Band 4 – Page 110

Algebra, STD2 A4 2014 HSC 3 MC

The diagram shows the graph of an equation.

Which of the following equations does the graph best represent?

`y = 3/x + 1`
`y = 3^x + 1`
`y = 3x^2 + 1`
`y = 3x^3 + 1`

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`C`

Show Worked Solution

`text(Graph is a parabola that passes through)\ (0, 1).`

`=> C`

Proof, EXT2 P2 EQ-Bank 10

Use mathematical induction to prove that

`sum_(r=1)^n r^3 = 1/4 n^2 (n + 1)^2`

`text(for integers)\ n>=1` (3 marks)

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

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

Show Worked Solution

`text(Need to prove)\ sum_(r=1)^n r^3 = 1/4 n^2 (n + 1)^2\ \ text(integral)\ n>=1 `

`text(i.e.)\ 1^3 + 2^3 + 3^3 + … + n^3 = 1/4 n^2 (n + 1)^2`

`text(When)\ n = 1`

`text(LHS) = 1^3 = 1`

`text(RHS) = 1/4 1^2 (1 + 1)^2 = 1`

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

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

`text(i.e.)\ 1^3 + 2^3 + … + k^3 = 1/4 k^2 (k + 1)^2`

`text(Need to prove true for)\ n = k + 1`

`1^3 + 2^3 + … + k^3 + (k + 1)^3 = 1/4 (k + 1)^2 (k + 2)^2`

`text(LHS)`	`= 1/4 k^2 (k + 1)^2 + (k + 1)^3`
	`= 1/4 (k + 1)^2 [k^2 + 4(k + 1)]`
	`= 1/4 (k + 1)^2 (k^2 + 4k + 4)`
	`= 1/4 (k + 1)^2 (k + 2)^2`
	`=\ text(RHS)`

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

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

Quadratic, EXT1 2014 HSC 13c

The point `P(2at, at^2)` lies on the parabola `x^2 = 4ay` with focus `S`.

The point `Q` divides the interval `PS` internally in the ratio `t^2 :1`.

Show that the coordinates of `Q` are
`x = (2at)/(1 + t^2)` and `y = (2at^2)/(1 + t^2)`. (2 marks)
Express the slope of `OQ` in terms of `t`. (1 mark)
Using the result from part (ii), or otherwise, show that `Q` lies on a fixed circle of radius `a`. (3 marks)

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`text(Proof)\ \ text{(See Worked Solutions)}`
`m_(OQ) = t`
`text(Proof)\ \ text{(See Worked Solutions)}`

Show Worked Solution

(i) `text(Show)\ \ Q = ((2at)/(1 + t^2), (2at^2)/(1 + t^2))`

`P (2at, at^2),\ S (0, a)`

`PS\ text(is divided internally in ratio)\ t^2: 1`

`Q`	`= ((nx_1 + mx_2)/(m + n), (ny_1 + my_2)/(m + n))`
	`= ((1(2at) + t^2(0))/(t^2 + 1), (1(at^2) + t^2 (a))/(t^2 + 1))`
	`= ((2at)/(1 + t^2), (2at^2)/(1 + t^2))\ \ text(… as required.)`

(ii)	`m_(OQ)`	`= (y_2\ – y_1)/(x_2\ – x_1)`
		`= ((2at^2)/(1 + t^2))/((2at)/(1 + t^2)) xx (1+t^2)/(1+t^2)`
		`= (2at^2)/(2at)`
		`= t`

(iii)	`text(Show)\ Q\ text(lies on a fixed circle radius)\ a`
	`text(S)text(ince)\ Q\ text(passes through)\ (0, 0)`

`=>\ text(If locus of)\ Q\ text(is a circle, it has)`

`text(diameter)\ QT\ text(where)\ T(0, 2a)`

`text(Show)\ \ QT _|_ OQ`

♦♦ Mean mark 22%

`text{(} text(angles on circum. subtended by)`

`text(a diameter are)\ 90^@ text{)}`

`m_(OQ) = t\ \ \ \ text{(see part (ii))}`

`text(Find)\ m_(QT),\ \ text(where:)`

`Q((2at)/(1 + t^2), (2at^2)/(1 + t^2)),\ \ \ \ \ T (0,2a)`

`m_(QT)`	`= (y_2\ – y_1)/(x_2\ – x_1)`
	`= ((2at^2)/(1 + t^2)\ – 2a)/((2at)/(1 + t^2)\ – 0)`
	`= (2at^2\ – 2a (1 + t^2))/(2at)`
	`= – (2a)/(2at)`
	`= – 1/t`

`m_(QT) xx m_(OT) = -1/t xx t = -1`

`=> QT _|_ OQ`

`=>O,\ T,\ Q\ text(lie on a circle.)`

`:.\ text(Locus of)\ Q\ text(is a fixed circle,)`

`text(centre)\ (0, a),\ text(radius)\ a`

Quadratic, EXT1 2009 HSC 2c

The diagram shows points `P(2t, t^2)` and `Q(4t, 4t^2)` which move along the parabola `x^2 = 4y`. The tangents to the parabola at `P` and `Q` meet at `R`.

Show that the equation of the tangent at `P` is `y = tx\ - t^2.` (2 marks)
Write down the equation of the tangent at `Q`, and find the coordinates of the point `R` in terms of `t`. (2 marks)
Find the Cartesian equation of the locus of `R`. (1 mark)

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`text(Proof)\ \ text{(See Worked Solutions)}`
`y = 2tx\ – 4t^2,\ R (3t, 2t^2)`
`y = 2/9 x^2`

Show Worked Solution

(i)	`x^2`	`= 4y`
	`y`	`= (x^2)/4`
	`dy/dx`	`= x/2`

IMPORTANT: Deriving this equation was specifically asked in 2009 and 2011, as well as being required in 2014. KNOW IT! Completing this in 60 seconds gains 2 full minutes on allocated time.

`text(At)\ \ x = 2t,`

`dy/dx = (2t)/2 = t`

`:.\ text(T)text(angent has)\ \ m = t\ \ text(and passes)`

`text(through)\ \ (2t, t^2).`

`y – t^2`	`= t (x – 2t)`
`y`	`= tx\ – 2t^2 + t^2`
	`= tx\ – t^2\ \ text(… as required)`

(ii)

`text(T)text(angent at)\ \ P (2t, t^2)\ \ text(is)\ \ y = tx\ – t^2`

`=>Q(4t, 4t^2) -= Q(2(2t), (2t)^2)`

`:.\ text(Equation of the tangent at)\ \ Q,`

`y`	`= 2tx\ – (2t)^2`
	`= 2tx\ – 4t^2`

`text(T)text(angents intersect at)\ \ R`

`y`	`= tx\ – t^2\ \ \ \ \ …\ text{(1)}`
`y`	`= 2tx\ – 4t^2\ \ \ \ \ …\ text{(2)}`

`text(Intersection when)\ text{(1)} = text{(2)}`

`tx\ – t^2`	`= 2tx\ – 4t^2`
`tx`	`= 3t^2`
`x`	`= 3t`

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

`y`	`= t (3t)\ – t^2`
	`= 2t^2`

`:.\ R (3t, 2t^2)`

(iii)

`text(Find locus of)\ \ R`

`x`	`= 3t`	`\ \ \ \ \ …\ text{(1)}`
`y`	`= 2t^2\ \ \ \ \ …\ text{(2)}`

`text{From (1),}\ \ \ \ t = x/3`

`text(Substitute)\ \ t = x/3\ text(into)\ text{(2)}`

`y`	`= 2 (x/3)^2`
	`= 2/9 x^2`

`:.\ text(Locus of)\ \ R\ \ text(is)\ \ y = 2/9 x^2`

Polynomials, EXT1 2013 HSC 14c

The equation `e^t = 1/t` has an approximate solution `t_0 = 0.5`

Use one application of Newton’s method to show that `t_1 = 0.56` is another approximate solution of `e^t = 1/t`. (2 marks)
Hence, or otherwise, find an approximation to the value of `r` for which the graphs `y = e^(rx)` and `y = log_e x` have a common tangent at their point of intersection. (3 marks)

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`text(Proof)\ \ text{(See Worked Solutions)}`
`0.097\ text{(to 3 d.p.)}`

Show Worked Solution

MARKER’S COMMENT: Many students had difficulty expressing a valid function while too many others carelessly used `t_0=0.56` rather than 0.5.

(i)	`e^t`	`= 1/t`
	`e^t\ – 1/t`	`= 0`

`text(Let)\ \ f(t)`	`= e^t\ – 1/t`
`f prime (t)`	`= e^t + 1/(t^2)`
`f(0.5)`	`=e^0.5-1/0.5`
	`=–0.3512…`
`f′(0.5)`	`=e^0.5 +1/0.5^2`
	`=5.6487…`

`text(Applying Newton where)\ \ t_0 = 0.5`

`t_1`	`= 0.5\ – (f(0.5))/(f prime (0.5)`
	`= 0.5\ – (–0.3512…)/(5.6487…)`
	`= 0.5\ – (- 0.062)`
	`= 0.56\ text{(2 d.p.)}\ \ text(… as required.)`

(ii)	`y_1`	`= e^(rx),`	`\ \ \ \ \ (dy_1)/(dx) = re^(rx)`
	`y_2`	`= log_e x,`	`\ \ \ \ \ (dy_2)/(dx) = 1/x`

♦♦♦ 2nd toughest question on 2013 paper with mean mark 15%.
MARKER’S COMMENT: Few students used the common tangent information to equate the derivative functions.

`text(S)text(ince tangent is common),\ \ (dy_1)/(dx)=(dy_2)/(dx)`

`re^(rx)`	`= 1/x`
`e^(rx)`	`= 1/(rx)`

`text(Using part)\ text{(i)},\ rx ~~ 0.56`

`text(At intersection of curves)\ \ y_1 = y_2`

`e^(rx)`	`= log_e x`
`e^0.56`	`= log_e x`
`x`	`= e^(e^0.56)`
	`= 5.758 …`

`text(S)text(ince)\ rx`	`~~ 0.56`
`=> r`	`~~ 0.56/(5.758 … )`
	`~~ 0.0972 …`
	`~~ 0.097\ text{(to 3 d.p.)}`

Binomial, EXT1 2013 HSC 14b

Write down the coefficient of `x^(2n)` in the binomial expansion of `(1 + x)^(4n)`. (1 mark)
Show that
1. `(1 + x^2 + 2x)^(2n) = sum_(k=0)^(2n) ((2n),(k)) x^(2n\ - k)(x + 2)^(2n\ - k)`. (2 marks)
It is known that
`x^(2n\ - k) (x + 2)^(2n\ - k) = ((2n\ - k),(0)) 2^(2n\ - k) x^(2n\ - k) + ((2n\ - k),(1)) 2^(2n\ - k\ - 1) x^(2n\ - k + 1)`
2. 1. `+ ... + ((2n\ - k),(2n\ - k)) 2^0 x^(4n\ - 2k)`. (Do NOT prove this.)
Show that
1. `((4n),(2n)) = sum_(k = 0)^(n) 2^(2n\ - 2k) ((2n),(k))((2n\ - k),(k))`. (3 marks)

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`((4n),(2n))`
`text(Proof)\ \ text{(See Worked Solutions)}`
`text(Proof)\ \ text{(See Worked Solutions)}`

Show Worked Solution

(i) `text(Find co-efficient of)\ \ x^(2n)`

`text(Expanding)\ \ (1+x)^(4n)`

`((4n),(0)) + ((4n),(1))x + ((4n),(2))x^2 + … + ((4n),(2n))x^(2n) + …`

`:.\ text(Co-efficient of)\ \ x^(2n)\ text(is)\ ((4n),(2n))`

(ii) `text(Show)\ (1 + x^2 + 2x)^(2n) = sum_(k=0)^(2n) ((2n),(k)) x^(2n\ – k) (x + 2)^(2n\ – k)`

♦♦ Mean mark 30%.

`text(Using)\ (1 + x^2 + 2x)^(2n) = [x(x + 2) + 1]^(2n)`

`[x (x + 2) + 1]^(2n)`

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

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

`= sum_(k=0)^(2n) ((2n),(k)) x^(2n\ – k) (x + 2)^(2n\ – k)\ text(… as required.)`

(iii) `((4n),(2n))\ text(is the co-eff of)\ x^(2n)\ text(in expansion)\ (1+x)^(4x)`

`text(S)text(ince)\ (1 + x^2 + 2x)^(2n) = ((x+1)^2)^(2n) = (1 + x)^(4n)`

`=> ((4n),(2n))\ text(is co-efficient of)\ x^(2n)\ text(in expansion)\ (1 + x^2 + 2x)^(2n)`

`text(Using part)\ text{(ii)}`

`(1 + x^2 + 2x)^(2n) = sum_(k=0)^(2n) ((2n),(k))\ x^(2n\ – k) (x + 2)^(2n\ – k)`

`text(Using the given identity,)\ x^(2n)\ text(co-efficients are)`

♦♦♦ Toughest question in the 2013 exam with Mean mark 12%!
MARKER’S COMMENT: Simply stating `(1+x^2+2x)^(2n)“=(1+x)^(4n)` received 1 full mark, showing that even initial working can gain marks.

`k = 0,`	`\ ((2n),(0))((2n\ – 0),(0)) 2^(2n\ – 0)`
`k = 1,`	`\ ((2n),(1))((2n\ – 1),(1)) 2^(2n\ – 1\ – 1)`
`vdots`
`k = n,`	`\ ((2n),(n))((2n\ – n),(n)) 2^(2n\ – n\ – n)`

`:.\ ((4n),(2n))`

`= ((2n),(0))((2n),(0))2^(2n) + ((2n),(1))((2n\ – 1),(1))2^(2n\ – 2) + … + ((2n),(n))((n),(n)) 2^0`

` = sum_(k=0)^(n)\ ((2n),(k))((2n\ – k),(k)) 2^(2n\ – 2k)\ \ \ \ text(… as required)`

Calculus, EXT1 C1 2013 HSC 13a

A spherical raindrop of radius `r` metres loses water through evaporation at a rate that depends on its surface area. The rate of change of the volume `V` of the raindrop is given by

`(dV)/(dt) = -10^(-4) A`,

where `t` is time in seconds and `A` is the surface area of the raindrop. The surface area and the volume of the raindrop are given by `A = 4pir^2` and `V = 4/3 pi r^3` respectively.

Show that `(dr)/(dt)` is constant. (1 mark)

--- 5 WORK AREA LINES (style=lined) ---
How long does it take for a raindrop of volume `10^(–6)` m³ to completely evaporate? (2 marks)

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

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

Show Worked Solution

i.	`text(Show)\ \ (dr)/(dt)\ \ text(is a constant)`

`(dV)/(dt) = (dV)/(dr) * (dr)/(dt)\ \ \ …\ text{(1)}`

`V`	`= 4/3 pi r^3`
`:. (dV)/(dr)`	`= 4 pi r^2`
`(dV)/(dt)`	`= -10^(-4) A\ \ text{(given)}`

`text(Substituting into)\ text{(1)}`

`-10^(-4) A`	`= 4 pi r^2 xx (dr)/(dt)`
	`= A xx (dr)/(dt)`
`:.\ (dr)/(dt)`	`= -10^(-4)\ \ text(… as required)`

ii.	`V`	`= 10^(-6)\ text(m³)`
	`4/3 pi r^3`	`= 10^(-6)`
	`r^3`	`= (3 xx 10^(-6))/(4pi)`
	`r`	`= root(3)((3 xx 10^(-6))/(4pi))`

`text(S)text(ince the radius decreases at a constant rate,)`

♦♦ Mean mark 31%

`t=(root(3)((3 xx 10^(-6))/(4pi)))/(10^(-4))`

`\ \ =62.035 …`

`\ \ =62\ text(seconds)\ text{(nearest whole)}`

`:.\ text(It takes 62 seconds for the raindrop)`

`text(to evaporate.)`

Mechanics, EXT2* M1 2013 HSC 12e

A particle moves along a straight line. The displacement of the particle from the origin is `x`, and its velocity is `v`. The particle is moving so that `v^2 + 9x^2 = k`, where `k` is a constant.

Show that the particle moves in simple harmonic motion with period `(2pi)/3`. (2 marks)

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

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

Show Worked Solution

`v^2 + 9x^2`	`= k`
`v^2`	`= k\ – 9x^2`
`1/2 v^2`	`= 1/2k\ – 9/2 x^2`

`text(For SHM,)\ \ ddot x = -n^2x`

`ddot x`	`= d/(dx) (1/2v^2)`
	`= -9x`
	`= -3^2 x \ \ \ text(… as required)`

`text(Period)\ (T)\ text(of SHM) = (2pi)/n`

`text(Here,)\ \ n=3`

`:.T= (2pi)/3\ \ \ text(… as required)`

Geometry and Calculus, EXT1 2013 HSC 12d

The point `P(t, t^2 + 3)` lies on the curve `y = x^2 + 3`. The line `l` has equation `y = 2x\ – 1`. The perpendicular distance from `P` to the line `l` is `D(t)`.

Show that
1. `D(t) = (t^2\ - 2t + 4)/sqrt5`. (2 marks)
Find the value of `t` when `P` is closest to `l`. (1 mark)
Show that, when `P` is closest to `l`, the tangent to the curve at `P` is parallel to `l`. (1 mark)

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`text(Proof)\ \ text{(See Worked Solutions)}`
`t=1`
`text(Proof)\ \ text{(See Worked Solutions)}`

Show Worked Solution

(i)

`text(Show)\ \ D(t) = (t^2\ – 2t + 4)/sqrt5`

`D(t)`	`= _\|_\ text(dist of)\ (t, t^2 + 3)\ text(from)\ 2x\ – y\ – 1 = 0`
`D`	`= \|(ax_1 + by_1 + c)/sqrt(a^2 + b^2)\|`
	`= \|(2(t)\ – 1 (t^2 + 3)\ – 1)/sqrt(2^2 + (-1)^2)\|`
	`= \|(2t\ – t^2\ – 3\ – 1)/sqrt5\|`
	`= \|(-(t^2\ – 2t + 4))/sqrt5\|`

Mean mark 40%
MARKER’S COMMENT: The biggest difficulty in part (i) was to explain the removal of the absolute value sign.

`text(S)text(ince)\ \ t^2\ – 2t + 4\ \ text(is positive definite)`

`text{(i.e.} \ Delta < 0\ text(and co-efficient of)\ t^2 > 0 text{)},`

`t^2\ – 2t + 4\ \ text(is positive for all)\ t.`

`:.\ D(t) = (t^2\ – 2t + 4)/sqrt5\ \ text(… as required)`

(ii)

`text(MAX or MIN when)\ \ (dD)/(dt) = 0`

`(dD)/(dt) = 1/sqrt5 (2t\ – 2)`	`= 0`
`2t\ – 2`	`= 0`
`t`	`= 1`

`(d^2D)/(dt^2) = 2/sqrt5 > 0\ =>\ text(MIN)`

`:.\ P\ text(is closest to)\ l\ text(when)\ t = 1`

(iii)	`y`	`= x^2 + 3`
	`dy/dx`	`= 2x`
`text(When)\ x = t`

`dy/dx = 2t`

`text(At)\ t = 1`

`dy/dx = 2,`

`=> m_text(tangent) = 2\ \ and\ \ m_l = 2`

`:.\ text(T)text(angent is parallel to)\ \ l\ \ \ \ text(… as required)`

Calculus, EXT1 C3 2013 HSC 12b

The region bounded by the graph `y = 3 sin\ x/2` and the `x`-axis between `x = 0` and `x = (3pi)/2` is rotated about the `x`-axis to form a solid.

Find the exact volume of the solid. (3 marks)

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

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`(9pi)/2 ((3pi)/2 + 1) text(u³)`

Show Worked Solution

`y`	`= 3 sin\ x/2`
`y^2`	`= 9 sin^2\ x/2`

`text(Using:)\ \ sin^2x= 1/2 (1 – cos 2x)`

`:. V`	`= pi int_0^((3pi)/2) 9 sin^2\ x/2\ dx`
	`= (9pi)/2 int_0^((3pi)/2) (1\ – cosx)\ dx`
	`= (9pi)/2 [x\ – sinx]_0^((3pi)/2)`
	`= (9pi)/2 [((3pi)/2\ – sin\ (3pi)/2)\ – 0]`
	`= (9pi)/2 ((3pi)/2 + 1)\ text(u³)`

Trigonometry, EXT1 T3 2013 HSC 12a

Write `sqrt3cos x - sin x` in the form `2 cos (x + alpha)`, where `0 < alpha < pi/2`. (1 mark)

--- 6 WORK AREA LINES (style=lined) ---
Hence, or otherwise, solve `sqrt3 cos x = 1 + sin x`, where `0 < x < 2pi`. (2 marks)

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

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`2cos (x + pi/6)`
`pi/6,\ (3pi)/2`

Show Worked Solution

i. `text(Write)\ sqrt3 cosx\ – sinx\ text(in form)`

`2cos (x + alpha),\ \ \ 0 < alpha < pi/2`

`2 (cosx cos alpha\ – sinx sin alpha)`	`= sqrt 3 cosx\ – sinx`
`cosx cos alpha\ – sinx sin alpha`	`= sqrt3/2 cos x\ – 1/2 sinx`

`=> cos alpha`	`= sqrt3/2`
`=> sin alpha`	`= 1/2`
`alpha`	`= pi/6`

`:.\ 2cos (x + pi/6) = sqrt3 cosx\ – sinx`

ii. `text(Solve)\ sqrt3 cosx = 1 + sinx,\ \ \ 0 < x < 2pi`

`sqrt3 cosx\ – sinx`	`= 1`
`2 cos (x + pi/6)`	`= 1\ \ \ text{(from part (i))}`
`cos (x + pi/6)`	`= 1/2`
`cos(pi/3)`	`=1/2`

`text(S)text(ince cos is positive in)\ 1^text(st) // 4^text(th)\ text(quadrants)`

`x + pi/6`	`= pi/3,\ 2pi\ – pi/3\ \ \ \ \ (0 < x < 2pi)`
`:.\ x`	`= pi/6,\ (3pi)/2`

Calculus, EXT1 C2 2013 HSC 11f

Use the substitution `u = e^(3x)` to evaluate `int_0^(1/3) (e^(3x))/(e^(6x) + 1)\ dx`. (3 marks)

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

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`1/3 (tan^(-1)e\ – pi/4)`

Show Worked Solution

MARKER’S COMMENT: Many students did not calculate in radians and incorrectly got an answer of 8.2. BE CAREFUL!
Note that converting your answer to 0.14 is also correct but not required.

`text(Let)\ \ u = e^(3x)`

`(du)/(dx)`	`= 3e^(3x)`
`:.dx`	`= (du)/(3e^(3x))`

`text(When)`	`\ x = 1/3,`	`\ u = e^(3 xx 1/3) = e`
	`\ x = 0,`	`\ u = e^0 = 1`

`:.int_0^(1/3) (e^(3x))/(e^(6x) + 1)\ dx`

`=int_1^e (e^(3x))/(u^2 + 1) xx (du)/(3e^(3x))`

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

`= 1/3 [tan^(-1)u]_1^e`

`= 1/3 [tan^(-1) e\ – tan^(-1) 1]`

`= 1/3 (tan^(-1)e\ – pi/4)`

Trig Calculus, EXT1 2013 HSC 11e

Find `lim_(x -> 0) (sin\ x/2)/(3x)`. (1 mark)

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`1/6`

Show Worked Solution

`lim_(x -> 0) (sin\ x/2)/(3x)`

`= lim_(x -> 0) (sin\ x/2)/(6 xx x/2)`

`= 1/6 xx lim_(x -> 0) (sin\ x/2)/(x/2)`

`= 1/6`

Geometry and Calculus, EXT1 2013 HSC 11d

Consider the function `f(x) = x/(4\ - x^2)`.

Show that `f prime (x) > 0` for all `x` in the domain of `f(x)`. (2 marks)
Sketch the graph `y = f(x)`, showing all asymptotes. (2 marks)

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

Show Worked Solution

(i) `f(x) = x/(4\ – x^2)`

`u`	`= x`	`\ \ \ \ \ v`	`= 4\ – x^2`
`u prime`	`= 1`	`\ \ \ \ \ \ \ \ \ \ v prime`	`= -2x`

`f prime (x)`	`= (u prime v\ – u v prime)/(v^2)`
	`= (1* (4\ – x^2)\ – x(-2x))/((4\ – x^2)^2)`
	`= (4 + x^2)/((4\ – x^2)^2)`

`text(Domain is all)\ x,\ x != +- 2`

`text(Consider)\ f prime (x)`

`4 + x^2`	`> 0\ \ \ text{(in domain)}`
`(4\ – x^2)^2`	`> 0\ \ \ text{(in domain)}`

`:.\ f prime (x) > 0\ text(for all)\ x,\ x != +- 2`

(ii)

`text(Asymptotes at)\ x = +- 2`

`text(Passes through)\ (0,0)`

COMMENT: Note that `x->2(–)` is a short way of writing as `x` approaches 2 from the negative (or left-hand) side. This notation can save time when required.

`text(As)`	`\ x -> 2 (-),`	`\ y -> oo`
	`\ x -> 2 (+),`	`\ y -> – oo`

`text(As)`	`\ x -> -2 (-),`	`\ y -> oo`
	`\ x -> -2 (+),`	`\ y -> -oo`

`text(As)`	`\ x -> oo,`	`\ y -> 0`
	`\ x -> – oo,`	`\ y -> 0`

Calculus, EXT1 C2 2013 HSC 11b

Find `int 1/sqrt (49 - 4x^2)\ dx`. (2 marks)

Show Answers Only

`1/2 sin^(-1) ((2x)/7) + c`

Show Worked Solution

`int 1/sqrt(49 – 4x^2)\ dx`

`= int 1/(2 sqrt(49/4 – x^2))\ dx`

`= 1/2 int 1/sqrt((7/2)^2 – x^2)\ dx`

`= 1/2 sin^(-1) ((2x)/7) + c`

Trigonometry, EXT1 T1 2013 HSC 9 MC

The diagram shows the graph of a function.

Which function does the graph represent?

`y = cos^(-1) x`
`y = pi/2 + sin^(-1) x`
`y = - cos^(-1) x`
`y = - pi/2\ - sin^(-1) x`

Show Answers Only

`B`

Show Worked Solution

`text(By elimination,)`

`text(The graph passes through)\ \ (1, pi)`

`text(The only equation to satisfy this point is)`

`y = pi/2 + sin^(-1) x`

`=> B`

Combinatorics, EXT1 A1 2013 HSC 7 MC

A family of eight is seated randomly around a circular table.

What is the probability that the two youngest members of the family sit together?

`(6!\ 2!)/(7!)`
`(6!)/(7!\ 2!)`
`(6!\ 2!)/(8!)`
`(6!)/(8!\ 2!)`

Show Answers Only

`A`

Show Worked Solution

`text(Fix youngest person in 1 seat,)`

`text(Total combinations around table) = 7!`

`text(Combinations with youngest side by side) =2!6!`

`:.\ text{P(sit together)} = (6!\ 2!)/(7!)`

`=> A`

Calculus, EXT1 C2 2013 HSC 5 MC

Which integral is obtained when the substitution `u = 1 + 2x` is applied to `int x sqrt(1 + 2x)\ dx`?

`1/4 int (u - 1) sqrt u\ du`
`1/2 int (u - 1) sqrt u\ du`
`int (u - 1) sqrt u\ du`
`2 int (u - 1) sqrt u\ du`

Show Answers Only

`A`

Show Worked Solution

`text(Let)\ \ u`	`= 1 + 2x`
`:.x`	`= 1/2 (u – 1)`
`(du)/(dx)`	`= 2`
`:.dx`	`= 1/2\ du`

`int x sqrt(1 + 2x)\ dx`

`=int 1/2 (u – 1) xx u^(1/2) xx 1/2\ du`

`= 1/4 int (u – 1) sqrt u\ du`

`=> A`

Functions, EXT1 F2 2013 HSC 4 MC

Which diagram best represents the graph `y = x (1- x)^3 (3- x)^2`?

Show Answers Only

`D`

Show Worked Solution

`y = x(1 – x)^3 (3 – x)^2`

`text(By elimination)`

`text(Consider when)\ \ x < 0,`

`y = text{(–ve)} xx text{(+ve)} xx text{(+ve)} < 0`

`:.\ text(Cannot be)\ A\ text(or)\ C`

`text(Consider the cubic factor)\ \ (1 – x)^3,`

`text(The graph must have a stationary point at)\ x = 1`

`:.\ text(Cannot be)\ B`

`=> D`

Binomial, EXT1 2010 HSC 7b

The binomial theorem states that

`(1 + x)^n = ((n),(0)) + ((n),(1))x + ((n),(2))x^2 + ((n),(3))x^3 + ... + ((n),(n))x^n.`

Show that
1. `2^n = sum_(k = 0)^n ((n),(k))`. (1 mark)
Hence, or otherwise, find the value of
1. `((100),(0)) + ((100),(1)) + ((100),(2)) + ... + ((100),(100))`. (1 mark)
Show that
1. `n2^(n\ - 1) = sum_(k = 1)^n k ((n),(k))`. (2 marks)

Show Answers Only

`text(Proof)\ \ text{(See Worked Solutions)}`
`2^100`
`text(Proof)\ \ text{(See Worked Solutions)}`

Show Worked Solution

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

`text(Let)\ \ x = 1`

`(1 + 1)^n`	`= ((n),(0)) + ((n),(1))1 + ((n),(2))1^2 + … + ((n),(n))1^n`
`2^n`	`= ((n),(0)) + ((n),(1)) + … + ((n),(n))`
	`= sum_(k=0)^n ((n),(k))\ text(… as required)`

(ii) `((100),(0)) + ((100),(1)) + ((100),(2)) + … + ((100),(100)) = 2^100`

`text{(} text(from part)\ text{(i)} text{)}`

♦ Mean mark 37%

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

`text(Differentiate both sides)`

`n(1 + x)^(n\ – 1) = 1((n),(1)) + 2x ((n),(2)) + 3x^2 ((n),(3)) + … + nx^(n\ – 1) ((n),(n))`

`text(Let)\ \ x = 1`

`n (1 + 1)^(n\ – 1)`	`= 1 ((n),(1)) + 2 ((n),(2)) + 3 ((n),(3)) + … + n ((n),(n))`
`n 2^(n\ – 1)`	`= sum_(k = 1)^n k ((n),(k))\ text(… as required)`

Trigonometry, EXT1 T3 2010 HSC 6a

Show that `cos(A - B) = cos A cos B(1 + tan A tan B)`. (1 mark)

--- 3 WORK AREA LINES (style=lined) ---
Suppose that `0 < B < pi/2` and `B < A < pi`.
Deduce that if `tan Atan B = − 1`, then `A\ - B = pi/2`. (1 mark)

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

Show Answers Only

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

Show Worked Solution

i. `text(Show)\ \ cos(A\ – B) = cosA cosB (1 + tanA tanB)`

`text(RHS)`	`= cosA cosB (1 + (sinA sinB)/(cosA cosB))`
	`= cosA cos B + sinA sin B`
	`= cos(A – B)\ text(… as required)`

ii. `text(Given)\ \ tanA tanB = -1`

`cos (A – B)`	`= cosA cosB (1\ – 1)`
`cos (A – B)`	`= 0`
`A – B`	`= cos^(-1) 0`
	`= pi/2, (3pi)/2, …`

`text(S)text(ince)\ \ \ 0 < B < pi/2\ \ text(and)\ \ \ B < A < pi,`

`=> A\ – B = pi/2`

Trig Ratios, EXT1 2010 HSC 5a

A boat is sailing due north from a point `A` towards a point `P` on the shore line.

The shore line runs from west to east.

In the diagram, `T` represents a tree on a cliff vertically above `P`, and `L` represents a landmark on the shore. The distance `PL` is 1 km.

From `A` the point `L` is on a bearing of 020°, and the angle of elevation to `T` is 3°.

After sailing for some time the boat reaches a point `B`, from which the angle of elevation to `T` is 30°.

Show that

`qquad BP = (sqrt3 tan 3°)/(tan20°)`. (3 marks)
Find the distance `AB`. Give your answer to 1 decimal place. (1 mark)

Show Answers Only

`text(Proof)\ \ text{(See Worked Solutions)}`
`2.5\ text(km)\ \ text{(to 1 d.p.)}`

Show Worked Solution

(i) `text(Show)\ \ BP = (sqrt3 tan 3°)/(tan 20°)`

`text(In)\ Delta ATP`

`tan 3°`	`= (TP)/(AP)`
`=> AP`	`= (TP)/(tan 3)`

`text(In)\ Delta APL`

`tan 20°`	`= 1/(AP)`
`=> AP`	`= 1/tan 20`

`:. (TP)/(tan3)`	`= 1/(tan20)`
`TP`	`= (tan3°)/(tan20°)\ \ \ text(…)\ text{(} text(1)text{)}`

`text(In)\ \ Delta BTP`

`tan 30°`	`= (TP)/(BP)`
`1/sqrt3`	`= (TP)/(BP)`
`BP`	`= sqrt3 xx TP\ \ \ \ \ text{(using (1) above)}`
	`= (sqrt3 tan3°)/(tan20°)\ \ \ text(… as required)`

(ii)	`AB`	`= AP\ – BP`
	`AP`	`= 1/(tan20°)\ \ \ text{(} text(from part)\ text{(i)} text{)}`

`:.\ AB`	`= 1/(tan 20°)\ – (sqrt3 tan3°)/(tan20°)`
	`= (1\ – sqrt3 tan 3)/(tan20°)`
	`= 2.4980…`
	`= 2.5\ text(km)\ text{(to 1 d.p.)`

Quadratic, EXT1 2010 HSC 4c

The diagram shows the parabola `x^2 = 4ay`. The point `P(2ap, ap^2)`, where `p != 0`, is on the parabola.

The tangent to the parabola at `P`, `y = px − ap^2`, meets the `y`-axis at `L`.

The point `M` is on the directrix, such that `PM` is perpendicular to the directrix.

Show that `SLMP` is a rhombus. (3 marks)

Show Answers Only

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

Show Worked Solution

`text(Solution 1)`

`text(Show)\ \ SLMP\ \ text(is a rhombus)`

`L\ \ text(occurs when tangent meets)\ x text(-axis)`

`y = 0 – ap^2`

`:.L(0,–ap^2),\ \ M(2ap, –a)`

`m_(PS)`	`= (y_2\ – y_1)/(x_2\ – x_1)`
	`= (ap^2\ – a)/(2ap\ – 0)`
	`= (p^2\ – 1)/(2p)`
`m_(ML)`	`= (-a + ap^2)/(2ap\ – 0`
	`= (p^2\ – 1)/(2p)`

`text(S)text(ince)\ \ `	`PS \ text(\|\|)\ ML\ \ \ text{(same gradient)}`
`text(and)\ \ `	`PM \ text(\|\|)\ SL\ \ \ text{(vertical)}`
`=>`	`\ \ SLMP\ text(is a parallelogram)`

`text(S)text(ince)\ \ `	`PS = PM\ \ \ text{(definition of a parabola)}`
`=>`	`\ \ SLMP\ text(is a rhombus … as required.)`

`text(Alternate Solution)`

`L(0,–ap^2),\ \ M(2ap, –a)`

`m_(SM)`	`= (y_2\ – y_1)/(x_2\ – x_1) = (-a\ -a)/(2ap\ – 0) = – 1/p`
`m_(LP)`	`= p`

`text(S)text(ince)\ \ m_(SM) xx m_(LP) = p xx – 1/p = -1`

`=>\ text(Diagonals are perpendicular)`

`text(Midpoint)\ LP`	`= ( (0 + 2ap)/2,\ (-ap^2 + ap^2)/2)`
	`= (ap, 0)`
`text(Midpoint)\ MS`	`= ((0 + 2ap)/2,\ (a\ – a)/2)`
	`= (ap,\ 0)`

`=>\ text(Diagonals bisect)`

`:.\ text(S)text(ince diagonals are)\ _|_\ text(bisectors,)`

`SLMP\ text(is a rhombus … as required.)`

Trigonometry, EXT1 T3 2010 HSC 4b

Express `2 cos theta + 2 cos (theta + pi/3)` in the form `R cos (theta + alpha)`,

where `R > 0` and `0 < alpha < pi/2`. (3 marks)

--- 12 WORK AREA LINES (style=lined) ---
Hence, or otherwise, solve `2 cos theta + 2 cos (theta + pi/3) = 3`,
for `0 < theta < 2pi`. (2 marks)

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

Show Answers Only

`2 sqrt 3 cos (theta + pi/6)`
`(5pi)/3`

Show Worked Solution

i.	`2 cos theta + 2 cos (theta + pi/3)`
	`= 2 cos theta + 2 (cos theta cos (pi/3)\ – sin theta sin (pi/3))`
	`= 2 cos theta + 2 cos theta xx 1/2\ – 2 sin theta xx sqrt3/2`
	`= 2 cos theta + cos theta\ – sqrt3 sin theta`
	`= 3 cos theta\ – sqrt3 sin theta`

`R cos (theta + alpha) = R cos theta cos alpha – R sin theta sin alpha`

`R cos alpha`	`= 3\ \ \ \ \ `	`R sin alpha`	`= sqrt3`
`cos alpha`	`= 3/R\ \ \ \ \ `	`sin alpha`	`= sqrt3/R`

`tan alpha`	`= sin alpha/cos alpha = sqrt3/3 = 1/sqrt3`
`tan\ pi/6`	`=1/sqrt3`
`:. alpha`	`=pi/6\ \ \ \ \ (0 < alpha < pi/2)`

`R^2`	`= 3^2 + (sqrt3)^2`
	`= 9 + 3`
`R`	`= sqrt 12 = 2 sqrt3`

`:.\ 2 cos theta + 2 cos (theta + pi/3) = 2 sqrt 3 cos (theta + pi/6)`

♦ Mean mark part (ii) 49%
MARKER’S COMMENT: Many students did not check their answers against the stated domain for `theta`.

ii.	`2 cos theta + 2 cos(theta + pi/3)`	`= 3`
	`2 sqrt 3 cos (theta + pi/6)`	`= 3`
	`cos (theta + pi/6)`	`= 3/(2sqrt3) = sqrt3/2`
	`cos^(-1) (sqrt3/2)`	`= pi/6`

`text(S)text(ince cos is positive in 1st and 4th quadrants,)`

`theta + pi/6`	`= pi/6,\ 2pi\ – pi/6`
`:. theta`	`= (5pi)/3\ \ \ \ \ (0 < theta < 2pi)`

Mechanics, EXT2* M1 2010 HSC 4a

A particle is moving in simple harmonic motion along the `x`-axis.

Its velocity `v`, at `x`, is given by `v^2 = 24 − 8x − 2x^2`.

Find all values of `x` for which the particle is at rest. (1 mark)

--- 4 WORK AREA LINES (style=lined) ---
Find an expression for the acceleration of the particle, in terms of `x`. (1 mark)

--- 2 WORK AREA LINES (style=lined) ---
Find the maximum speed of the particle. (2 marks)

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

Show Answers Only

`x = –6\ \ text(or)\ \ 2`
`-4\ – 2x`
`4 sqrt 2`

Show Worked Solution

(i)

`v^2 = 24\ – 8x\ – 2x^2`

`text(Find)\ \ x\ \ text(when)\ \ v=0`

`24 – 8x – 2x^2`	`= 0`
`x^2 + 4x – 12`	`= 0`
`(x + 6)(x – 2)`	`= 0`
`x= -6\ \ text(or)\ \ 2`

`:.\ text(Particle at rest when)\ \ x = –6\ \ text(or)\ \ 2`

(ii)	`ddot x`	`= d/(dx) (1/2 v^2)`
		`= d/(dx) (12 – 4x – x^2)`
		`= -4 – 2x`

(iii) `text(Solution 1)`

`text(Max speed when)\ \ ddot x = 0`

`-4 – 2x`	`= 0`
`2x`	`= -4`
`x`	`= -2`

`text(At)\ \ x = -2`

MARKER’S COMMENT: While most students found `x=-2` correctly, too many made errors substituting this back in or failed to finish the answer by taking the square root. BE CAREFUL! .

`v^2`	`= 24 – 8(–2) – 2(–2)^2`
	`= 24 + 16 – 8`
	`= 32`
`v`	`= +- sqrt32`
	`= +- 4 sqrt2`

`:.\ text(Maximum speed is)\ \ 4 sqrt2`

`text(Alternate Answer)`

`text(Maximum speed occurs when)`

`x = (-6+2)/2 = –2`

`text(At)\ \ x = –2`

`v^2`	`= 32\ \ \ text{(see working above)}`
`v`	`= +- 4 sqrt 2`

`:.\ text(Maximum speed is)\ 4sqrt2`

Inverse Functions, EXT1 2010 HSC 3b

Let `f(x) = e^(-x^2)`. The diagram shows the graph `y = f(x)`.

The graph has two points of inflection.
Find the `x` coordinates of these points. (3 marks)
Explain why the domain of `f(x)` must be restricted if `f(x)` is to have an inverse function. (1 mark)
Find a formula for `f^(-1) (x)` if the domain of `f(x)` is restricted to `x ≥ 0`. (2 marks)
State the domain of `f^(-1) (x)`. (1 mark)
Sketch the curve `y = f^(-1) (x)`. (1 mark)
(1) Show that there is a solution to the equation `x = e^(-x^2)` between `x = 0.6` and `x = 0.7`. (1 mark)
(2) By halving the interval, find the solution correct to one decimal place. (1 mark)

Show Answers Only

`x = +- 1/sqrt2`
`text(There can only be 1 value of)\ y`
`text(for each value of)\ x.`
`f^(-1)x = sqrt(ln(1/x))`
`0 <= x <= 1`
(1) `text(Proof)\ \ text{(See Worked Solutions)}`
(2) `0.7\ text{(1 d.p.)}`

Show Worked Solution

(i)	`y`	`= e^(-x^2)`
	`dy/dx`	`= -2x * e^(-x^2)`
	`(d^2y)/(dx^2)`	`= -2x (-2x * e^(-x^2)) + e ^(-x^2) (-2)`
		`= 4x^2 e^(-x^2)\ – 2e^(-x^2)`
		`= 2e^(-x^2) (2x^2\ – 1)`

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

`2e^(-x^2) (2x^2\ – 1)`	`= 0`
`2x^2\ – 1`	`= 0`
`x^2`	`= 1/2`
`x`	`= +- 1/sqrt2`

COMMENT: It is also valid to show that `f(x)` is an even function and if a P.I. exists at `x=a`, there must be another P.I. at `x=–a`.

`text(When)\ \ `	`x < 1/sqrt2,`	`\ (d^2y)/(dx^2) < 0`
	`x > 1/sqrt2,`	`\ (d^2y)/(dx^2) > 0`

`=>\ text(Change of concavity)`

`:.\ text(P.I. at)\ \ x = 1/sqrt2`

`text(When)\ \ `	`x < – 1/sqrt2,`	`\ (d^2y)/(dx^2) > 0`
	`x > – 1/sqrt2,`	`\ (d^2y)/(dx^2) < 0`

`=>\ text(Change of concavity)`

`:.\ text(P.I. at)\ \ x = – 1/sqrt2`

(ii)	`text(In)\ f(x), text(there are 2 values of)\ y\ text(for)`
	`text(each value of)\ x.`
	`:.\ text(The domain of)\ f(x)\ text(must be restricted)`
	`text(for)\ \ f^(-1) (x)\ text(to exist).`

(iii)

`y = e^(-x^2)`

`text(Inverse function can be written)`

`x`	`= e^(-y^2),\ \ \ x >= 0`
`lnx`	`= ln e^(-y^2)`
`-y^2`	`= lnx`
`y^2`	`= -lnx`
	`=ln(1/x)`
`y`	`= +- sqrt(ln (1/x))`

`text(Restricting)\ \ x>=0,\ \ =>y>=0`

`:. f^(-1) (x)=sqrt(ln (1/x))`

♦ Parts (iv) and (v) were poorly answered with mean marks of 39% and 49% respectively.

(iv)

`f(0) = e^0 = 1`

`:.\ text(Range of)\ \ f(x)\ \ text(is)\ \ 0 < y <= 1`

`:.\ text(Domain of)\ \ f^(-1) (x)\ \ text(is)\ \ 0 < x <= 1`

(v)

(vi)(1) `x = e^(-x^2)`

`text(Let)\ g(x) = x\ – e^(-x^2)`

`g(0.6)`	`=0.6\ – e^(-0.6^2)`
	`=0.6\ – 0.6977 < 0`
`g(0.7)`	`=0.7\ – e^(-0.7^2)`
	`=0.7\ – 0.6126 > 0`
`=>g(x)\ text(changes sign)`

`:.\ g(x)\ \ text(has a root between 0.6 and 0.7)`

`:.\ x = e^(-x^2)\ \ text(has a solution between 0.6 and 0.7)`

♦ Mean mark 37%.
MARKER’S COMMENT: Better responses showed the change in sign between `g(0.65)` and `g(0.7)` as shown in the solution.

(vi)(2)	`g(0.65)`	`=0.65\ – e^(-0.65^2)`
		`=0.65\ – 0.655 < 0`

`:.\ text(A solution lies between 0.65 and 0.7)`

`:.\ x = 0.7\ \ text{(1 d.p.)}`

Combinatorics, EXT1 A1 2010 HSC 3a

At the front of a building there are five garage doors. Two of the doors are to be painted red, one is to be painted green, one blue and one orange.

How many possible arrangements are there for the colours on the doors? (1 mark)

--- 1 WORK AREA LINES (style=lined) ---
How many possible arrangements are there for the colours on the doors if the two red doors are next to each other? (1 mark)

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

Show Answers Only

`60`
`24`

Show Worked Solution

i.	`text(# Arrangements)`	`= (5!)/(2!)`
		`= 60`

♦ Mean mark 50%
MARKER’S COMMENT: Drawing a diagram was a successful strategy for many students in this part.

ii.

`text(When 2 red doors are side-by-side,)`

`text(# Arrangements)`	`= 4!`
	`= 24`

Calculus, EXT1 C1 2010 HSC 2d

A radio transmitter `M` is situated 6 km from a straight road. The closest point on the road to the transmitter is `S`.

A car is travelling away from `S` along the road at a speed of `text(100 km h)`⁻¹. The distance from the car to `S` is `x\ text(km)` and from the car to `M` is `r\ text(km)`.

Find an expression in terms of `x` for `(dr)/(dt)`, where `t` is time in hours. (3 marks)

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

Show Answers Only

`(100x)/sqrt(x^2 + 36)\ \ text(km/hr)`

Show Worked Solution

`text(Using Pythagoras,)`

`r^2`	`= x^2 + 6^2`
`r`	`= sqrt(x^2 + 36),\ r > 0`
`(dr)/(dt)`	`= (dx)/(dt) * (dr)/(dx)\ \ \ \ …\ (1)`
`(dx)/(dt)`	`= 100\ \ \ text{(given)}`
`(dr)/(dx)`	`= 1/2 (x^2 + 36)^(-1/2) xx d/(dx) (x^2 + 36)`
	`= x/sqrt(x^2 + 36)`

`text{Substituting into (1)}`
`:.\ (dr)/(dt)`	`= (100x)/sqrt(x^2 + 36)\ \ text(km/hr)`

Functions, EXT1 F2 2010 HSC 2c

Let `P(x) = (x + 1)(x-3) Q(x) + ax + b`,

where `Q(x)` is a polynomial and `a` and `b` are real numbers.

The polynomial `P(x)` has a factor of `x-3`.

When `P(x)` is divided by `x + 1` the remainder is `8`.

Find the values of `a` and `b`. (2 marks)

--- 6 WORK AREA LINES (style=lined) ---
Find the remainder when `P(x)` is divided by `(x + 1)(x-3)`. (1 mark)

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

Show Answers Only

`a = -2,\ b = 6`
` -2x + 6`

Show Worked Solution

i. `P(x) = (x+1)(x-3)Q(x) + ax + b`

`(x-3)\ \ text{is a factor (given)}`

`:. P (3)`	`= 0`
`3a + b`	`= 0\ \ \ …\ text{(1)}`

`P(x) ÷ (x+1)=8\ \ \ text{(given)}`

`:.P(-1)`	`= 8`
`-a + b`	`= 8\ \ \ …\ text{(2)}`

`text{Subtract (1) – (2)}`

`4a`	`= -8`
`a`	`= -2`

`text(Substitute)\ \ a = -2\ \ text{into (1)}`

`-6 + b`	`= 0`
`b`	`= 6`

`:. a= – 2, \ b=6`

ii. `P(x) -: (x + 1)(x-3)`

`= ((x+1)(x-3)Q(x)-2x + 6)/((x+1)(x-3))`

`= Q(x) + (-2x + 6)/((x+1)(x-3))`

`:.\ text(Remainder is)\ \ -2x + 6`

COMMENT: This question requires a fundamental understanding of the remainder theorem.

Statistics, EXT1 S1 2010 HSC 1f

Five ordinary six-sided dice are thrown.

What is the probability that exactly two of the dice land showing a four?

Leave your answer in unsimplified form. (1 mark)

Show Answers Only

`\ ^5C_2 (1/6)^2 (5/6)^3`

Show Worked Solution

COMMENT: Surprisingly, half of students sitting the exam got this wrong. Easily the most poorly answered part of Q1 in 2010.

`P(4) = 1/6`

`P(bar4) = 5/6`

`text(# Combinations of two 4’s) =\ ^5C_2`

`:.\ P text{(exactly two 4s)} =\ ^5C_2 (1/6)^2 (5/6)^3`

L&E, EXT1 2010 HSC 1c

Solve `ln(x + 6) = 2 ln x`. (3 marks)

Show Answers Only

`x = 3`

Show Worked Solution

`ln(x + 6)`	`= 2 lnx`
`ln(x + 6)`	`= ln x^2`
`x^2`	`= x + 6`
`x^2\ – x\ – 6`	`= 0`
`(x\ – 3)(x + 2)`	`= 0`
`:. x = 3\ \ text(or)\ \ -2`

`text(S)text(ince RHS)\ = 2lnx,\ \ x>0`

`:. x=3`

Trigonometry, EXT1 T1 2010 HSC 1b

Let `f(x) = cos^(-1) (x/2)`. What is the domain of `f(x)`? (1 mark)

Show Answers Only

`-2 <= x <= 2`

Show Worked Solution

`f(x) = cos^(-1) (x/2)`

`–1`	`<= x/2`	`<= 1`
`–2`	`<= x`	`<= 2`

`:.\ text(Domain of)\ \ f(x)\ \ text(is)\ \ \ –2 <= x <= 2`

Calculus, EXT1 C1 2011 HSC 7a

The diagram shows two identical circular cones with a common vertical axis. Each cone has height `h` cm and semi-vertical angle 45°.

The lower cone is completely filled with water. The upper cone is lowered vertically into the water as shown in the diagram. The rate at which it is lowered is given by

`(dl)/(dt) = 10`,

where `l` cm is the distance the upper cone has descended into the water after `t` seconds.

As the upper cone is lowered, water spills from the lower cone. The volume of water remaining in the lower cone at time `t` is `V` cm³.

Show that `V = pi/3(h^3\ - l^3)`. (1 mark)

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Find the rate at which `V` is changing with respect to time when `l = 2`. (2 marks)

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Find the rate at which `V` is changing with respect to time when the lower cone has lost `1/8` of its water. Give your answer in terms of `h`. (2 marks)

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

Show Answers Only

`text(Proof)\ \ text{(See Worked Solutions)}`
`-40 pi\ \ text(cm³)// text(sec)`
`(-5pih^2)/2\ text(cm³)// text(sec)`

Show Worked Solution

i. `text(Show that)\ V = pi/3 (h^3\ – l^3)`

♦ Mean mark 42%

`text(S)text(ince)\ \ tan45° = r/h=1`

`=>r=h`

`=>\ text(Radius of lower cone) = h`

`:.\ V text{(lower cone)}`	`= 1/3 pi r^2 h`
	`= 1/3 pi h^3`

`text(Similarly,)`

`V text{(submerged upper cone)} = 1/3 pi l^3`

`V text{(water left)}`	`= 1/3 pi h^3\ – 1/3 pi l^3`
	`= pi/3 (h^3\ – l^3)\ \ \ text(… as required)`

ii. `text(Find)\ (dV)/(dt)\ text(at)\ l = 2`

`(dV)/(dt)= (dV)/(dl) xx (dl)/(dt)\ …\ text{(1)}`

`=>(dl)/(dt)`

`= 10\ text{(given)}`

`text(Using)\ \ V`	`= pi/3 (h^3\ – l^3)\ \ \ \ text(from part)\ text{(i)}`
`=>(dV)/(dl)`	`= -3 xx pi/3 l^2`
	`= -pi l^2`

`text(At)\ \ l = 2,`

`text{Substitute into (1) above}`

`(dV)/(dt)`	`= -pi xx 2^2 xx 10`
	`= -40 pi\ \ \ text(cm³)//text (sec)`

iii. `text(Find)\ \ (dV)/(dt)\ \ text(when lower cone has lost)\ 1/8 :`

♦♦♦ Mean mark 12%
MARKER’S COMMENT: Many unsuccessful answers attempted to find an alternate version of `(dV)/(dt)`. Part (ii) directed students directly toward the correct strategy.

`text(Find)\ \ l\ \ text(when)\ \ V = 7/8 xx 1/3 pi h^3`

`pi/3 (h^3\ – l^3)`	`= 7/8 xx 1/3 pi h^3`
`h^3 -l^3`	`= 7/8 h^3`
`l^3`	`= 1/8 h^3`
`l`	`= h/2`

`text(When)\ \ l = h/2 ,`

`(dV)/(dt)`	`= -pi (h/2)^2 xx 10\ \ …\ text{(*)}`
	`= (-5pih^2)/2\ text(cm³)// text(sec)`

`:. V\ text(is decreasing at the rate of)\ \ (5 pi h^2)/2\ text(cm³)//text(sec).`

Geometry and Calculus, EXT1 2011 HSC 4a

Consider the function `f(x) = e^(-x)\ - 2e^(-2x)`.

Find `f prime (x)`. (1 mark)
The graph `y = f(x)` has one maximum turning point.
Find the coordinates of the maximum turning point. (2 marks)
Evaluate `f(ln2)`. (1 mark)
Describe the behaviour of `f(x)` as `x -> oo`. (1 mark)
Find the `y`-intercept of the graph `y = f(x)`. (1 mark)
Sketch the graph `y = f(x)` showing the features from parts (ii) - (v).
You are not required to find any points of inflection. (2 marks)

Show Answers Only

`-e^(-x) + 4e^(-2x)`
`text(Max T.P. at)\ (ln4, 1/8)`
`0`
`text(As)\ x -> oo, f(x) -> 0`
`-1`

Show Worked Solution

(i)	`f(x)`	`= e^(-x)\ – 2e^(-2x)`
	`f prime (x)`	`= -e^(-x) + 4e^(-2x)`

(ii)

`text(T.P. when)\ f prime (x) = 0`

`-e^(-x) + 4e^(-2x) = 0`

`text(Let)\ e^(-x) = X`

`- X + 4X^2`	`= 0`
`X (4X\ – 1)`	`= 0`

`X = 0\ \ text(or)\ \ 1/4`

`text(If)\ \ e^(-x) = 0,\ \ \ text(No solution)`

`text(If)\ \ e^(-x) = 1/4`

`ln e^(-x)`	`= ln (1/4)`
`-x`	`= ln 4^(-1)`
	`= -ln 4`
`x`	`= ln4`

ALGEBRA TIP: Note that `e^ln x=x` can often help calculations. This is easily proven by simply taking the `ln` of both sides of the equation.

`f″(x)`	`= e^(-x)\ – 8e^(-2x)`
`f″(ln4)`	`< 0 => text(MAX)`
`f (ln4)`	`= e^(-ln4)\ – 2e^(-2ln4)`
	`= e^(ln(1/4))\ – 2 e^(ln(1/16))`
	`= 1/4\ – 2(1/16)`
	`= 1/8`

`:.\ text(Maximum T.P. at)\ \ (ln4, 1/8)`

(iii)	`f(ln2)`	`= e^(-ln2)\ – 2e^(-2ln2)`
		`= e^(ln(1/2))\ – 2e^(ln(1/4))`
		`= 1/2\ – 2 xx 1/4`
		`= 0`

(iv)	`f(x)`	`= e^(-x)\ – 2e^(-2x)`
		`= e^(-x) (1\ – 2e^(-x))`
	`text(As)\ x -> oo, \ e^(-x) ->0,\ f(x) -> 0`

(v)

`y text(-intercept at)\ \ f(0)`

`f(0)`	`= e^0\ – 2e^0`
	`= -1`

(vi)

Quadratic, EXT1 2011 HSC 3b

The diagram shows two distinct points `P(t, t^2)` and `Q(1\ - t, (1\ - t)^2)` on the parabola `y = x^2`. The point `R` is the intersection of the tangents to the parabola at `P` and `Q`.

Show that the equation of the tangent to the parabola at `P` is `y = 2tx\ – t^2`. (2 marks)
Using part `text{(i)}`, write down the equation of the tangent to the parabola at `Q`. (1 mark)
Show that the tangents at `P` and `Q` intersect at
`R (1/2, t\ - t^2)`. (2 marks)
Describe the locus of `R` as `t` varies, stating any restriction on the `y`-coordinate. (2 marks)

Show Answers Only

`text(Proof)\ \ text{(See Worked Solutions)}`
`y = 2 (1 -t)x\ – (1\ – t)^2`
`text(Proof)\ \ text{(See Worked Solutions)}`
`text(Locus of)\ R\ text(is vertical line)`
`x = 1/2,\ y<1/4`

Show Worked Solution

(i)

`text(Show tangent at)\ P\ text(is)\ y = 2tx\ – t^2`

`y`	`= x^2`
`dy/dx`	`= 2x`

`x=t\ \ \ \ text(at)\ P`

`dy/dx = 2t`

`text(Find equation with)\ \ m = 2t\ \ text(through)\ \ P(t, t^2)`

`y\ – y_1`	`= m(x\ – x_1)`
`y\ – t^2`	`= 2t ( x\ – t)`
`y`	`= 2tx\ – 2t^2 + t^2`
	`= 2tx\ – t^2\ text(… as required)`

(ii) `text(T)text(angent at)\ Q\ text(has equation)`

MARKER’S COMMENT: Many students derived this equation rather than substituting the new parameter, costing them valuable time. This is a benefit of using the parametric approach.

`y = 2(1\ – t)x\ – (1\ – t)^2`

(iii) `text(Need to show)\ R(1/2,\ t\ – t^2)`

`R\ text(is at intersection of tangents)`

`2tx\ – t^2`	`= 2(1\ – t)x\ – (1\ – t)^2`
`2tx\ – t^2`	`= 2x\ – 2tx\ – 1 + 2t\ – t^2`
`4tx\ – 2x`	`= -1 + 2t\ – t^2 + t^2`
`2x(2t\ – 1)`	`= 2t\ – 1`
`2x`	`= 1`
`x`	`= 1/2`

`text(Using)\ \ y = 2tx – t^2\ \ text(when)\ \ x = 1/2`

`y`	`= 2t(1/2)\ – t^2`
	`= t\ – t^2`

`:.\ R(1/2, t\ – t^2)\ text(… as required)`

(iv) `text(Locus of)\ R`

♦♦ Mean mark of 22%.
MARKER’S COMMENT: Many students stated the locus as `y=t-t^2` rather than realising it had to be a straight line since `x=½`, and that `y=t-t^2` simply restricted the values of `y`.

`text(S)text(ince)\ x = 1/2\ text(is a constant)`

`R\ text(is a vertical line)`

`text(Now,)\ y = t\ – t^2 = t(1\ – t)`

`text(Graphically,)\ \ y\ \ text(has a maximum at)\ \ t = 1/2`

`text(Max)\ \ y = 1/2\ – (1/2)^2 = 1/4`

`=> y < 1/4\ \ text{(tangents can’t meet on parabola)}`

`:.\ text(Locus of)\ R\ text(is vertical line)\ x = 1/2,\ \ y<1/4`

Mechanics, EXT2* M1 2011 HSC 3a

The equation of motion for a particle undergoing simple harmonic motion is

`(d^2x)/(dt^2) = -n^2 x`,

where `x` is the displacement of the particle from the origin at time `t`, and `n` is a positive constant.

Verify that `x = A cos nt + B sin nt`, where `A` and `B` are constants, is a solution of the equation of motion. (1 mark)

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The particle is initially at the origin and moving with velocity `2n`.

Find the values of `A` and `B` in the solution `x = A cos nt + B sin nt`. (2 marks)

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When is the particle first at its greatest distance from the origin? (1 mark)

--- 5 WORK AREA LINES (style=lined) ---
What is the total distance the particle travels between `t = 0` and `t = (2pi)/n`? (1 mark)

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

Show Answers Only

`text(Proof)\ \ text{(See Worked Solutions)}`
`A = 0,\ B = 2`
`t = pi/(2n)`
`text(8 units)`

Show Worked Solution

i.	`x`	`= A cos nt + B sin nt`
	`(dx)/(dt)`	`= – An sin nt + Bn cos nt`
	`(d^2x)/(dt^2)`	`= – An^2 cos nt\ – Bn^2 sin nt`
		`= -n^2 (A cos nt + B sin nt)`
		`= -n^2 x\ \ \ text(… as required)`

ii. `text(At)\ \ t=0, \ x=0, \ v=2n:`

`x`	`= Acosnt + Bsinnt`
`0`	`= A cos 0 + B sin 0`
`:.A`	`= 0`

`text(Using)\ \ (dx)/(dt) = Bn cos nt`

`2n`	`= Bn cos 0`
`Bn`	`= 2n`
`:.B`	`= 2`

♦♦ Mean mark part (iii) 47%

iii. `text(Max distance from origin when)\ (dx)/(dt) = 0`

`(dx)/(dt)`	`= 2n cos nt`
`0`	`= 2n cos nt`
`cos nt`	`= 0`
`nt`	`= pi/2,\ (3pi)/2,\ (5pi)/2`
`t`	`= pi/(2n),\ (3pi)/(2n), …`

`:.\ text(Particle is first at greatest distance)`

`text(from)\ O\ text(when)\ t = pi/(2n).`

iv. `text(Solution 1)`

`text(Find the distance travelled from)\ \ t=0\ \→\ \ t=(2pi)/n`

`text{(i.e. 1 full period)}`

♦♦ Mean mark 22%
MARKER’S COMMENT: Many students found the displacement at `t` rather than the distance travelled while another common error found distance as 2 x amplitude.

`text(S)text(ince)\ \ x=2 sin (nt)`

`=> text(Amplitude)=2`

`:.\ text(Distance travelled)=4 xx2=8\ text(units)`

`text(Solution 2)`

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

`text(At)\ t`	`= pi/(2n)`
`x`	`= 2 sin (n xx pi/(2n)) = 2`

`text(At)\ t`	`= (3pi)/(2n)\ \ \ text{(i.e. the next time}\ \ (dx)/(dt) = 0 text{)}`
`x`	`= 2 sin (n xx (3pi)/(2n)) = -2`

`text(At)\ t`	`= (2pi)/n`
`x`	`= 2 sin (n xx (2pi)/n) = 0`

`:.\ text(Total distance travelled)`

`= 2 + 4 + 2`

`= 8\ \ text(units)`

Combinatorics, EXT1 A1 2011 HSC 2e

Alex’s playlist consists of 40 different songs that can be arranged in any order.

How many arrangements are there for the 40 songs? (1 mark)

--- 1 WORK AREA LINES (style=lined) ---
Alex decides that she wants to play her three favourite songs first, in any order.
How many arrangements of the 40 songs are now possible? (1 mark)

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

Show Answers Only

`40!`
`6 xx 37!`

Show Worked Solution

i.	`#\ text(Arrangements) = 40!`

ii.	`#\ text(Arrangements)`	`= 3! xx 37!`
		`= 6 xx 37!`

Trigonometry, EXT1 T1 2011 HSC 2d

Sketch the graph of the function `f(x) = 2arccos x`. Clearly indicate the domain and range of the function. (2 marks)

Show Answers Only

Show Worked Solution

`y`	`= 2 cos^(-1) x`
`y/2`	`= cos^(-1)x`

`text(Domain)\ -1 <= x <= 1`

`text(S)text(ince)\ \ 0 <= y/2 <= pi`

`text(Range)\ 0 <= y <= 2pi`

Combinatorics, EXT1 A1 2011 HSC 2c

Find an expression for the coefficient of `x^2` in the expansion of `(3x - 4/x)^8`. (2 marks)

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

Show Answers Only

`-870\ 912`

Show Worked Solution

`(3x – 4/x)^8 = sum_(k=0)^8\ ^8C_k *(3x)^(8\ – k) *(–4/x)^k`

`text(General term)`	`=\ ^8C_k * 3^(8\ – k) * x^(8\ – k) * (–1)^k4^k x^(-k)`
	`=\ ^8C_k(–1)^k 3^(8\ – k) * 4^k * x^(8\ – 2k)`

`text(Co-efficient of)\ \ x^2=2\ \ text(occurs when,)`

` 8 – 2k`	`= 2`
`2k`	`= 6`
`k`	`= 3`

`:.\ text(Co-efficient of)\ \ x^2`

`=\ ^8C_3 * (–1)^3*3^5 * 4^3`

`= -56 xx 243 xx 64`

`= -870\ 912`

Polynomials, EXT1 2011 HSC 2b

The function `f(x) = cos2x\ - x` has a zero near `x = 1/2`.

Use one application of Newton’s method to obtain another approximation to this zero. Give your answer correct to two decimal places. (3 marks)

Show Answers Only

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

Show Worked Solution

`f(x)`	`= cos2x\ – x`
`f prime (x)`	`= -2 sin 2x\ – 1`
`f(0.5)`	`=cos (2xx0.5) – 0.5`
	`=0.0403…`
`f′(0.5)`	`=-2 sin (2 xx 0.5) – 1`
	`=-2.6829…`

`x_1`	`= x_0\ – (f(0.5))/(f prime (0.5))`
	`= 0.5\ – (0.0403…)/(-2.6829…)`
	`= 0.5\ – (-0.01502…)`
	`= 0.51502…`
	`= 0.52\ \ \ text{(to 2 d.p.)}`

Mechanics, EXT2* M1 2012 HSC 14b

A firework is fired from `O`, on level ground, with velocity `70` metres per second at an angle of inclination `theta`. The equations of motion of the firework are

`x = 70t cos theta\ \ \ \ `and`\ \ \ y = 70t sin theta\ – 4.9t^2`. (Do NOT prove this.)

The firework explodes when it reaches its maximum height.

Show that the firework explodes at a height of `250 sin^2 theta` metres. (2 marks)

--- 6 WORK AREA LINES (style=lined) ---
Show that the firework explodes at a horizontal distance of `250 sin 2 theta` metres from `O`. (1 mark)

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For best viewing, the firework must explode at a horizontal distance between 125 m and 180 m from `O`, and at least 150 m above the ground.

For what values of `theta` will this occur? (3 mark)

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

Show Answers Only

`text(Proof)\ \ text{(See Worked Solutions)}`
`text(Proof)\ \ text{(See Worked Solutions)}`
`66.97^@… <= theta <= 75^@\ text(for best viewing)`

Show Worked Solution

i. `text(Show max height is)\ 250 sin^2 theta\ text(metres)`

`text(Firework explodes at max height)`

`text(Find)\ \ t\ \ text(when)\ dot y = 0`

`y = 70t sin theta\ – 4.9t^2`

`dot y = 70 sin theta\ – 9.8t`

`text(When)\ \ dot y = 0`

`70 sin theta\ – 9.8t`	`= 0`
`9.8t`	`=70 sin theta`
`t`	`= 70/9.8 sin theta`

`:.\ y_max`	`= 70 * 70/9.8 sin theta * sin theta\ – 4.9 * (70^2)/(9.8^2) sin^2 theta`
	`= 500 sin^2 theta\ – 250 sin^2 theta`
	`= 250 sin^2 theta\ text(metres)\ \ \ \ \ text(… as required)`

ii. `x = 70t cos theta`

`text(At)\ \ t = 70/9.8 sin theta`

`x`	`= 70 * 70/9.8 sin theta cos theta`
	`= 250 * 2 sin theta cos theta`
	`= 250 sin 2 theta\ text(metres)\ \ \ \ \ text(… as required)`

iii. `text(Best viewing when)`

♦♦ Mean mark 28%
MARKER’S COMMENT: Many students ignored the findings in earlier parts and tried unsuccessfully to solve inequalities that still contained `t`.

`125 <= x <= 180\ \ text(and)\ \ y >= 150`

`text(S)text(ince)\ x = 250 sin 2 theta`

`125`	`<= 250 sin 2 theta <= 180`
`1/2`	`<= sin 2 theta <= 18/25`

`text(In the 1st quadrant)`

`30^@ <= 2 theta <= 46.054…^@`

`15^@ <= theta <= 23.0^@`

`text(In the 2nd quadrant)`

`133.94…^@`	`<= 2 theta <= 150^@`
`67.0^@`	`<= theta <= 75^@`

`text{Using part (i)}`

`text(When)\ theta = 23.0^@`

`y_(max)`	`= 250 xx sin^2 23.0^@`
	`~~ 38 < 150text(m)`

`=> text(“Highest” max height for)\ \ 15°<=theta<=23.0\ \ text(does not satisfy.)`

`text(When)\ theta = 67.0…^@`

`y_text(max)`	`= 250 xx sin^2 67.0^@`
	`~~ 212 > 150 text(m)`

`=> text(“Lowest” max height for)\ \ 67°<=theta<=75°\ \ text(satisfies.)`

`:.\ 67.0^@ <= theta <= 75^@\ text( for best viewing)`.

Plane Geometry, EXT1 2012 HSC 14a

The diagram shows a large semicircle with diameter `AB` and two smaller semicircles with diameters `AC` and `BC`, respectively, where `C` is a point on the diameter `AB`. The point `M` is the centre of the semicircle with diameter `AC`.

The line perpendicular to `AB` through `C` meets the largest semicircle at the point `D`. The points `S` and `T` are the intersections of the lines `AD` and `BD` with the smaller semicircles. The point `X` is the intersection of the lines `CD` and `ST`.

Copy or trace the diagram into your writing booklet.

Explain why `CTDS` is a rectangle. (1 mark)
Show that `Delta MXS` and `Delta MXC` are congruent. (2 marks)
Show that the line `ST` is a tangent to the semicircle with diameter `AC`. (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)

`/_SDT = 90°\ text{(angle in semi-circle)}`

♦ Mean mark 49%

`/_ ASC = 90°\ \ text{(angle in semi-circle)}`

`=> /_ CSD = 90°\ \ text{(} /_ ASD\ text{is a straight angle)}`

`text(Similarly)`

`/_CTB = /_CTD=90°`

`/_SCT = 90°\ \ text{(angle sum of quadrilateral}\ CTDS text{)}`

`text(S)text(ince all angles are right angles,)`

`CTDS\ text(is a rectangle)`

(ii) `ST\ text(and)\ DC\ text(are diagonals of rectangle)`

`text(S)text(ince they bisect and are equal)`

`=> XS = XC`

`SM = CM\ text{(radii)}`

`MX\ text(is common)`

`:.\ Delta MXS -= Delta MXC\ text{(SSS)}`

♦ Mean mark part (iii) 41%
STRATEGY: The congruency proof in part (ii) provided the critical information to answer this efficiently. Keep previous parts of questions front and centre of your mind when working on a solution.

(iii) `/_ XSM = /_ XCM = 90°`

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

`text(S)text(ince)\ MS _|_ ST\ text(at circumference)`

`text(and)\ MS\ text(is a radius,)`

`=> ST\ text(is a tangent)`

Geometry and Calculus, EXT1 2012 HSC 13d

The concentration of a drug in the blood of a patient `t` hours after it was administered is given by

`C(t) = 1.4te^(–0.2t),`

where `C(t)` is measured in `text(mg/L)`.

Initially the concentration of the drug in the blood of the patient increases until it reaches a maximum, and then it decreases. Find the time when this maximum occurs. (3 marks)
Taking `t = 20` as a first approximation, use one application of Newton’s method to find approximately when the concentration of the drug in the blood of the patient reaches `0.3\ text(mg/L)`. (2 marks)

Show Answers Only

`t = 5\ \ text(hours.)`
`text(The concentration of the blood will be)`
`text(approx 0.3 mg/L after 22.8 hours.)`

Show Worked Solution

(i)	`\ \ C(t)`	`= 1.4 te^(-0.2t)`
	`(dC)/(dt)`	`= 1.4t (-0.2)e^(-0.2t) + 1.4 e^(-0.2t)`
		`= 1.4 e^(-0.2t) (1\ – 0.2t)`

`text(Max or min when)\ (dC)/(dt) = 0`

`1.4e^(-0.2t)`	`= 0 => text(has no solution)`
`(1 – 0.2t)`	`= 0`
`0.2t`	`= 1`
`t`	`= 5\ \ text(hours)`

`text(When)\ \ t < 5,\ (dC)/(dt)>0`

`text(When)\ \ t > 5,\ (dC)/(dt)<0`

`:.\ text(Maximum when)\ \ t = 5\ text(hours.)`

♦ Mean mark 45%
MARKER’S COMMENT: Only a minority of students could successfully apply Newton’s method in this instance. A must know area asked every year!

(ii)

`text(Taking)\ C = 0.3`

`0.3`	`= 1.4te^(-0.2t)`
`f(t)`	`= 1.4te^(-0.2t)\ – 0.3`
`f′(t)`	`= 1.4t(-0.2)e^(-0.2t) + 1.4e^(-0.2t)`
	`= 1.4 e^(-0.2t) (1\ – 0.2t)`
`f(20)`	`=1.4 xx 20 xx e^(-0.2 xx 20)\ – 0.3`
	`=0.21283…`
`f′(20)`	`= 1.4 e^(-0.2 xx 20) (1-0.2 xx20)`
	`=-0.0769…`
`t_2`	`= 20\ – f(20)/(f prime (20))`
	`= 20\ – (0.21283…)/(-0.0769…)`
	`= 22.766…`
	`= 22.8\ text(hours)\ \ \ text{(to 1 d.p.)}`

`:.\ text(The concentration of the blood will be)`

`text(approx 0.3 mg/L after 22.8 hours.)`

Geometry and Calculus, EXT1 2012 HSC 13b

Find the horizontal asymptote of the graph

`qquad qquad y=(2x^2)/(x^2 + 9)`. (1 mark)
Without the use of calculus, sketch the graph

`qquad qquad y=(2x^2)/(x^2 + 9)`,

showing the asymptote found in part (i). (2 marks)

Show Answers Only

`text(Horizontal asymptote at)\ y = 2`

Show Worked Solution

(i)	`y`	`= (2x^2)/(x^2 +9)`
		`= 2/(1 + 9/(x^2))`

`text(As)\ \ x -> oo,\ y ->2`

`text(As)\ \ x -> – oo,\ y -> 2`

`:.\ text(Horizontal asymptote at)\ y = 2`

(ii)

`text(At)\ \ x = 0,\ y = 0`

`f(x) = (2x^2)/(x^2 + 9) >= 0\ text(for all)\ x`

`f(–x) = (2(–x)^2)/((–x)^2 + 9) = (2x^2)/(x^2 + 9) = f(x)`

`text(S)text(ince)\ \ f(x) = f(–x) \ \ =>\ text(EVEN function)`

Trigonometry, EXT1 T1 2012 HSC 13a

Write `sin(2 cos ^(-1) (2/3))` in the form `a sqrtb`, where `a` and `b` are rational. (2 mark)

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

Show Answers Only

`4/9 sqrt5`

Show Worked Solution

TIP: This question is made less complicated by reminding yourself that `cos^(-1) (2/3)` is simply an angle.

`sin (2 cos^(-1) (2/3))`

`text(Let)\ \ x = cos ^(-1) (2/3)`

`:.\ cos x = 2/3`

`sin x = sqrt5/3`

`sin(2 cos^(-1) (2/3))`	`= sin 2x`
	`= 2 sin x cos x`
	`= 2 xx sqrt5/3 xx 2/3`
	`= 4/9 sqrt5`

Quadratic, EXT1 2012 HSC 12d

Let `A(0, –k)` be a fixed point on the `y`-axis with `k > 0`. The point `C(t, 0)` is on the `x`-axis. The point `B(0, y)` is on the `y`-axis so that `Delta ABC` is right-angled with the right angle at `C`. The point `P` is chosen so that `OBPC` is a rectangle as shown in the diagram.

Show that `P` lies on the parabola given parametrically by (2 marks)
1. `x = t\ \ ` and`\ \ y = (t^2)/k`.
Write down the coordinates of the focus of the parabola in terms of `k`. (1 mark)

Show Answers Only

`text(Proof) text{(See Worked Solutions)}`
`text(Focus is)\ (0, k/4)`

Show Worked Solution

♦♦ Mean mark 30%.
IMPORTANT: The highlighted right-angle in this question should flag the potential use of `m_1 xx m_2=–1`.

`A(0,–k)\ \ \ B(0,y)\ \ \ C(t,0)`

`m`	`= (y_2\ – y_1)/(x_2\ – x_1)`
`m_(AC)`	`= (0 + k)/(t\ – 0) = k/t`
`m_(BC)`	`= (y\ – 0)/(0\ – t) = -y/t`

`text(S)text(ince)\ Delta ABC\ text(is right-angled)`

`m_(AC) xx m_(BC)`	`= -1`
`k/t xx (-y)/t`	`= -1`
`-yk`	`= -t^2`
`y`	`= (t^2)/k`

`text(S)text(ince)\ OBPC\ text(is a rectangle)`

`=>P\ text(has the same)\ x text(-coordinate as)\ C`

`text(and the same)\ y text(-coordinate as)\ B`

`:.P\ \ text(has coordinates)\ \ (t, (t^2)/k)`

`:.P\ text(lies on the parabola where)`

`x = t,\ \ y = (t^2)/k`

(ii)	`y`	`= (x^2)/k`
	`x^2`	`= ky`

`text(Using)\ \ x^2 = 4ay`

`4a`	`=k`
`a`	`=k/4`

`text(Parabola has vertex)\ \ (0,0)\ \ text(and)\ \ a = k/4`

`:.\ text(Focus is)\ \ (0, k/4)`

Statistics, EXT1 S1 2012 HSC 12c

Kim and Mel play a simple game using a spinner marked with the numbers 1, 2, 3, 4 and 5.

The game consists of each player spinning the spinner once. Each of the five numbers is equally likely to occur.

The player who obtains the higher number wins the game.

If both players obtain the same number, the result is a draw.

Kim and Mel play one game. What is the probability that Kim wins the game? (1 mark)

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Kim and Mel play six games. What is the probability that Kim wins exactly three games? (2 marks)

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

Show Answers Only

`2/5`
`864/3125`

Show Worked Solution

i. `text(Method 1)`

`P text{(Kim wins game)}`

`= P text{(K spins 5)} xx P text{(M<5)} + P text{(K spins 4)} xx P text{(M<4)} +\ …`

`= (1/5 xx 4/5) + (1/5 xx 3/5) + (1/5 xx 2/5) + (1/5 xx 1/5)`

`= 10/25`

`= 2/5`

♦ Mean mark 37%

`text(Method 2)`

`P text{(Draw)} = 1/5`

`:.\ P text{(Not a draw)}= 1\ – 1/5=4/5`

`text(S)text(ince Kim and Mel have equal chance)`

`P text{(K wins)}`	`= 1/2 xx 4/5`
	`= 2/5`

ii. `P text{(Kim wins)} = 2/5`

`P text{(Kim doesn’t win)} = 3/5`

`text(After 6 games,)`

`P text{(Kim wins exactly 3)}`

`=\ ^6C_3 (2/5)^3 (3/5)^3`

`= (6!)/(3!3!) xx 8/125 xx 27/125`

`= 864/3125`

Functions, EXT1 F1 2012 HSC 12b

Let `f(x) = sqrt(4x-3)`

Find the domain of `f(x)`. (1 mark)

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Find an expression for the inverse function `f^(-1) (x)`. (2 marks)

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Find the points where the graphs `y = f(x)` and `y=x` intersect. (1 mark)

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On the same set of axes, sketch the graphs `y = f(x)` and `y = f^(-1) (x)` showing the information found in part (iii). (2 marks)

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

Show Answers Only

`x >= 3/4`
`f^(-1) (x) = 1/4 x^2 + 3/4,\ x >= 0`
`(1,1),\ (3,3)`

Show Worked Solution

i. `f(x) = sqrt(4x-3)`

`text(Domain exists when)\ \ 4x-3`	`>= 0`
`4x`	`>= 3`
`x`	`>= 3/4`

ii. `text(Inverse function when)`

`x`	`= sqrt(4y-3)`
`x^2`	`= 4y-3`
`4y`	`= x^2 + 3`
`y`	`= 1/4 x^2 + 3/4`

`text(S)text(ince range of)\ \ f(x)\ \ text(is)\ \ y >= 0`

`=> text(Domain of)\ \ f^(-1) (x)\ \ text(is)\ \ x >= 0`

`:.\ f^(-1) (x) = 1/4 x^2 + 3/4,\ \ \ \ x >= 0`

iii. `text(Find intersection)`

`y`	` = sqrt(4x-3)\ \ …\ text{(1)}`
`y `	`=x\ \ …\ text{(2)}`

`text{Intersection occurs when}`

`x`	`= sqrt(4x-3)`
`x^2`	`= 4x-3`
`x^2-4x + 3`	`= 0`
`(x-3)(x-1)`	`= 0`
`x`	`=1\ \ text(or 3)`

`:.\ text(Intersection at)\ \ (1,1)\ \ text(and)\ \ (3,3)`

iv.

`qquad`

Proof, EXT1 P1 2012 HSC 12a

Use mathematical induction to prove that `2^(3n)\ – 3^n` is divisible by `5` for `n >= 1`. (3 marks)

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

Show Answers Only

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

Show Worked Solution

`text(Prove by induction)`

`2^(3n)\ – 3^n\ text(is divisible by)\ 5\ text(for)\ n >= 1`

`text(If)\ n = 1`

`2^(3n)\ – 3^n`	`= 2^3\ – 3^1`
	`= 5\ \ \ \ text{(divisible by 5)}`

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

IMPORTANT: Making `3^k` or `2^(3k)` the subject helps for the substitution required in the proof for `n+1`. Important to state that `P` is an integer.

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

`text(i.e.)\ \ 2^(3k)\ – 3^k`	`= 5P\ \ \ \ text{(} P\ text{integer)}`
`3^k`	`= 2^(3k)\ – 5P\ \ …\ text{(1)}`

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

`2^(3(k+1))\ – 3^(k+1)`	`= 2^(3k + 3)\ – 3 * 3^k`
	`= 2^3 * 2^(3k)\ – 3(2^(3k)\ – 5P)\ \ \ text{(from (1) above)}`
	`= 8 * 2^(3k)\ – 3 * 2 ^(3k) + 15P`
	`= 5 * 2^(3k) + 15P`
	`= 5 (2^(3k) + 3P)`

`=>\ 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 2012 HSC 11f

Use the binomial theorem to find an expression for the constant term in the expansion of

`(2x^3 - 1/x)^12`. (2 marks)

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For what values of `n` does `(2x^3 - 1/x)^n` have a non-zero constant term? (1 mark)

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

Show Answers Only

`-1760`
`n\ text(must be a multiple of 4)`

Show Worked Solution

i. `text(General term)`

`\ ^12C_k * (2x^3)^(12\ – k) (-1/x)^k`

`=\ ^12C_k * (–1)^k *2^(12\ – k) * x^(36\ – 3k) * x^(-k)`

`=\ ^12C_k * (–1)^k*2^(12\ – k) * x^(36\ – 4k)`

`text(Constant term occurs when)`

`36\ – 4k`	`= 0`
`k`	`= 9`

`:.\ text(Constant term)`	`=\ ^12C_9 * (–1)^9*2^3`
	`= – (12!)/(3!9!) xx 8`
	`= – 1760`

ii. `text(General term of)\ (2x^3\ – 1/x)^n`

♦♦♦ Mean mark 16%.

`\ ^nC_k * (2x^3)^(n\ – k) (–1/x)^k`

`=\ ^nC_k * 2^(n\ – k) * x^(3n\ – 3k) * (–1)^k * x^(-k)`

`=\ ^nC_k * (–1)^k*2^(n\ -k) * x^(3n\ – 4k)`

`text(Constant term when)\ \ 3n\ – 4k = 0.`

`text(i.e.)\ \ k=3/4n`

`text(S)text(ince)\ n\ text(and)\ k\ text(must be integers,)\ \ n\ \ text(must)`

`text(be a multiple of 4.)`

Combinatorics, EXT1 A1 2012 HSC 11e

In how many ways can a committee of 3 men and 4 women be selected from a group of 8 men and 10 women? (1 mark)

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

Show Answers Only

`11\ 760`

Show Worked Solution

`text(# Combinations)`	`=\ ^8C_3 xx\ ^10C_4`
	`= (8!)/(5!3!) xx (10!)/(6!4!)`
	`= 56 xx 210`
	`= 11\ 760`

Plane Geometry, EXT1 2012 HSC 10 MC

The points `A`, `B` and `P` lie on a circle centred at `O`. The tangents to the circle at `A` and `B` meet at the point `T`, and `/_ATB = theta`.

What is `/_APB` in terms of `theta`?

`theta/2`
`90^@-theta/2`
`theta`
`180^@-theta`

Show Answers Only

`B`

Show Worked Solution

`/_ BOA= 2 xx /_ APB`

`text{(angles at centre and circumference on arc}\ AB text{)}`

`/_TAO = /_ TBO = 90^@\ text{(angle between radius and tangent)}`

`:.\ theta + /_BOA`	`= 180^@\ text{(angle sum of quadrilateral}\ TAOB text{)}`
`theta + 2 xx /_APB`	`= 180^@`
`2 xx /_APB`	`= 180^@-theta`
`/_APB`	`= 90^@-theta/2`

`=> B`

Functions, EXT1 F2 2012 HSC 8 MC

When the polynomial `P(x)` is divided by `(x + 1)(x-3)`, the remainder is `2x + 7`.

What is the remainder when `P(x)` is divided by `x-3`?

`1`
`7`
`9`
`13`

Show Answers Only

`D`

Show Worked Solution

`text(Let)\ \ P(x) =A(x) * Q(x) + R(x)`

`text(where)\ \ A(x) = (x + 1)(x-3),\ text(and)\ \ R(x)=2x+7`

`text(When)\ \ P(x) -: (x-3),\ text(remainder) = P(3)`

`P(3)`	`= 0 + R(3)`
	`= (2 xx 3) + 7`
	`= 13`

`=> D`

Mechanics, EXT2* M1 2012 HSC 6 MC

A particle is moving in simple harmonic motion with displacement `x`. Its velocity `v` is given by

`v^2 = 16(9 − x^2)`.

What is the amplitude, `A`, and the period, `T`, of the motion?

`A = 3\ \ \ text(and)\ \ \ T = pi/2`
`A = 3\ \ \ text(and)\ \ \ T = pi/4`
`A = 4\ \ \ text(and)\ \ \ T = pi/3`
`A = 4\ \ \ text(and)\ \ \ T = (2pi)/3`

Show Answers Only

`A`

Show Worked Solution

`v^2 = 16(9 – x^2)`

`text(Find amplitude and period of motion)`

`v^2`	`= n^2(A^2 – x^2)`
`A^2`	`= 9`
`:.\ A`	`=3,\ \ \ (A > 0)`
`n^2`	`= 16`
`n`	`=4,\ \ \ (n>0)`

`:. T`	`= (2pi)/n`
	`= pi/2`

`=> A`

Combinatorics, EXT1 A1 2012 HSC 5 MC

How many arrangements of the letters of the word `OLYMPIC` are possible if the `C` and the `L` are to be together in any order?

`5!`
`6!`
`2 xx 5!`
`2 xx 6!`

Show Answers Only

`D`

Show Worked Solution

`text(S)text(ince)\ C\ text(and)\ L\ text(must be kept together, they)`

`text(act as 1 letter with 2 possible combinations.)`

`:.\ text(Total combinations)`

`= 2 xx 6 xx 5 xx 4 xx 3 xx 2 xx 1`

`= 2 xx 6!`

`=> D`

Trigonometry, EXT1 T1 2012 HSC 4 MC

Which function best describes the following graph?

`y = 3sin^(−1) 2x`
`y = 3/2 sin^(−1) 2x`
`y = 3sin^(−1)\ x/2`
`y = 3/2 sin^(−1)\ x/2`

Show Answers Only

`C`

Show Worked Solution

`text{Domain:}`	`\ \ -2 <= x <= 2`
	`\ \ -1 <= x/2 <= 1`

`:.\ text(Graph is)\ \ y = a sin^(-1)\ x/2`

`text(When)\ \ x = 2,\ \ y = (3pi)/2:`

`(3pi)/2`	`=a sin^(-1) 1`
`(3pi)/2`	`=a xx pi/2`
`a`	`= 3`
`:.\ y`	`= 3 sin ^(-1)\ x/2`

`=> C`

Calculus, 2ADV C3 2009 HSC 10

`text(Let)\ \ f(x) = x - (x^2)/2 + (x^3)/3`

Show that the graph of `y = f(x)` has no turning points. (2 marks)

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Find the point of inflection of `y = f(x)`. (1 mark)

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i. Show that `1 - x + x^2 - 1/(1 + x) = (x^3)/(1 + x)` for `x != -1`. (1 mark)

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ii. Let `g(x) = ln (1 + x)`.

Use the result in part c.i. to show that `f prime (x) >= g prime (x)` for all `x >= 0`. (2 marks)

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

Sketch the graphs of `y = f(x)` and `y = g(x)` for `x >= 0`. (2 marks)

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Show that `d/(dx) [(1 + x) ln (1 + x) - (1 + x)] = ln (1 + x)`. (2 marks)

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Find the area enclosed by the graphs of `y = f(x)` and `y = g(x)`, and the straight line `x = 1`. (2 marks)

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

Show Answers Only

`text{Proof (See Worked Solutions)}`
`(1/2, 5/12)`
i. `text{Proof (See Worked Solutions)}`

ii. `text{Proof (See Worked Solutions)}`
`text{Proof (See Worked Solutions)}`
`1 5/12\ – 2ln2\ \ text(u²)`

Show Worked Solution

a.	`f(x) = x\ – (x^2)/2 + (x^3)/3`

♦♦ Mean mark 28% for all of Q10 (note that data for each question part is not available).

`text(Turning points when)\ f prime (x) = 0`

`f prime (x) = 1\ – x + x^2`

`x^2\ – x + 1 = 0`

`text(S)text(ince)\ \ Delta`	`= b^2\ – 4ac`
	`= (–1)^2\ – 4 xx 1 xx 1`
	`= -3 < 0 => text(No solution)`

`:.\ f(x)\ text(has no turning points)`

b.	`text(P.I. when)\ f″(x) = 0`

`f″(x)`	`= -1 + 2x = 0`
`2x`	`= 1`
`x`	`= 1/2`

`text(Check for change in concavity)`

`f″(1/4)`	`= -1/2 < 0`
`f″(3/4)`	`= 1/2 > 0`

`=>\ text(Change in concavity)`

`:.\ text(P.I. at)\ \ x = 1/2`

`f(1/2)`	`= 1/2\ – ((1/2)^2)/2 + ((1/2)^3)/3`
	`= 1/2\ – 1/8 + 1/24`
	`= 5/12`

`:.\ text(Point of Inflection at)\ (1/2, 5/12)`

c.i.

`text(Show)\ 1\ – x + x^2\ – 1/(1 + x) = (x^3)/(1 + x),\ \ \ x != -1`

`text(LHS)`	`= (1+x)/(1+x)\ – (x(1+x))/(1+x) + (x^2(1+x))/((1+x))\ – 1/(1+x)`
	`= (1 + x\ – x\ – x^2 + x^2 + x^3\ – 1)/(1+x)`
	`= (x^3)/(1+x)\ \ \ text(… as required)`

c.ii.	`text(Let)\ g(x) = ln(1+x)`
	`g prime (x) = 1/(1 + x)`

`f prime (x)\ – g prime (x)`	`= 1\ – x + x^2\ – 1/(1+x)`
	`= (x^3)/(1 + x)\ \ text{(using part (i))}`

`text(S)text(ince)\ (x^3)/(1 + x) >= 0\ text(for)\ x >= 0`

`f prime (x)\ – g prime (x) >= 0`

`f prime (x) >= g prime (x)\ text(for)\ x >= 0`

MARKER’S COMMENT: When 2 graphs are drawn on the same set of axes, you must label them.

e.	`text(Show)\ d/(dx) [(1 + x) ln (1 + x)\ – (1 + x)] = ln (1 + x)`
	`text(Using)\ d/(dx) uv=uv′+vu′`

`text(LHS)`	`= (1+x) xx 1/(1 + x) + ln(1+x)xx1 + – 1`
	`= 1+ ln(1+x)\ – 1`
	`= ln(1+x)`
	`=\ text(RHS … as required)`

f.	`text(Area)`	`= int_0^1 f(x)\ – g(x)\ dx`
		`= int_0^1 (x\ – (x^2)/2 + (x^3)/3\ – ln(x+1))\ dx`
		`= [x^2/2\ – x^3/6 + (x^4)/12\ – (1 + x) ln (1+x) + (1+x)]_0^1`
		`text{(using part (e) above)}`
		`= [(1/2 – 1/6 + 1/12 – (2)ln2 + 2) – (ln1 + 1)]`
		`= 5/12\ – 2ln2 + 2\ – 1`
		`= 1 5/12\ – 2 ln 2\ \ text(u²)`

Trigonometry, 2ADV T3 2009 HSC 7b

Between 5 am and 5 pm on 3 March 2009, the height, `h`, of the tide in a harbour was given by

`h = 1 + 0.7 sin(pi/6 t)\ \ \ text(for)\ \ 0 <= t <= 12`

where `h` is in metres and `t` is in hours, with `t = 0` at 5 am.

What is the period of the function `h`? (1 mark)

--- 2 WORK AREA LINES (style=lined) ---
What was the value of `h` at low tide, and at what time did low tide occur? (2 marks)

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A ship is able to enter the harbour only if the height of the tide is at least 1.35 m.

Find all times between 5 am and 5 pm on 3 March 2009 during which the ship was able to enter the harbour. (3 marks)

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

Show Answers Only

`12\ text(hours)`
`text(2pm)\ \ text{(5am + 9 hours)}`
`text(6am to 10am)`

Show Worked Solution

i. `h = 1 + 0.7 sin (pi/6 t)\ \ text(for)\ 0 <= t <= 12`

`T`	`= (2pi)/n\ \ text(where)\ n = pi/6`
	`= 2 pi xx 6/pi`
	`= 12\ text(hours)`

`:.\ text(The period of)\ h\ text(is 12 hours.)`

ii. `text(Find)\ h\ text(at low tide)`

IMPORTANT: Using `sin x=–1` for a minimum here is very effective and time efficient. This property of trig functions is often very useful in harder questions.

`=> h\ text(will be a minimum when)`

`sin(pi/6 t) = -1`

`:.\ h_text(min)`	`= 1 + 0.7(-1)`
	`= 0.3\ text(metres)`

`text(S)text(ince)\ \ sinx = -1\ \ text(when)\ \ x = (3pi)/2`

`pi/6 t`	`= (3pi)/2`
`t`	`= (3pi)/2 xx 6/pi`
	`= 9\ text(hours)`

`:.\ text{Low tide occurs at 2pm (5 am + 9 hours)}`

iii. `text(Find)\ \ t\ \ text(when)\ \ h >= 1.35`

`1 + 0.7 sin (pi/6 t)`	`>= 1.35`
`0.7 sin (pi/6 t)`	`>= 0.35`
`sin (pi/6 t)`	`>= 1/2`

`sin (pi/6 t)`	`= 1/2\ text(when)`
`pi/6 t`	`= pi/6,\ (5pi)/6,\ (13pi)/6,\ text(etc …)`

`t`	`= 1,\ 5\ \ \ \ \ \ (0 <= t <= 12)`

`text(From the graph,)`

`sin(pi/6 t) >= 1/2\ \ \ text(when)\ \ 1 <= t <= 5`

`:.\ text(Ship can enter the harbour between 6 am and 10 am.)`

Calculus, EXT1* C3 2009 HSC 6a

The diagram shows the region bounded by the curve `y = sec x`, the lines `x = pi/3` and `x = -pi/3`, and the `x`-axis.

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

Show Answers Only

`2 sqrt 3 pi\ text(u³)`

Show Worked Solution

`V`	`= pi int_(-pi/3)^(pi/3) y^2\ dx`
	`= pi int_(-pi/3)^(pi/3) sec^2x\ dx`
	`= pi [tanx]_(-pi/3)^(pi/3)`
	`= pi[tan(pi/3) – tan(-pi/3)]`
	`= pi [sqrt3\ – (-sqrt3)]`
	`= 2 sqrt3 pi\ text(u³)`

`D(t)`	`= _\|_\ text(dist of)\ (t, t^2 + 3)\ text(from)\ 2x\ – y\ – 1 = 0`
`D`	`= \|(ax_1 + by_1 + c)/sqrt(a^2 + b^2)\|`
	`= \|(2(t)\ – 1 (t^2 + 3)\ – 1)/sqrt(2^2 + (-1)^2)\|`
	`= \|(2t\ – t^2\ – 3\ – 1)/sqrt5\|`
	`= \|(-(t^2\ – 2t + 4))/sqrt5\|`