - y/1000).` (2 marks)

Show Answers Only

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

Show Worked Solution

`log_e y – log_e (1000 – y) = x/50 – log_e 3`

`1/y* (dy)/(dx) – (-1)/((1000 – y))*(dy)/(dx)`	`= 1/50 – 0`
`(dy)/(dx)(1/y + 1/(1000 – y))`	`= 1/50`
`(dy)/(dx) ((1000 – y + y)/(y (1000 – y)))`	`= 1/50`
`(dy)/(dx) (1000/(y (1000 – y)))`	`= 1/50`
`(dy)/(dx)`	`= (y (1000 – y))/(50 xx 1000)`
`(dy)/(dx)`	`= y/50 (1 – y/1000)`

Calculus, EXT2 C1 2013 HSC 12a

Using the substitution ` t = tan\ x/2`, or otherwise, evaluate

`int_0^(pi/2) 1/(4 + 5 cos x)\ dx.` (4 marks)

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

Show Answers Only

`1/3 ln 2`

Show Worked Solution

`t = tan\ x/2`

`=>cos x = (1 – t^2)/(1 + t^2)\ ,\ \ \ dx = (2\ dt)/(1 + t^2)`

`text(When)\ \ x = 0\ ,\ t = 0\ ;\ \ x = pi/2\ ,\ t = 1`

`int_0^(pi/2) (dx)/(4 + 5 cos x)`	`= int_0^1 1/{4 + (5(1 – t^2))/(1 + t^2)} xx (2\ dt)/(1 + t^2)`
	`= int_0^1 (2\ dt)/(4 + 4t^2 + 5 – 5t^2)`
	`= int_0^1 (2\ dt)/(9 – t^2)`
	`=2 int_0^1 1/((3-t)(3+t))`
	`= 1/3 int_0^1 (1/(3 – t) + 1/(3 + t))\ dt`
	`= 1/3 [-ln (3 – t) + ln (3 + t)]_0^1`
	`= 1/3 [(-ln 2 + ln 4) – (-ln 3 + ln 3)]`
	`= 1/3 ln2`

Complex Numbers, EXT2 N1 2013 HSC 11c

Factorise `z^2 + 4iz + 5.` (2 marks)

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

Show Answers Only

`(z – i) (z + 5i)`

Show Worked Solution

`text(Solution 1)`

`alpha + beta = 4i,\ \ \ alpha beta = 5 = -5i^2`

`:. z^2 + 4iz + 5 = (z – i) (z + 5i)`

`(alpha + beta = 4i,\ \ \ alpha beta = 5 = -5i^2)`

MARKER’S COMMENT: Best practice includes stating the general quadratic formula before substituting in values.

`text(Solution 2)`

`z`	`=(-b+- sqrt(b^2 – 4ac))/(2a)`
	`=(-4i +- sqrt((4i)^2-4 xx 5))/2`
	`=(-4i+-sqrt(-16-20))/2`
	`=(-4i +- 6i)/2`
	`=i\ \ text(or)\ \ -5i`

`:.z^2 + 4iz + 5 = (z – i) (z + 5i)`

Harder Ext1 Topics, EXT2 2014 HSC 16a

The diagram shows two circles `C_1` and `C_2` . The point `P` is one of their points of intersection. The tangent to `C_2` at `P` meets `C_1` at `Q`, and the tangent to `C_1` at `P` meets `C_2` at `R`.

The points `A` and `D` are chosen on `C_1` so that `AD` is a diameter of `C_1` and parallel to `PQ`. Likewise, points `B` and `C` are chosen on `C_2` so that `BC` is a diameter of `C_2` and parallel to `PR`.

The points `X` and `Y` lie on the tangents `PR` and `PQ`, respectively, as shown in the diagram.

Copy or trace the diagram into your writing booklet.

Show that `∠APX = ∠DPQ`. (2 marks)
Show that `A`, `P` and `C` are collinear. (3 marks)
Show that `ABCD` is a cyclic quadrilateral. (1 mark)

Show Answers Only

`text(See Worked Solutions.)`
`text(See Worked Solutions.)`
`text(See Worked Solutions.)`

Show Worked Solution

(i)

`∠APX`	`= ∠ADP\ \ \ text{(angle in alternate segment)`βαγ
`∠ADP`	`= ∠DPQ\ \ \ text{(alternate angles,}\ AD\ text(\|\|)\ PQ)`
`:.∠APX`	`= ∠DPQ`

(ii) `text(Join)\ PC\ text(and)\ PB.`

♦♦ Mean mark 31%.
MARKER’S COMMENT: Be vigilant that your reasoning does not implicitly assume `APC` or `DPB` are straight lines.

`text{Let}\ \ ∠APX = ∠DPQ=α\ \ \ text{(from part (i))}`

`text{Similarly,}\ \ ∠YPB = ∠CPR=β`

`∠XPY = ∠RPQ=γ\ \ \ text{(vertically opposite angles.)}`

`∠APD`	`= 90^@\ \ \ text{(angle in a semicircle in}\ C_1)`
`∠CPB`	`= 90^@\ \ \ text{(angle in a semicircle in}\ C_2)`

`90^@ + 90^@ + 2(α+β+γ)`	`= 360^@ \ \ text{(sum of angles at a point}\ Ptext{)}`
`:. (α+β+γ)`	`= 90^@`

`∠APC`

`=90+(α+β+γ)=180^@`

`:.\ A, P\ text(and)\ C\ text(are collinear.)`

♦ Mean mark 44%.

(iii) `:.∠APX = ∠CPR\ text{(vertically opposite angles,}\ APC\ text{is a straight line)}`

`:.α=β\ \ =>∠BCA = ∠BDA`

`text(S)text(ince chord)\ AB\ text(subtends a pair of equal angles at)\ C and D`

`=>ABCD\ text(is a cyclic quadrilateral.)`

Functions, EXT1′ F2 2014 HSC 14a

Let `P(x) =x^5-10x^2 +15x-6`.

Show that `x = 1` is a root of `P(x)` of multiplicity three. (2 marks)

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

Show Answers Only

`text(See Worked Solutions.)`

Show Worked Solution

`P(x) =x^5-10x^2 +15x-6`

`P(1) = 1-10 + 15-6 = 0`

`P^{′}(x)`	`= 5x^4-20x + 15`
`P^{′}(1)`	`= 5-20 + 15 = 0`

`P^{″}(x)`	`= 20x^3-20`
`P^{″}(1)`	`= 20-20 = 0`

`P^{‴}(x)`	`= 60x^2`
`P^{‴}(1)`	`= 60 ≠ 0`

`:.x = 1\ text(is a root of)\ P(x),text(of multiplicity 3.)`

Conics, EXT2 2014 HSC 13c

The point `S(ae, 0)` is the focus of the hyperbola `(x^2)/(a^2)-(y^2)/(b^2) = 1` on the positive `x`-axis.

The points `P(at, bt)` and `Q(a/t, −b/t)` lie on the asymptotes of the hyperbola, where `t > 0`.

The point `M((a(t^2 + 1))/(2t), (b(t^2 – 1))/(2t))` is the midpoint of `PQ`.

Show that `M` lies on the hyperbola. (1 mark)
Prove that the line through `P` and `Q` is a tangent to the hyperbola at `M`. (3 marks)
Show that `OP xx OQ = OS^2`. (2 marks)
If `P` and `S` have the same `x`-coordinate, show that `MS` is parallel to one of the asymptotes of the hyperbola. (2 marks)

Show Answers Only

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

Show Worked Solution

(i) `text(Substitute)\ \ \ M((a(t^2 + 1))/(2t), (b(t^2 − 1))/(2t))\ \ text(into)`

`(x^2)/(a^2)-(y^2)/(b^2) = 1`

`text(LHS)`	`=1/(a^2) xx (a^2(t^2 + 1)^2)/(4t^2) − 1/(b^2) xx (b^2(t^2 − 1)^2)/(4t^2)`
	`=1/(4t^2)[t^4 + 2t^2 + 1 – (t^4 − 2t^2 + 1)]`
	`=1/(4t^2) xx 4t^2`
	`=1`
	`=\ text(RHS)`

`:.M\ text(lies on the hyperbola).`

(ii) `P(at, bt), \ \ Q(a/t, −b/t)`

`m_(PQ)`	`= (bt − (-b/t))/(at − a/t) xx t/t`
	`=(bt^2+b)/(at^2-a)`
	`= b/a ((t^2 + 1)/(t^2 − 1))`

`text(Differentiate the hyperbola)`

`(2x)/(a^2) − (2y*dy/dx)/(b^2)`	`= 0`
`(dy)/(dx)`	`= (b^2x)/(a^2y)`

`text(At)\ \ M((a(t^2 + 1))/(2t), (b(t^2 − 1))/(2t)),`

`(dy)/(dx)`	`= (b^2)/(a^2) xx (a(t^2 + 1))/(2t) xx (2t)/(b(t^2 – 1))`
	`= b/a ((t^2 + 1)/(t^2 -1))`

`:. m_(PQ)=m_(at\ M)`

`:.text(S)text(ince the tangent to the hyperbola has the same gradient as)\ \ PQ`

`text(and both pass through)\ M,text(they are the same line.)`

(iii)	`OP`	`= sqrt(a^2t^2 + b^2t^2) = t sqrt(a^2 + b^2)`
	`OQ`	`= sqrt((a^2)/(t^2) + (b^2)/(t^2)) = 1/tsqrt(a^2 + b^2)`

`OP xx OQ`	`= tsqrt(a^2 + b^2) xx 1/tsqrt(a^2 + b^2)`
	`= a^2 + b^2`
	`=a^2+a^2(e^2 − 1)`
	`= a^2e^2`
	`=OS^2\ \ \ …\ text(as required)`

(iv) `text(Given)\ \ P and S\ \ text(have the same)\ x text(-coordinate)`

`at = ae\ \ => t = e`

`S(ae,0),\ \ M((a(e^2 + 1))/(2e), (b(e^2 − 1))/(2e))`

`m_(MS)`	`= ((b(e^2 − 1))/(2e) − 0)/((a(e^2 + 1))/(2e) − ae)`
	`= (b(e^2 − 1))/(ae^2 + a − 2ae^2)`
	`= (b(e^2 − 1))/(a − ae^2)`
	`= (b(e^2 − 1))/(-a(e^2 − 1))`
	`= – b/a`

`:.text(S)text(ince the gradients of the hyperbola asymptotes are)\ ±b/a.`

`=>MS\ text(is parallel to one … as required)`

Graphs, EXT2 2014 HSC 12c

The point `P(x_0, y_0)` lies on the curves `x^2 − y^2 = 5` and `xy = 6`. Prove that the tangents to these curves at `P` are perpendicular to one another. (3 marks)

Show Answers Only

`text{Proof (See Worked Solutions)}`

Show Worked Solution

`text(Differentiating)\ \ \ x^2 − y^2 = 5`

`2x − 2y*(dy)/(dx)`	`= 0`
`(dy)/(dx)`	`=x/y`
`text(At)\ P(x_0, y_0),\ \ (dy)/(dx)`	`= x_0/y_0`

`text(Differentiating)\ \ \ xy = 6`

`y + x*(dy)/(dx)`	`= 0`
`(dy)/(dx)`	`= -y/x`
`text(At)\ P(x_0, y_0),\ \ (dy)/(dx)`	`= -y_0/x_0`

`text(S)text(ince)\ \ \ x_0/y_0 xx (-y_0/x_0) = -1`

`:.\ text(The tangents to the curves at)\ P(x_0, y_0)\ text(are)`

`text(perpendicular to one another.)`

Functions, EXT1′ F1 2014 HSC 12a

The diagram shows the graph of a function `f(x)`.

Draw a separate half-page graph for each of the following, showing all asymptotes and intercepts.

`y = f(|\ x\ |)` (2 marks)

--- 10 WORK AREA LINES (style=lined) ---
`y = 1/(f(x))` (2 marks)

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

Show Answers Only

`text(See Worked Solutions.)`
`text(See Worked Solutions.)`

Show Worked Solution

ii.

Volumes, EXT2 2014 HSC 11e

The region enclosed by the curve `x = y(6 − y)` and the `y`-axis is rotated about the `x`-axis to form a solid.

Using the method of cylindrical shells, or otherwise, find the volume of the solid. (3 marks)

Show Answers Only

`216pi\ \ text(u³)`

Show Worked Solution

`x = y(6 − y)\ \ =>text(Rotate about)\ xtext(-axis.)`

MARKER’S COMMENT: The best responses included a sketch that clearly showed the axis of rotation and an understanding of the use of cylindrical shells, as per the worked solution.

`δV`	`=2 pi r h delta y`
	`= 2 pi y*y (6-y) delta y`
	`=2 pi(6y^2 – y^3) delta y`

`:.V`	`= 2pi int_0^6(6y^2 − y^3)\ dy`
	`= 2pi[2y^3 −(y^4)/4]_0^6`
	`= 2pi[(432 − 324) − 0]`
	`= 216pi\ \ text(u³)`

Calculus, EXT2 C1 2014 HSC 11b

Evaluate `int_0^(1/2)(3x-1)\ cos\ (pix)\ dx`. (3 marks)

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

Show Answers Only

`(pi-6)/(2pi²)`

Show Worked Solution

`text(Integrating by parts:)`

`u`	`=3x-1`	`u^{′}`	`=3`
`v^{′}`	`=cos(pi x)`	`v`	`=1/pi sin(pi x)`

`int_0^(1/2)(3x-1)\ cos\ (pix)\ dx`

`= [(3x-1)(sin\ (pix))/pi]_0^(1/2)-int_0^(1/2) 3 xx (sin\ (pix))/pi\ dx`

`= (1/(2pi)\ sin\ pi/2-0) − 3/pi[(-cos\ (pix))/pi]_0^(1/2)`

`= 1/(2pi) + 3/(pi^2)(cos\ pi/2-cos\ 0)`

`= 1/(2pi) + 3/(pi^2)(0-1)`

`= 1/(2pi)-3/(pi^2)`

Complex Numbers, EXT2 N1 2014 HSC 11a

Consider the complex numbers `z = -2- 2i` and `w = 3 + i`.

Express `z + w` in modulus–argument form. (2 marks)

--- 4 WORK AREA LINES (style=lined) ---
Express `z/w` in the form `x + iy`, where `x` and `y` are real numbers. (2 marks)

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

Show Answers Only

`sqrt2\ text(cis)(-pi/4)`
`−4/5 −2/5i`

Show Worked Solution

i.	`z + w`	`= −2 − 2i + 3 + i`
		`= 1 − i`
	`\|\ z+w\ \|`	`= sqrt2`
	`text(arg)\ (z+w)`	`=- pi/4`
	`:. z+w`	`= sqrt2\ text(cis)(-pi/4)`

ii.	`z/w`	`= (−2 − 2i)/(3 + i)`
		`= ((−2 − 2i)(3 − i))/((3 + i)(3 − i))`
		`= (−6 − 6i + 2i − 2)/(9 + 1)`
		`= −8/10 −4/10i`
		`= −4/5 − 2/5i`

Statistics, EXT1 S1 2007 HSC 4a

In a large city, 10% of the population has green eyes.

What is the probability that two randomly chosen people both have green eyes? (1 mark)

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What is the probability that exactly two of a group of 20 randomly chosen people have green eyes? Give your answer correct to three decimal places. (1 mark)

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What is the probability that more than two of a group of 20 randomly chosen people have green eyes? Give your answer correct to two decimal places. (2 marks)

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Show Answers Only

`0.01`
`0.285\ \ \ text{(to 3 d.p.)}`
`0.32\ \ \ text{(to 2 d.p.)}`

Show Worked Solution

i.	`P(text(G))`	`= 0.1`
	`P(text(GG))`	`= 0.1 xx 0.1`
		`= 0.01`

ii. `P(text(not G)) = 1 − 0.1 = 0.9`

`:. P(text(2 out of 20 have green eyes))`

`= \ ^(20)C_2 · (0.1)^2 · (0.9)^(18)`

`= 0.2851…`

`= 0.285\ \ \ text{(to 3 d.p.)}`

iii. `P(text(more than 2 have green eyes))`

`= 1 − [P(0) + P(1) + P(2)]`

`= 1 − [0.9^20 + \ ^20C_1(0.1)^1(0.9)^19 + \ ^20C_2(0.1)^2(0.9)^(18)]`

`= 1 − [0.1215… + 0.2701… + 0.2851…]`

`= 1 − 0.6769…`

`= 0.3230`

`= 0.32\ \ \ text{(to 2 d.p.)}`

Mechanics, EXT2* M1 2007 HSC 2d

A skydiver jumps from a hot air balloon which is 2000 metres above the ground. The velocity, `v` metres per second, at which she is falling `t` seconds after jumping is given by `v =50(1 - e^(-0.2t))`.

Find her acceleration ten seconds after she jumps. Give your answer correct to one decimal place. (2 marks)

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Find the distance that she has fallen in the first ten seconds. Give your answer correct to the nearest metre. (2 marks)

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Show Answers Only

`1.4\ text(ms)^(−2)\ \ text{(to 1 d.p.)}`
`284\ text{m (nearest m)}`

Show Worked Solution

i.	`v`	`= 50(1 − e^(-0.2t))`
		`=50-50e^(-0.2t)`
	`ddot x`	`= (dv)/(dt)`
		`= −0.2 xx 50 xx −e^(−0.2t)`
		`= 10 e^(−0.2t)`

`text(When)\ \ t = 10:`

`ddot x`	`= 10 e^(−0.2 xx 10)`
	`= 10 e ^(−2)`
	`= 1.353…`
	`= 1.4\ text(ms)^(−2)\ \ \ text{(to 1 d.p.)}`

ii. `text(Distance travelled)`

`= int_0^10 v\ dt`

`= 50 int_0^10 1 − e^(−0.2t) \ dt`

`= 50 [t + 1/0.2 · e^(−0.2t)]_0^10`

`= 50 [t + 5e^(−0.2t)]_0^10`

`= 50 [(10 + 5e^(−2)) − (0 + 5e^0)]`

`= 50 [10 + 5e^(−2) − 5]`

`= 50 [5 + 5e^(−2)]`

`= 283.833…`

`= 284\ text{m (nearest m)}`

Functions, EXT1 F2 2007 HSC 2c

The polynomial `P(x) = x^2 + ax + b` has a zero at `x = 2`. When `P(x)` is divided by `x + 1`, the remainder is `18`.

Find the values of `a` and `b`. (3 marks)

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Show Answers Only

`a = -7\ \ text(and)\ \ b = 10`

Show Worked Solution

`P(x) = x^2 + ax + b`

`text(S)text(ince there is a zero at)\ \ x = 2,`

`P(2)`	`=0`
`2^2 + 2a + b`	`= 0`
`2a + b`	`= -4`	`…\ (1)`

`P(-1) = 18,`

`(-1)^2-a + b`	`= 18`
`-a + b`	`= 17`	`…\ (2)`

`text(Subtract)\ \ (1)-(2),`

`3a`	`= -21`
`a`	`= -7`

`text(Substitute)\ \ a = -7\ \ text{into (1),}`

`2(-7) + b`	`= -4`
`b`	`= 10`

`:.a = -7\ \ text(and)\ \ b = 10`

Trigonometry, EXT1 T1 2007 HSC 2b

Let `f(x) = 2 cos^(-1)x`.

Sketch the graph of `y = f(x)`, indicating clearly the coordinates of the endpoints of the graph. (2 marks)

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State the range of `f(x)`. (1 mark)

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Show Answers Only

`text(See Worked Solutions.)`
`0 ≤ y ≤ 2pi`

Show Worked Solution

i. `y= 2\ cos^(-1)x`

`text(Domain:)\ -1 ≤ x ≤ 1`

`text{Range:}\ \0 ≤`	`y/2 ≤ pi`
`0 ≤`	`y ≤ 2pi`

ii. `text(Range:)\ \ 0 ≤ y ≤ 2pi`

Linear Functions, EXT1 2007 HSC 1d

The graphs of the line `x - 2y + 3= 0` and the curve `y = x^3+ 1` intersect at `(1, 2)`. Find the exact value, in radians, of the acute angle between the line and the tangent to the curve at the point of intersection. (3 marks)

Show Answers Only

`pi/4\ text(radians)`

Show Worked Solution

`x − 2y + 3`	`= 0`
`2y`	`= x + 3`
`y`	`= 1/2x + 3/2`

`:. m_1 = 1/2`

`y`	`= x^3 + 1`
`(dy)/(dx)`	`= 3x^2`

`text(When)\ x = 1,`

`(dy)/(dx) = 3`

`:. m_2 = 3`

`tan\ theta`	`= \|\ (m_1 − m_2)/(1 + m_1m_2)\ \|`
	`= \|\ (1/2 − 3)/(1 + 1/2 xx 3)\ \|`
	`= \|\ (−2 1/2)/(2 1/2)\|`
	`= 1`
`:. theta`	`= pi/4\ text(radians)`

Calculus, EXT1 C2 2007 HSC 1c

Differentiate `tan^(–1)(x^4)` with respect to `x`. (2 marks)

Show Answers Only

`(4x^3)/(1 + x^8)`

Show Worked Solution

`y`	`= tan^(−1)(x^4)`
`(dy)/(dx)`	`= 1/(1 + (x^4)^2) xx d/(dx) (x^4)`
	`= (4x^3)/(1 + x^8)`

Linear Functions, EXT1 2007 HSC 1b

The interval `AB`, where `A` is `(4, 5)` and `B` is `(19, text(−5))`, is divided internally in the ratio `2\ :\ 3` by the point `P(x,y)`. Find the values of `x` and `y`. (2 marks)

Show Answers Only

`(10, 1)`

Show Worked Solution

`A(4, 5), \ B(19, text(−5))`

`text(Internal division in ratio 2 : 3 at)\ \ P(x, y)`

`P`	`= ((nx_1 + mx_2)/(m + n) , (ny_1 + my_2)/(m + n))`
	`= ((3 xx 4 + 2 xx 19)/(2 + 3) , (3 xx 5 + 2 xx (text(−5)))/(2+ 3))`
	`= ((12 + 38)/5 , (15 − 10)/5)`
	`= (10, 1)`

Mechanics, EXT2* M1 2004 HSC 6b

A fire hose is at ground level on a horizontal plane. Water is projected from the hose. The angle of projection, `theta`, is allowed to vary. The speed of the water as it leaves the hose, `v` metres per second, remains constant. You may assume that if the origin is taken to be the point of projection, the path of the water is given by the parametric equations

`x = vt\ cos\ theta`

`y = vt\ sin\ theta − 1/2 g t^2`

where `g\ text(ms)^(−2)` is the acceleration due to gravity. (Do NOT prove this.)

Show that the water returns to ground level at a distance`(v^2\ sin\ 2theta)/g` metres from the point of projection. (2 marks)

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This fire hose is now aimed at a 20 metre high thin wall from a point of projection at ground level 40 metres from the base of the wall. It is known that when the angle `theta` is 15°, the water just reaches the base of the wall.

Show that `v^2 = 80g`. (1 mark)

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Show that the cartesian equation of the path of the water is given by

`y = x\ tan\ theta − (x^2\ sec^2\ theta)/160`. (2 marks)

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Show that the water just clears the top of the wall if

`tan^2\ theta − 4\ tan\ theta + 3 = 0`. (2 marks)

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Find all values of `theta` for which the water hits the front of the wall. (2 marks)

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Show Answers Only

`text(See Worked Solutions)`
`text(See Worked Solutions)`
`text(See Worked Solutions)`
`text(See Worked Solutions)`
`15^@ ≤ theta ≤ 45^@ \ \ text(and)\ \ 71.6^@ ≤ theta ≤ 75^@`

Show Worked Solution

i.	`x`	`= vt\ cos\ theta`
	`y`	`= vt\ sin\ theta − 1/2 g t^2`

`text(Find)\ \ t\ \ text(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, \ \ 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 · \ 2\ sin\ theta\ cos\ theta)/g`
	`= (v^2\ sin\ 2theta)/g`

`:.\ text(When)\ \ x = (v^2\ sin\ 2theta)/g,\ \ \text(the water returns)`

`text(to ground level … as required.)`

ii. `text(Show)\ \ v^2 = 80\ text(g)`

`text(When)\ \ theta = 15^@, \ x = 40`

`text{From part (i)}`

`40`	`= (v^2\ sin\ 30^@)/g`
`v^2 xx 1/2`	`= 40g`
`v^2`	`= 80g\ \ \ …text(as required)`

iii. `text(Show)\ \ y = x\ tan\ theta − (x^2\ sec^2\ theta)/160`

`x`	`= vt\ cos\ theta`
`:.t`	`= x/(v\ cos\ theta)`

`text(Substitute into)`

`y`	`= vt\ sin\ theta − 1/2 g t^2`
	`= v · x/(v\ cos\ theta) · sin\ theta −1/2 g (x/(v\ cos\ theta))^2`
	`= x\ tan\ theta − 1/2 g (x^2/(v^2\ cos^2\ theta))`
	`= x\ tan\ theta − 1/2 g · x^2/(80g\ cos^2\ theta)`
	`= x\ tan\ theta − (x^2\ sec^2\ theta)/160\ \ \ …text(as required.)`

iv. `text(Water clears the top if)\ \ y = 20\ \ text(when)\ \ x = 40`

`text{Substitute into equation from (iii)}`

`40\ tan\ theta − (40^2\ sec^2\ theta)/160`	`= 20`
`40\ tan\ theta − 10\ sec^2\ theta`	`= 20`
`40\ tan\ theta − 10(1 + tan^2\ theta)`	`= 20`
`40\ tan\ theta − 10 − 10\ tan^2\ theta`	`= 20`
`10\ tan^2\ theta − 40\ tan\ theta\ + 30`	`= 0`
`tan^2\ theta − 4\ tan\ theta + 3`	`= 0\ \ \ text(… as required)`

`text(Water hits the bottom of the wall when)`

`x = 40\ \ text(and)\ \ y = 0`

`40\ tan\ theta − (40^2\ sec^2\ theta)/160`	`= 0`
`40\ tan\ theta − 10\ sec^2\ theta`	`= 0`
`4\ tan\ theta − (1 + tan^2\ theta)`	`= 0`
`tan^2\ theta − 4\ tan\ theta + 1`	`= 0`

`text(Using the quadratic formula)`

`tan\ theta`	`= (+4 ± sqrt(16 − 4 · 1 · 1))/ (2 xx 1)`
	`= (4 ± sqrt12)/2`
	`= 2 ± sqrt3`
`theta`	`= 15^@\ \ text(or)\ \ 75^@`

`text(Water hits the top of the wall when)`

`x = 40\ text(and)\ \ y = 20`

`tan^2\ theta − 4\ tan\ theta + 3`	`= 0\ \ \ text{from (iv)}`
`(tan\ theta − 1)(tan\ theta − 3)`	`= 0`

`tan\ theta`	`= 1`	`text(or)`	`tan\ theta`	`= 3`
`theta`	`= 45^@`		`theta`	`= tan^(−1)\ 3`
				`= 71.565…`
				`= 71.6^@\ \ \text{(to 1 d.p.)}`

`:.\ text(Following the motion of the water as)\ \ theta\ \ text(increases,)`

`text(water hits the wall when)`

`15^@ ≤ theta ≤ 45^@`	`\ text(and)`
`71.6^@ ≤ theta ≤ 75^@`

Inverse Functions, EXT1 2004 HSC 5b

The diagram below shows a sketch of the graph of `y = f(x)`, where `f(x) = 1/(1 + x^2)` for `x ≥ 0`.

Copy or trace this diagram into your writing booklet.
On the same set of axes, sketch the graph of the inverse function, `y = f^(−1)(x)`. (1 mark)
State the domain of `f^(−1)(x)`. (1 mark)
Find an expression for `y = f^(−1)(x)` in terms of `x`. (2 marks)
The graphs of `y = f(x)` and `y = f^(−1)(x)` meet at exactly one point `P`.
Let `α` be the `x`-coordinate of `P`. Explain why `α` is a root of the equation
1. `x^3 + x − 1 = 0`. (1 mark)
Take 0.5 as a first approximation for `α`. Use one application of Newton’s method to find a second approximation for `α`. (2 marks)

Show Answers Only

`text(See Worked Solutions)`
`0 < x ≤ 1`
`y = sqrt((1 − x)/x), y > 0`
`text(See Worked Solutions)`
`0.714\ \ \ text{(to 3 d.p.)}`

Show Worked Solution

(i)

(ii) `text(Domain of)\ \ f^(−1)(x)\ text(is)`

`0 < x ≤ 1`

(iii) `f(x) = 1/(1 + x^2)`

`text(Inverse: swap)\ \ x↔y`

`x`	`= 1/(1 + y^2)`
`x(1 + y^2)`	`= 1`
`1 + y^2`	`= 1/x`
`y^2`	`= 1/x − 1`
	`= (1 − x)/x`
`y`	`= ± sqrt((1 − x)/x)`

`:.y = sqrt((1 − x)/x), \ \ y >= 0`

(iv) `P\ \ text(occurs when)\ \ f(x)\ \ text(cuts)\ \ y = x`

`text(i.e. where)`

`1/(1 + x^2)`	`= x`
`1`	`= x(1 + x^2)`
`1`	`= x + x^3`
`x^3 + x − 1`	`= 0`

`=>\ text(S)text(ince)\ α\ text(is the)\ x\ text(-coordinate of)\ P,`

`text(it is a root of)\ \ \ x^3 + x − 1 = 0`

(v)	`f(x)`	`= x^3 + x − 1`
	`f′(x)`	`= 3x^2 + 1`
	`f(0.5)`	`= 0.5^3 + 0.5 − 1`
		`= −0.375`
	`f′(0.5)`	`= 3 xx 0.5^2 + 1`
		`= 1.75`

`α_2`	`= α_1 − (f(0.5))/(f′(0.5))`
	`= 0.5 − ((−0.375))/1.75`
	`= 0.5 − (−0.2142…)`
	`= 0.7142…`
	`= 0.714\ \ \ text{(to 3 d.p.)}`

Quadratic, EXT1 2004 HSC 4b

The two points `P(2ap, ap^2)` and `Q(2aq, aq^2)` are on the parabola `x^2 = 4ay`.

The equation of the tangent to `x^2 = 4ay` at an arbitrary point `(2at, at^2)` on the parabola is `y = tx − at^2`. (Do not prove this.)
Show that the tangents at the points `P` and `Q` meet at `R`, where `R` is the point `(a(p + q), apq)`. (2 marks)
As `P` varies, the point `Q` is always chosen so that `∠POQ` is a right angle, where `O` is the origin.
Find the locus of `R`. (2 marks)

Show Answers Only

`text{Proof (See Worked Solutions)}`
`y = −4a`

Show Worked Solution

(i)

`text(Show)\ \ R(a(prq), apq)`

`text(T)text(angent equations)`

`y`	`= px − ap^2`	`\ \ …\ (1)`
`y`	`= qx − aq^2`	`\ \ …\ (2)`

`text(Substitute)\ \y = px − ap^2\ \text(into)\ (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 xx a(p + q) − ap^2`
	`= ap^2 + apq − ap^2`
	`= apq`

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

(ii) `text(If)\ \ ∠POQ\ text(is a right angle)`

`M_(PO) xx M_(OQ) = −1`

`M_(PO)`	`= (ap^2 − 0)/(2ap − 0)`
	`= p/2`
`M_(OQ)`	`= (aq^2 − 0)/(2aq − 0)`
	`= q/2`

`:.p/2 xx q/2`	`= −1`
`pq`	`= −4`

`⇒R\ \ text(has coordinates)\ \ (a(p+q), −4a)`

`:.\ text(Locus of)\ R\ text(is)\ \ y = −4a`

Calculus, EXT1 C1 2004 HSC 3c

A ferry wharf consists of a floating pontoon linked to a jetty by a 4 metre long walkway. Let `h` metres be the difference in height between the top of the pontoon and the top of the jetty and let `x` metres be the horizontal distance between the pontoon and the jetty.

Find an expression for `x` in terms of `h`. (1 mark)

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When the top of the pontoon is 1 metre lower than the top of the jetty, the tide is rising at a rate of 0.3 metres per hour.

At what rate is the pontoon moving away from the jetty? (3 marks)

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Show Answers Only

`x=sqrt(16 − h^2)`
`0.077\ text(m/hr)`

Show Worked Solution

i. `text(Using Pythagoras,)`

`x^2 + h^2`	`= 4^2`
`x^2`	`= 16 − h^2`
`x`	`= sqrt(16 − h^2)`

ii. `text(Find)\ \ (dx)/(dt)\ \ text(when)\ \ h = 1`

`(dx)/(dt)`	`= (dx)/(dh)·(dh)/(dt)`
`x`	`= (16 − h^2)^(1/2)`
`(dx)/(dh)`	`= 1/2 xx (16 − h^2)^(−1/2) xx d/(dh)(16 − h^2)`
	`= 1/2 (16 − h^2)^(−1/2) xx −2h`
	`= (−h)/(sqrt(16 − h^2))`

`text(When)\ \ h = 1, (dh)/(dt)= −0.3\ text(m/hr)`

`text{(}h\ \ text{decreases when the tide is rising)}`

`(dx)/(dt)`	`= (−h)/(sqrt(16 − h^2)) xx −0.3`
	`= (−1)/sqrt(16 − 1^2) xx −0.3`
	`= 0.3/sqrt15`
	`= 0.0774…`
	`= 0.077\ \ \ text{metres per hr (to 2 d.p.)}`

`:.\ text(When)\ \ h = 1,\ text(the pontoon is moving away)`

`text(at 0.077 metres per hr.)`

Functions, EXT1 F2 2004 HSC 3b

Let `P(x) = (x + 1) (x − 3)Q(x) + a(x + 1) + b`, where `Q(x)` is a polynomial and `a` and `b` are real numbers.

When `P(x)` is divided by `(x + 1)` the remainder is `−11`.

When `P(x)` is divided by `(x − 3)` the remainder is `1`.

What is the value of `b`? (1 mark)

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What is the remainder when `P(x)` is divided by `(x + 1)(x − 3)`? (2 marks)

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Show Answers Only

`−11`
`3x − 8`

Show Worked Solution

i. `P(x)= (x + 1) (x − 3)Q(x) + a(x + 1) + b`

`P(-1)=-11`

`-11=(−1 + 1)(−1 − 3)Q(x) + a(−1 + 1) + b`

`-11=(0)(-4)Q(x)+a(0)+b`

`:.b = −11`

(ii) `P(3) = 1`

`:.(3 + 1)(3 − 3)Q(x) + a(3 + 1) −11 = 1`

`4a`	`= 12`
`a`	`= 3`

`text(When)\ \ P(x)\ \ text(is divided by)\ \ (x + 1)(x − 3)`

`R(x)`	`= a(x + 1) + b`
	`= 3(x + 1) − 11`
	`= 3x + 3 − 11`
	`= 3x − 8`

Combinatorics, EXT1 A1 2004 HSC 2e

A four-person team is to be chosen at random from nine women and seven men.

In how many ways can this team be chosen? (1 mark)

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What is the probability that the team will consist of four women? (1 mark)

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Show Answers Only

`1820`
`9/130`

Show Worked Solution

i. `text(# Team combinations)`

`=\ ^(16)C_4`

`= (16!)/((16 − 4)!\ 4!)`

`= 1820`

ii. `text{P(4 women)}`

`= (\ ^9C_4)/1820`

`= 126/1820`

`= 9/130`

Calculus, EXT1 C2 2004 HSC 1e

Use the substitution `u = x − 3` to evaluate

`int_3^4 xsqrt(x − 3)\ dx.` (3 marks)

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Show Answers Only

`2 2/5`

Show Worked Solution

`text(Let)\ \ u = x − 3`

`=> x = u + 3`

`(du)/dx = 1`

`=> dx = du`

`text(When)\ \ `	`x = 4,`	`u = 1`
	`x = 3,`	`u = 0`

`int_3^4 xsqrt(x − 3\ dx)`

`= int_0^1(u + 3)\ u^(1/2)\ du`

`= int_0^1u^(3/2) + 3u^(1/2)\ du`

`=[2/5u^(5/2) + 3 xx 2/3u^(3/2)]_0^1`

`= [2/5u^(5/2) + 2u^(3/2)]_0^1`

`= [(2/5 + 2) − 0]`

`= 2 2/5`

Calculus, EXT1 C2 2004 HSC 1d

Find `int_0^1(dx)/(sqrt(4 − x^2))`. (2 marks)

Show Answers Only

`pi/6`

Show Worked Solution

`int_0^1(dx)/(sqrt(4 − x^2))`	`= [sin^(−1)\ x/2]_0^1`
	`= sin^(−1)(1/2) − sin^(−1)(0)`
	`= pi/6`

Linear Functions, EXT1 2004 HSC 1c

Let `A` be the point `(3, text(−1))` and `B` be the point `(9, 2)`.

Find the coordinates of the point `P` which divides the interval `AB` externally in the ratio `5:2`. (2 marks)

Show Answers Only

`(13, 4)`

Show Worked Solution

`A(3, text(−1))\ \ B(9, 2)`

`P\ \ text(divides)\ \ AB\ \ text(externally in ratio)\ \ 5:2`

`:.P`	`= ((nx_1 + mx_2)/(m + n),(ny_1 + my_2)/(m + n))`
	`= ((text(−2)(3) + 5(9))/(5 + (text(−2))), (text(−2)(text(−1)) + 5(2))/(5 + text{(−2)}))`
	`= ((text(−6)\ + 45)/3, (+2 + 10)/3)`
	`= (13, 4)`

Functions, EXT1 F1 2004 HSC 1b

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

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Show Answers Only

`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)

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`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)

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(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`