Band 5 – Page 79

Plane Geometry, 2UA 2009 HSC 4c

In the diagram, `Delta ABC` is a right-angled triangle, with the right angle at `C`. The midpoint of `AB` is `M`, and `MP _|_ AC`.

Copy or trace the diagram into your writing booklet.

Prove that `Delta AMP` is similar to `Delta ABC`. (2 marks)
What is the ratio of `AP` to `AC`? (1 mark)
Prove that `Delta AMC` is isosceles. (2 marks)
Show that `Delta ABC` can be divided into two isosceles triangles. (1 mark)
Copy or trace this triangle into your writing booklet and show how to divide it into four isosceles triangles. (1 mark)

Show Answers Only

`text(Proof)\ \ text{(See Worked Solutions)}`
`text(Ratio)\ AP:AC = 1:2`
`text(Proof)\ \ text{(See Worked Solutions)}`
`text(Proof)\ \ text{(See Worked Solutions)}`

Show Worked Solution

(i)	`text(Need to prove)\ Delta AMP\ text(\|\|\|)\ Delta ABC`
	`/_ PAM\ text(is common)`
	`/_ MPA = /_BCA = 90°\ \ \ text{(given)}`
	`:.\ Delta AMP \ text(\|\|\|) \ Delta ABC\ \ \ text{(equiangular)}`

(ii)	`(AP)/(AC) = (AM)/(AB)\ \ \ `	`text{(corresponding sides of}`
		`text{similar triangles)}`

`text(S)text(ince)\ \ AB = 2 xx AM`

`(AP)/(AC) = 1/2`

`:.\ text(Ratio)\ AP:AC = 1:2`

(iii)

`AP = PC \ \ \ text{(from part (ii))}`

`PM\ text(is common)`

`/_APM = /_CPM = 90°\ \ text{(∠}\ APC\ text{is a straight angle)}`

`:.\ Delta AMP ~= Delta CMP\ text{(SAS)}`

`=> AM = CM\ \ `	`text{(corresponding sides of}`
	`text{congruent triangles)}`

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

(iv)	`text(S)text(ince)\ \ AM`	`= MC\ \ \ text{(part (iii))}`
	`AM`	`= MB\ text{(given)}`
	`:. MC`	`= MB`

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

`:. Delta ABC\ text(can be divided into 2 isosceles)`

`text(triangles)\ (Delta AMC\ text(and)\ Delta MCB text{)}`

(v)

♦♦ Mean mark 25%.
MARKER’S COMMENT: Use a ruler, draw large diagrams and always pay careful attention to previous parts of the question!

`/_AFB = /_CFB = 90°`

`DF\ text(bisects)\ AB`

`EF\ text(bisects)\ BC`

`text(From part)\ text{(iv)}\ text(we get 4 isosceles)`

`text(triangles as shown.)`

Plane Geometry, 2UA 2010 HSC 10a

In the diagram `ABC` is an isosceles triangle with `AC = BC = x`. The point `D` on the interval `AB` is chosen so that `AD = CD`. Let `AD = a`, `DB = y` and `/_ADC = theta`.

Show that `Delta ABC` is similar to `Delta ACD`. (2 marks)
Show that `x^2 = a^2 + ay`. (1 mark)
Show that `y = a(1 − 2 cos theta )`. (2 marks)
Deduce that `y <= 3a`. (1 mark)

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

Show Worked Solution

MARKER’S COMMENT: When dealing with multiple triangles, the best performing students label all angles clearly and unambiguously.

(i)

`/_CAD\ text(is common)`

`/_CAD = /_ACD = /_DBC`

`text{(Angles opposite equal sides in}`

`\ \ text{isosceles}\ Delta ACD\ text(and)\ Delta ABC text{)}`

`/_ADC = /_ACB\ \ (180^@\ text(in)\ Delta text{)}`

`:.\ Delta ABC\ text(|||) \ Delta ACD\ \ \ text{(AAA)}\ \ \ text(… as required)`

♦♦ Mean mark 25%.
MARKER’S COMMENT: The similarity proof in part (i) should have immediately alerted students to using similar ratios of sides to solve part (ii). Many did not follow this obvious hint.

(ii) `text(Using similarity)`

`(AC)/(AD)`	`= (AB)/(CB)`	`\ \ text{(corresponding sides of}` `\ \ \ \ text{similar triangles)}`
`x/a`	`= (a + y)/x`

`:.\ x^2 = a^2 + ay\ \ \ text(… as required)`

(iii) `text(Show)\ \ y = a(1\ – 2 cos theta)`

`text(Using Cosine Rule:)`

`cos theta`	`= (a^2 + a^2\ – x^2)/(2 xx a xx a)`
`2a^2costheta`	`= 2a^2\ – x^2`
	`= 2a^2\ – (a^2 + ay)\ \ text{(from part (ii))}`
	`= a^2\ – ay`
`ay`	`= a^2\ – 2a^2cos theta`
`y`	`= a\ – 2a cos theta`
	`= a (1\ – 2 cos theta)\ \ \ \ text(… as required)`

(iv) `text(S)text(ince)\ 1 <= cos theta <= 1`

♦♦♦ Parts (iii) and (iv) mean marks – 18% and just 4% respectively.
IMPORTANT: Limits of trig functions are often extremely useful in proving inequalities (see Worked Solutions).

` -2 <= 2 cos theta <= 2`

` -1 <= 1\ – 2 cos theta <= 3`

`y`	`=a(1\ – 2 cos theta)\ \ \ text{(from part (iii))}`
`:.\ y`	`<= a(3)`
	`<= 3a\ \ …\ text(as required)`

Calculus, 2ADV C3 2010 HSC 9b

Let `y=f(x)` be a function defined for `0 <= x <= 6`, with `f(0)=0`.

The diagram shows the graph of the derivative of `f`, `y = f^{′}(x)`.

The shaded region `A_1` has area 4 square units. The shaded region `A_2` has area 4 square units.

For which values of `x` is `f(x)` increasing? (1 mark)

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What is the maximum value of `f(x)`? (1 mark)

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Find the value of `f(6)`. (1 mark)

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Draw a graph of `y =f(x)` for `0 <= x <= 6`. (2 marks)

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Show Answers Only

`f(x)\ text(is increasing when)\ 0 <= x < 2`
`text(MAX value of)\ f(x) = 4`
`-6`

Show Worked Solution

i.	`f(x)\ text(is increasing when)\ \ f^{′}(x) > 0`
	`text(From the graph)`
	`f(x)\ text(is increasing when)\ 0 <= x < 2`

ii.	`f^{′}(x) = 0\ \ text(when)\ \ x=2`
	`:.\ text(MAX at)\ \ x = 2`
	`int_0^2 f^{′}(x)\ dx = 4\ \ \ (text(given since)\ A_1 = 4 text{)}`
	`text(We also know)`

`int_0^2\ f^{′}(x)\ dx`	`= [f(x)]_0^2`
	`= f(2)-f(0)`
	`= f(2)\ \ \ \ text{(since}\ f(0) = 0 text{)}`

`=> f(2) = 4`

♦♦♦ Parts (ii) and (iii) proved particularly difficult for students with mean marks of 12% and 11% respectively.

`:.\ text(MAX value of)\ \ f(x) = 4`

iii.	`int_0^4 f^{′}(x)`	`= A_1-A_2`
		`=0`

`text(We also know)`

`int_0^4 f^{′}(x)\ dx`	`= int_2^4 f^{′}(x)\ dx + int_0^2 f^{′}(x)\ dx`
	`=[f(x)]_2^4 + 4`
	`= f(4)-f(2) + 4\ \ \ (text(note)\ f(2)=4)`
	`=f(4)`

`=> f(4) = 0`

`text(Gradient = – 3 from)\ \ x = 4\ \ text(to)\ \ x = 6`

`:.\ f(6)`	`= -3 (6\ – 4)`
	`= -6`

♦♦ Mean mark 28%
EXAM TIP: Clearly identify THE EXTREMES when given a defined domain. In this case, the origin is obvious graphically, and the other extreme at `x=6`, is CLEARLY LABELLED!

iv.

Calculus, 2ADV C3 2010 HSC 8d

Let `f(x) = x^3-3x^2 + kx + 8`, where `k` is a constant.

Find the values of `k` for which `f(x)` is an increasing function. (2 marks)

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`k>3`

Show Worked Solution

`f(x)`	`= x^3-3x^2 + kx + 8`
`f^{′}(x)`	`= 3x^2-6x + k`

`f(x)\ text(is increasing when)\ \ f^{′}(x) > 0`

`=> 3x^2-6x + k > 0`

♦♦ Mean mark 28%.
MARKER’S COMMENT: The arithmetic required to solve `36-12k<0` proved the undoing of many students.

`f^{′}(x)\ text(is always positive)`

`=> f^{′}(x)\ text(is a positive definite.)`

`text(i.e. when)\ \ a > 0\ text(and)\ Delta < 0`

`a=3>0`

`Delta = b^2-4ac`

`:. (-6)^2-(4 xx 3 xx k)`	`<0`
`36-12k`	`<0`
`12k`	`>36`
`k`	`>3`

`:.\ f(x)\ text(is increasing when)\ \ k > 3.`

Probability, 2ADV S1 2010 HSC 8b

Two identical biased coins are tossed together, and the outcome is recorded.

After a large number of trials it is observed that the probability that both coins land showing heads is 0.36.

What is the probability that both coins land showing tails? (2 marks)

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`0.16`

Show Worked Solution

♦♦ Mean mark 28%.
NOTE: The most common error was `P(T_1T_2)“=1-0.36“=0.64`. Ensure you understand why this does not apply.

`P(H_1 H_2)`	`=P(H_1) xx P(H_2)`
	`=0.36`

`text(S)text(ince coins are identical:)`

`P(H)`	`= sqrt 0.36`
	`= 0.6`

`P(T)`	`= 1 – P(H)`
	`=0.4`

`:.\ P(T_1 T_2)`	`= 0.4 xx 0.4`
	`=0.16`

Calculus, 2ADV C1 2010 HSC 7b

The parabola shown in the diagram is the graph `y = x^2`. The points `A (–1,1)` and `B (2, 4)` are on the parabola.

Find the equation of the tangent to the parabola at `A`. (2 marks)

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Let `M` be the midpoint of `AB`.

There is a point `C` on the parabola such that the tangent at `C` is parallel to `AB`.

Show that the line `MC` is vertical. (2 marks)

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The tangent at `A` meets the line `MC` at `T`.

Show that the line `BT` is a tangent to the parabola. (2 marks)

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`2x + y + 1 = 0`
`text(Proof) text{(See Worked Solutions)}`
`text(Proof) text{(See Worked Solutions)}`

Show Worked Solution

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

`text(At)\ \ A text{(–1,1)}\ => dy/dx = -2`

`text(T)text(angent has)\ \ m=text(–2),\ text(through)\ text{(–1,1):}`

`y – y_1`	`= m(x\ – x_1)`
`y – 1`	`= -2 (x + 1)`
`y – 1`	`= -2x -2`
`2x + y + 1`	`= 0`

`:.\ text(T)text(angent at)\ A\ text(is)\ \ 2x + y + 1 = 0`

♦ Mean mark 37%.
IMPORTANT: The critical understanding required for this question is that the gradient of `AB` needs to be equated to the gradient function (i.e. `dy/dx`).

ii. `Atext{(–1,1)}\ \ \ B(2,4)`

`M`	`= ((-1+2)/2 , (1+4)/2)`
	`= (1/2, 5/2)`

`m_(AB)`	`= (y_2 – y_1)/(x_2 – x_1)`
	`= (4 – 1)/(2 + 1)=1`

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

`:.\ C \ (1/2, 1/4)`

`=>M\ text(and)\ C\ text(both have)\ x text(-value)=1/2`

`:. MC\ text(is vertical … as required)`

iii. `T\ text(is point on tangent when) \ x=1/2`

♦♦ Mean mark 29%.

`text(T)text(angent)\ \ \ 2x + y + 1 = 0`

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

`2 xx (1/2) + y + 1=0`

`=> y=–2`

`:.\ T (1/2, –2)`

`text (Given)\ \ B (2, 4)`

`m_(BT)`	`= (4+2)/(2\ – 1/2)`
	`=4`

`text(At)\ \ B(2,4),\ text(find gradient of tangent:)`

`dy/dx = 2x=2 xx2=4`

`:.m_text(tangent) = 4=m_(BT)`

`:.BT\ text(is a tangent)`

Trigonometry, 2ADV T1 2010 HSC 6b

The diagram shows a circle with centre `O` and radius 5 cm.

The length of the arc `PQ` is 9 cm. Lines drawn perpendicular to `OP` and `OQ` at `P` and `Q` respectively meet at `T`.

Find `/_POQ` in radians. (1 mark)

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Prove that `Delta OPT` is congruent to `Delta OQT`. (2 marks)

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Find the area of the shaded region. (2 marks)

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`9/5\ text(radians)`
`text(Proof) text{(See Worked Solutions)}`
`6.3\ text(cm)\ \ \ text{(to 1 d.p.)}`
`9.0\ text(cm²)\ \ \ text{(to 1 d.p.)}`

Show Worked Solution

i.	`text(Length of Arc)`	`= r theta`
	`9`	`= 5 xx /_POQ`
	`:.\ /_ POQ`	`= 9/5\ text(radians)`

ii.

`text(Prove)\ Delta OPT ~= Delta OQT`

`OT\ text(is common)`

MARKER’S COMMENT: Know the difference between the congruency proof of `RHS` and `SAS`. Incorrect identification will lose a mark.

`/_OPT = /_OQT = 90°\ \ \ text{(given)}`

`OP = OQ\ \ \ text{(radii)}`

`:.\ Delta OPT ~= Delta OQT\ \ \ text{(RHS)}`

iii.

`/_POT `	`= 1/2 xx /_POQ\ \ \ text{(from part (ii))}`
	`=1/2 xx 9/5`
	`= 9/10\ text(radians)`

♦♦ Mean mark below 30%.
MARKER’S COMMENT: Many students struggled to work in radians. Make sure you understand this concept.

`tan /_ POT`	`= (PT)/(OP)`
`tan (9/10)`	`= (PT)/5`
`PT`	`= 5 xx tan(9/10)`
	`=6.3007…`
	`=6.3\ text(cm)\ \ text{(to 1 d.p.)}`

iv.

`text(Shaded Area = Area)\ OQTP\ – text(Area Sector)\ OQP`

♦ Mean mark 35%.

`text(Area)\ OQTP`	`= 2 xx text(Area)\ Delta OPT`
	`=2 xx 1/2 xx OP xx PT`
	`= 5 xx 6.3007`
	`~~ 31.503…`
	`~~31.5\ text(cm²)`

`text(Area Sector)\ OQP`	` = 1/2 r^2 theta`
	`= 1/2 xx 25 xx 9/5`
	`= 22.5\ text(cm²)`

`:.\ text(Shaded Area)`	`= 31.503\ – 22.5`
	`=9.003…`
	`=9.0\ text(cm²)\ \ \ text{(to 1 d.p.)}`

Calculus, 2ADV C3 2010 HSC 6a

Let `f(x) = (x + 2)(x^2 + 4)`.

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

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Find the values of `x` for which the graph `y=f(x)` is concave down, and the values for which it is concave up. (2 marks)

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Sketch the graph `y=f(x)`, indicating the values of the `x` and `y` intercepts. (2 marks)

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`text(Proof) text{(See Worked Solutions)}`
`f(x)\ text(is concave down when)\ x < -2/3`

`f(x)\ text(is concave up when)\ x > -2/3`

Show Worked Solution

i. `text(Need to show no S.P.’s)`

`f(x)`	`= (x+2)(x^2 + 4)`
	`=x^3 + 2x^2 + 4x + 8`
`f prime (x)`	`= 3x^2 + 4x + 4`

`text(S.P.s occur when)\ \ f prime (x) =0,`

`3x^2 + 4x + 4 =0`

`Delta`	`= b^2\ – 4ac`
	`=4^2\ – (4 xx 3 xx 4)`
	`=16\ – 48`
	`= -32 < 0`

`text(S)text(ince)\ \ Delta < 0,\ \ text(No Solution)`

`:.\ text(No S.P.’s for)\ \ f(x)`

ii. `f(x)\ text(is concave down when)\ f″(x) < 0`

MARKER’S COMMENT: The significance of the sign of the second derivative was not well understood by most students.

`f″(x) = 6x + 4`

`=> 6x + 4`	`< 0`
`6x`	`< -4`
`x`	`< -2/3`

`:.\ f(x)\ text(is concave down when)\ x < -2/3`

`f(x)\ text(is concave up when)\ f″(x) > 0`

`f″(x) = 6x + 4`

`=> 6x + 4`	`> 0`
`6x`	`> -4`
`x`	`> -2/3`

`:. f(x)\ text(is concave up when)\ x > -2/3`

♦♦ Mean mark 33%.
MARKER’S COMMENT: Students are reminded to bring a ruler to the exam and use it to draw the axes for graphing and to help with an appropriate scale.

iii. `y text(-intercept) =2 xx4=8`

`x text(-intercept)=–2`

Calculus, 2ADV C4 2010 HSC 5b

Prove that `sec^2 x + secx tanx = (1 + sinx)/(cos^2x)`. (1 mark)

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Hence prove that `sec^2 x + secx tanx = 1/(1 - sinx)`. (1 mark)

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Hence, use the identity `int sec ax tan ax\ dx=1/a sec ax` to find the exact value of

`int_0^(pi/4) 1/(1 - sinx)\ dx`. (2 marks)

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`text(Proof) text{(See Worked Solutions)}`
`text(Proof) text{(See Worked Solutions)}`
`sqrt2`

Show Worked Solution

i. `text(Need to prove)`

`sec^2x + secxtanx = (1 + sinx)/(cos^2x)`

`text(LHS)`	`=sec^2x + secx tanx`
	`=1/(cos^2x) + 1/(cosx) xx (sinx)/cosx`
	`=1/(cos^2x) + (sinx)/(cos^2x)`
	`=(1 + sinx)/(cos^2x)`
	`= text(RHS)\ \ \ \ text(… as required)`

ii. `text(Need to prove)`

♦♦ Mean mark 31%.

`sec^2x + secx tanx`	`= 1/(1\ – sinx)`
`text(i.e.)\ \ (1 + sinx)/(cos^2x)`	`= 1/(1\ – sin x)\ \ \ \ \ text{(part (i))}`

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

iii. `int_0^(pi/4) 1/(1\ – sinx)\ dx`

♦ Mean mark 37%.

`= int_0^(pi/4) (sec^2x + secx tanx)\ dx`

`= [tanx + secx]_0^(pi/4)`

`= [(tan(pi/4) + sec(pi/4)) – (tan0 + sec0)]`

`= [(1 + 1/(cos(pi/4)))\ – (0 + 1/(cos0))]`

`= 1 + sqrt2\ – 1`

`= sqrt2`

Calculus, 2ADV C3 2011 HSC 9c

The graph `y = f(x)` in the diagram has a stationary point when `x = 1`, a point of inflection when `x = 3`, and a horizontal asymptote `y = –2`.

Sketch the graph `y = f^{′}(x)` , clearly indicating its features at `x = 1` and at `x = 3`, and the shape of the graph as `x -> oo`. (3 marks)

Show Answers Only

Show Worked Solution

♦ Mean mark 43%
IMPORTANT: Examiners regularly ask questions that require the graphing of an `f ^{′}(x)` given the `f(x)` graph and vice-versa. KNOW IT!

Plane Geometry, 2UA 2012 HSC 16a

The diagram shows a triangle `ABC` with sides `BC = a` and `AC = b`. The points `D`, `E` and `F` lie on the sides `AC`, `AB` and `BC`, respectively, so that `CDEF` is a rhombus with sides of length `x`.

Prove that `Delta EBF` is similar to `Delta AED`. (2 marks)
Find an expression for `x` in terms of `a` and `b`. (2 marks)

Show Answers Only

`text(Proof)\ \ text{(See Worked Solutions)}`
`x = (ab)/(a + b)`

Show Worked Solution

(i)

`text(S)text(ince)\ CDEF\ text(is a rhombus)`

♦ Mean mark 47%.

`=> BC\ |\| \ ED\ \ text(and)\ \ EF \ |\|\ CD`

`/_ FCD`	`=/_ EDA\ text{(corresponding angles,}\ BC\ text(\|\|)\ ED text{)}`
`/_FCD`	`=/_BFE\ text{(corresponding angles,}\ EF\ text(\|\|)\ CD text{)}`
`=>/_EDA`	`=/_BFE`
`/_FBE`	`=/_DEA\ text{(corresponding angles,}\ BC\ text(\|\|)\ ED text{)}`

`:.\ Delta EBF\ text(|||)\ Delta AED\ \ text{(equiangular) … as required}`

(ii) `text(In)\ Delta AED,\ DA = b\ – x\ \ \ text(and)\ \ \ ED = x`

♦ Mean mark 31%.

`text(In)\ Delta EBF,\ FE = x\ \ text(and)\ \ FB = a\ – x`

`text(Using similarity)`

`(DA)/(ED) = (EF)/(FB)`

`(b\ – x)/x`	`= x/(a\ – x)`
`x^2`	`= (b\ – x)(a\ – x)`
`x^2`	`= ab\ – bx\ – ax + x^2`
`0`	`= ab\ – x (a + b)`
`x (a + b)`	`= ab`
`x`	`= (ab)/(a+b)`

Calculus, 2ADV C3 2012 HSC 14a

A function is given by `f(x) = 3x^4 + 4x^3-12x^2`.

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

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Hence, sketch the graph `y = f(x)` showing the stationary points. (2 marks)

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For what values of `x` is the function increasing? (1 mark)

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For what values of `k` will `f(x) = 3x^4 + 4x^3-12x^2 + k = 0` have no solution? (1 mark)

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`text{MAX at (0,0), MINs at (1, –5) and (–2, –32)}`
`f(x)\ text(is increasing for)\ -2 < x < 0\ text(and)\ x > 1`
`text(No solution when)\ k > 32`

Show Worked Solution

i.	`f(x)`	`= 3x^4 + 4x^3-12x^2`
	`f^{′}(x)`	`= 12x^3 + 12x^2-24x`
	`f^{″}(x)`	`= 36x^2 + 24x-24`

`text(Stationary points when)\ f^{′}(x) = 0`

`12x^3 + 12x^2-24x`	`=0`
`12x(x^2 + x-2)`	`=0`
`12x (x+2) (x-1)`	`=0`

`:.\ text(Stationary points at)\ x=0,\ 1\ text(or)\ -2`

`text(When)\ x=0,\ \ \ \ f(0)=0`
`f^{″}(0)`	`= -24 < 0`
`:.\ text{MAX at (0,0)}`

`text(When)\ x=1`

`f(1)`	`= 3+4-12 = -5`
`f^{″}(1)`	`= 36 + 24-24 = 36 > 0`
`:.\ text{MIN at}\ (1,-5)`

`text(When)\ x=–2`

`f(-2)`	`=3(-2)^4 + 4(-2)^3-12(-2)^2`
	`= 48-32-48`
	`= -32`
`f^{″}(-2)`	`= 36(-2)^2 + 24(-2)-24`
	`=144-48-24 = 72 > 0`
`:.\ text{MIN at (–2, –32)}`

ii.

♦ Mean mark 42%
MARKER’S COMMENT: Be careful to use the correct inequality signs, and not carelessly include ≥ or ≤ by mistake.

iii.	`f(x)\ text(is increasing for)`
	`-2 < x < 0\ text(and)\ x > 1`

iv. `text(Find)\ k\ text(such that)`

♦♦♦ Mean mark 12%.

`3x^4 + 4x^3-12x^2 + k = 0\ text(has no solution)`

`k\ text(is the vertical shift of)\ \ y = 3x^4 + 4x^3-12x^2`

`=>\ text(No solution if it does not cross the)\ x text(-axis.)`

`:.\ text(No solution when)\ k > 32`

Calculus, 2ADV C4 2011 HSC 6c

The diagram shows the graph `y = 2 cos x` .

State the coordinates of `P`. (1 mark)

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

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Indicate which area in the diagram, `A`, `B`, `C` or `D`, is represented by the integral

`int_((3pi)/2)^(2pi) 2 cos x\ dx`. (1 mark)

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Using parts (ii) and (iii), or otherwise, find the area of the region bounded by the curve `y = 2 cos x` and the `x`-axis, between `x = 0` and `x = 2pi` . (1 mark)

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Using the parts above, write down the value of

`int_(pi/2)^(2pi) 2 cos x\ dx`. (1 mark)

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Show Answers Only

`P(0,2)`
`2`
`C`
`8\ text(u²)`
`-2`

Show Worked Solution

i. `y = 2 cos x`

`text(At)\ x = 0`

`y = 2 cos 0 = 2`

`:.\ P(0,2)`

ii. `int_0^(pi/2) 2 cos x\ dx`

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

`= [2 sin (pi/2)\ – 2 sin 0]`

`= 2\ – 0`

`= 2`

iii. `C`

iv.	`text(S)text(ince Area)\ A`	`=\ text(Area)\ C,\ \ text(and)`
	`text(Area)\ B`	`= 2 xx text(Area)\ A`

`:.\ text(Total Area)`	`= 2 + (2xx2) + 2`
	`= 8\ text(u²)`

MARKER’S COMMENT: “Using the parts above” in part (v) was ignored by many students. Important to know when finding an area and evaluating an integral may differ.

v. `int_(pi/2)^(2pi) 2 cos x\ dx`

`= text(Area)\ C\ – text(Area)\ B`

`= 2\ – (2 xx 2)`

`= -2`

Real Functions, 2UA 2011 HSC 6b

A point `P(x, y)` moves so that the sum of the squares of its distance from each of the points `A(–1, 0)` and `B(3, 0)` is equal to 40.

Show that the locus of `P(x, y)` is a circle, and state its radius and centre. (3 marks)

Show Answers Only

`text(Centre)\ (1,0),\ text(radius)\ 4\ text(units).`

Show Worked Solution

♦ Mean mark 39%.
MARKER’S COMMENT: Challenging question with many students unable to handle the algebra in expanding and completing the squares.

`text(Find locus of)\ \ P(x,y)`

`P(x,y)\ \ \ \ \ Atext{(-1,0)}\ \ \ \ \ B(3,0)`

`text(Dist)\ PA^2`	`= (y_2\ – y_1)^2 + (x_2\ – x_1)^2`
	`= y^2 + (x + 1)^2`

`text(Dist)\ PB^2`	`= (y\ – 0)^2 + (x\ – 3)^2`
	`= y^2 + (x\ – 3)^2`

`text(S)text(ince)\ \ PA^2 + PB^2 = 40`

`=> y^2 + (x + 1)^2 + y^2 + (x\ – 3)^2`	`=40`
`2y^2 + x^2 + 2x + 1 + x^2\ – 6x + 9`	`=40`
`2y^2 + 2x^2\ – 4x + 10`	`=40`
`y^2 + x^2\ – 2x + 5`	`=20`
`y^2 + (x -1)^2 + 4`	`=20`
`y^2 + (x -1)^2`	`=16`

`:.\ P(x,y)\ text(is a circle, centre)\ \ (1,0),\ text(radius 4 units.)`

Statistics, 2ADV 2011 HSC 5b

Kim has three red shirts and two yellow shirts. On each of the three days, Monday, Tuesday and Wednesday, she selects one shirt at random to wear. Kim wears each shirt that she selects only once.

What is the probability that Kim wears a red shirt on Monday? (1 mark)
What is the probability that Kim wears a shirt of the same colour on all three days? (1 mark)
What is the probability that Kim does not wear a shirt of the same colour on consecutive days? (2 marks)

Show Answers Only

`3/5`
`1/10`
`3/10`

Show Worked Solution

i.	`P (R\ text(on Monday) text{)}`	`= text(# Red)/text(# Shirts)`
		`= 3/5`

MARKER’S COMMENT: Students could also have drawn a probability tree setting out this problem over 3 different days (stages).

ii.	`text(S)text(ince not enough yellow shirts)`
	`P text{(same colour each day)}`

`= P (R,R,R)`

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

`= 1/10`

iii.

`P text{(not wearing same colour 2 days in a row)}`

`= P (Y,R,Y) + P (R,Y,R)`

♦ Mean mark 47%.
MARKER’S COMMENT: Students should clearly write what probability they are finding in words or symbols before showing calculations.

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

`= 6/60 + 12/60 `

`= 3/10`

Functions, EXT1* F1 2011 HSC 4e

The diagram shows the graphs `y = |\ x\ |\ - 2` and `y = 4- x^2`.

Write down inequalities that together describe the shaded region. (2 marks)

Show Answers Only

`text(Inequalities are)`

`y <= 4\ – x^2`

`y >= |\ x\ |\ – 2`

Show Worked Solution

♦ Mean mark 46%.

`text(Inequalities are)`

`y <= 4 – x^2`

`y >= |\ x\ |\ – 2`

Calculus, 2ADV C4 2011 HSC 4d

Differentiate `y=sqrt(9 - x^2)` with respect to `x`. (2 marks)

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Hence, or otherwise, find `int (6x)/sqrt(9 - x^2)\ dx`. (2 marks)

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Show Answers Only

`- x/sqrt(9\ – x^2)`
`-6 sqrt(9\ – x^2) + C`

Show Worked Solution

IMPORTANT: Some students might find calculations easier by rewriting the equation as `y=(9-x^2)^(1/2)`.

i.	`y`	`= sqrt(9 – x^2)`
		`= (9 – x^2)^(1/2)`

`dy/dx`	`=1/2 xx (9 – x^2)^(-1/2) xx d/dx (9 – x^2)`
	`= 1/2 xx (9 – x^2)^(-1/2) xx -2x`
	`= – x/sqrt(9 – x^2)`

ii.	`int (6x)/sqrt(9 – x^2)\ dx`	`= -6 int (-x)/sqrt(9 – x^2)\ dx`
		`= -6 (sqrt(9 – x^2)) + C`
		`= -6 sqrt(9 – x^2) + C`

Calculus, 2ADV C4 2011 HSC 2e

Find `int 1/(3x^2)\ dx`. (2 marks)

Show Answers Only

`-1/(3x) + C`

Show Worked Solution

♦ Mean mark 46%.
MARKER’S COMMENT: Students who took the `1/3` out the front before integrating made less errors.

`int 1/(3x^2)\ dx`	`= 1/3 int x^-2\ dx`
	`= 1/3 xx 1/-1 xx x^-1 + C`
	`= – 1/(3x) + C`

Functions, EXT1* F1 2012 HSC 8 MC

The diagram shows the region enclosed by `y = x- 2` and `y^2 = 4-x`.

Which of the following pairs of inequalities describes the shaded region in the diagram?

`y^2 <= 4-x\ \ and\ \ y <= x-2`
`y^2 <= 4-x\ \ and\ \ y >= x-2`
`y^2 >= 4-x\ \ and\ \ y<= x-2`
`y^2 >= 4-x\ \ and\ \ y >= x-2`

Show Answers Only

`A`

Show Worked Solution

♦ Mean mark 44%.

`text(Using information from diagram)`

`(3,0)\ text(is in the shaded region)`

`text{Substituting (3,0) into}\ \ \ y^2<=4-x,\ \ \ 0 <= 4-3 => text(true)`

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

`text(Similarly)`

`(3,0)\ text(must satisfy other inequality)`

`text(i.e.)\ \ y <= x-2\ \ text(becomes)\ \ 0<= 3-2 =>\ text(true)`

`=> A`

Functions, 2ADV F2 2013 HSC 15c

Sketch the graph `y = |\ 2x-3\ |`. (1 mark)

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Using the graph from part (i), or otherwise, find all values of `m` for which the equation `|\ 2x-3\ | = mx + 1` has exactly one solution. (2 marks)

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Show Answers Only

`text(When)\ m = -2/3,\ m >= 2\ text(or)\ m<-2`

Show Worked Solution

♦ Mean mark 49%
MARKER’S COMMENT: Many students drew diagrams that were “too small”, didn’t use rulers or didn’t use a consistent scale on the axes!

ii.

`text(Line of intersection)\ \ y=mx + 1\ \ text(passes through)\ \ (0,1)`

♦♦ Mean mark 25%.
COMMENT: Students need a clear graphical understanding of what they are finding to solve this very challenging, Band 6 question.

`text(If it also passes through)\ \ (1.5, 0) => text(1 solution)`

`m`	`=(y_2-y_1)/(x_2-x_1)`
	`= (1 -0)/(0- 3/2)`
	`=-2/3`

`text(Gradients of)\ \ y=|\ 2x-3\ |\ \ text(are)\ \ 2\ text(or)\ -2`

`text(Considering a line through)\ \ (0,1):`

`text(If)\ \ m >= 2\ text(, only intersects once.)`

`text(Similarly,)`

`text(If)\ \ m<-2 text(, only intersects once.)`

`:.\ text(Only one solution when)\ \ m = -2/3,\ \ m >= 2\ \ text(or)\ \ m<-2`

Calculus, EXT1* C3 2013 HSC 15b

The region bounded by the `x`-axis, the `y`-axis and the parabola `y = (x-2)^2` is rotated about the `y`-axis to form a solid.

Find the volume of the solid. (4 marks)

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Show Answers Only

`text(Volume) = (8 pi)/3\ text(u³)`

Show Worked Solution

`text(S)text(ince rotation about the)\ y text(-axis,)`

♦ Mean mark 39%.

`(x -2)^2`	`= y`
`x\-2`	` = ± y^(1/2)`
`x `	`= 2 +- y^(1/2)`

`text(When)\ \ x=0,\ y=4`

`:. x = 2-y^(1/2)`

`:.\ text(Volume)`	`= pi int_0^4 x^2 dy`
	`= pi int_0^4 (2-y^{1/2})^2 dy`
	`= pi int_0^4 (4-4 y^{1/2} + y) dy`
	`= pi [4y-(4 xx 2/3 xx y^(3/2)) + (1/2 y^2)]_0^4`
	`= pi [4y-8/3 y^(3/2) + 1/2 y^2]_0^4`
	`= pi [(4 xx 4)-(8/3 xx 4^(3/2)) + (1/2 xx 4^2)]`
	`= pi [16\-64/3 + 8]`
	`= (8 pi)/3\ text(u³)`

Trigonometry, 2ADV T1 2013 HSC 14c

The right-angled triangle `ABC` has hypotenuse `AB = 13`. The point `D` is on `AC` such that `DC = 4`, `/_DBC = pi/6` and `/_ABD = x`.

Using the sine rule, or otherwise, find the exact value of `sin x`. (3 marks)

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Show Answers Only

`(7sqrt3)/26 text(.)`

Show Worked Solution

`text(Find)\ \ /_ADB`

`/_ADB`	`= pi/6 + pi/2 \ \ \ text{(exterior angle of}\ Delta BDC text{)}`
	`= (2pi)/3\ text(radians)`

`text(Find)\ \ AD`

♦ Mean mark 36%.
STRATEGY TIP: The hint to use the sine rule should flag to students that they will be dealing in non-right angled trig (i.e. `Delta ABD`) and to direct their energies at initially finding `/_ADB` and `AD`.

`tan (pi/6)`	`= 4/(BC)`
`1/sqrt3`	`=4/(BC)`
`BC`	`=4 sqrt3`

`text(Using Pythagoras:)`

`AC^2 + BC^2`	`= AB^2`
`AC^2 + (4sqrt3)^2`	`= 13^2`
`AC^2`	`= 169\-48`
	`= 121`
`=>AC`	`= 11`

`:.AD`	`=AC\-DC`
	`= 11 -4`
	`=7`

`text(Using sine rule:)`

`(AB)/(sin /_BDA)`	`= (AD)/(sinx)`
`13/(sin ((2pi)/3))`	`=7/(sinx)`
`13 xx sinx`	`= 7 xx sin ((2pi)/3)`
`sinx`	`= 7/13 xx sin((2pi)/3)`
	`= 7/13 xx sqrt3/2`
	`= (7 sqrt3)/26`

`:.\ text(The exact value of)\ sinx = (7sqrt3)/26 text(.)`

Trigonometry, 2ADV T3 2013 HSC 13a

The population of a herd of wild horses is given by

`P(t) = 400 + 50 cos (pi/6 t)`

where `t` is time in months.

Find all times during the first 12 months when the population equals 375 horses. (2 marks)

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Sketch the graph of `P(t)` for `0 <= t <= 12`. (2 marks)

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

Show Answers Only

`t=4\ text(months and 8 months)`
`text(See Worked Solutions.)`

Show Worked Solution

i. `P(t) = 400 + 50 cos (pi/6 t)`

`text(Need to find)\ t\ text(when)\ P(t) = 375`

`375`	`= 400 + 50 cos (pi/6 t)`
`50 cos (pi/6 t)`	`=-25`
`cos (pi/6 t)`	`= – 1/2`

`text(S)text(ince)\ \ cos(pi/3)=1/2, text(and cos is)`
`text(negative in)\ 2^text(nd) // 3^text(rd)\ text(quadrants:)`

`=>pi/6 t`	`= (pi\ – pi/3),\ (pi + pi/3),\ (3pi\ – pi/3)`
	`= (2pi)/3,\ (4pi)/3,\ (8pi)/3,\ …`
`:.t`	`= 4,\ 8,\ 16,\ …`

`:.\ text(In the 1st 12 months,)\ P(t) = 375\ text(when)`

`t=4\ text(months and 8 months.)`

♦ Mean mark 39%

ii.

Calculus, 2ADV C3 2013 HSC 8 MC

The diagram shows points `A`, `B`, `C` and `D` on the graph `y = f(x)`.

At which point is `f^{′}(x) > 0` and ` f^{″}(x)= 0`?

Show Answers Only

`B`

Show Worked Solution

♦ Mean mark 48%

`text(At)\ A,\ \ \ f^{′}(x) <0`

`text(At)\ C,\ \ \ f^{′}(x) =0`

`:.\ text(It cannot be)\ A\ text(or)\ C.`

`text(At)\ D,\ \ \ f^{″}(x) >0\ \ \ text{(concave up)}`

`text(At)\ B,\ \ \ f^{″}(x) =0\ \ \ text{(concavity changes)}`

`=> B`

Trigonometry, 2ADV T3 2013 HSC 6 MC

Which diagram shows the graph `y=sin(2x + pi/3)`?

Show Answers Only

`D`

Show Worked Solution

♦♦ Mean mark 34%

`text(At)\ x = 0 text(,)\ \ y = sin (pi/3) = sqrt3/2`

`=>\ text(It cannot be A or C)`

`text(Find)\ x\ text(when)\ y = 0,`

`sin (2x + pi/3)`	`= 0`
`:.\ 2x + pi/3`	`= 0\ \ \ \ \ text{(sin 0 = 0)}`
`2x`	`=-pi/3`
`x`	`= -pi/6`

`=> D`

Financial Maths, STD2 F4 2011 HSC 28b

Norman and Pat each bought the same type of tractor for $60 000 at the same time. The value of their tractors depreciated over time.

The salvage value `S`, in dollars, of each tractor, is its depreciated value after `n` years.

Norman drew a graph to represent the salvage value of his tractor.

Find the gradient of the line shown in the graph. (1 mark)

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What does the value of the gradient represent in this situation? (1 mark)

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Write down the equation of the line shown in the graph. (1 mark)

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Find all the values of `n` that are not suitable for Norman to use when calculating the salvage value of his tractor. Explain why these values are not suitable. (2 marks)

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Pat used the declining balance formula for calculating the salvage value of her tractor. The depreciation rate that she used was 20% per annum.

What did Pat calculate the salvage value of her tractor to be after 14 years? (2 marks)

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Using Pat’s method for depreciation, describe what happens to the salvage value of her tractor for all values of `n` greater than 15. (1 mark)

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

Show Answers Only

`text(Gradient) =-4000`
`text(The amount the tractor depreciates each year.)`
`S = 60\ 000\-4000n`
`text(It is unsuitable to use)`
`n<0\ text(because time must be positive)`
`n>15\ text(because the tractor has no more value after 15 years and)`
`text(therefore can’t depreciate further.)`
`text(After 14 years, the tractor is worth $2638.83)`
`text(As)\ n\ text(increases above 15 years,)\ S\ text(decreases but remains)>0.`

Show Worked Solution

♦♦♦ Mean mark 14%
COMMENT: The intercepts of both axes provide points where the gradient can be quickly found.

i.	`text(Gradient)`	`= text(rise)/text(run)`
		`= (- 60\ 000)/15`
		`=-4000`

♦ Mean mark 37%

ii. `text(The amount the tractor depreciates each year)`

♦♦ Mean mark 28%
COMMENT: Using the general form `y=mx+b` is quick here because you have the gradient (from part (i)) and the `y`-intercept is obviously `60\ 000`.

iii.	`text(S)text(ince)\ \ S = V_0\-Dn`
	`:.\ text(Equation of graph:)`
	`S = 60\ 000-4000n`

iv. `text(It is unsuitable to use)`

♦♦♦ Mean mark 20%

`n<0,\ text(because time must be positive:)`

`n>15,\ text(because it has no more value after 15)`

`text(years and therefore can’t depreciate further.)`

v.	`text(Using)\ S = V_0 (1-r)^n\ \ text(where)\ r = text(20%,)\ n = 14`

`S`	`= 60\ 000 (1\-0.2)^14`
	`= 60\ 000 (0.8)^14`
	`= 2\ 638.8279…`

`:.\ text(After 14 years, the tractor is worth $2638.83`

♦ Mean mark 37%

vi.	`text(As)\ n\ text(increases above 15 years,)\ S\ text(decreases)`
	`text(but remains > 0.)`

Algebra, STD2 A4 2011 HSC 28a

The air pressure, `P`, in a bubble varies inversely with the volume, `V`, of the bubble.

Write an equation relating `P`, `V` and `a`, where `a` is a constant. (1 mark)

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It is known that `P = 3` when `V = 2`.

By finding the value of the constant, `a`, find the value of `P` when `V = 4`. (2 marks)

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Sketch a graph to show how `P` varies for different values of `V`.

Use the horizontal axis to represent volume and the vertical axis to represent air pressure. (2 marks)

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

Show Answers Only

`P = a/V`
`P = 1 1/2`

Show Worked Solution

♦ Mean mark (i) 39%
COMMENT: Expressing the proportional relationship `P prop 1/V` as the equation `P=k/V` is a core skill here.

i.	`P`	`prop 1/V`
		`= a/V`

ii.

`text(When)\ P=3,\ V = 2`

`3`	`= a/2`
`a`	`=6`

`text(Need to find)\ P\ text(when)\ V = 4`

♦ Mean mark (ii) 47%

`P`	`=6/4`
	`= 1 1/2`

♦♦ Mean mark (iii) 26%
COMMENT: An inverse relationship is reflected by a hyperbola on the graph.

iii.

Financial Maths, STD2 F5 2011 HSC 27d

Josephine invests $50 000 for 15 years, at an interest rate of 6% per annum, compounded annually.

Emma invests $500 at the end of each month for 15 years, at an interest rate of 6% per annum, compounded monthly.

Financial gain is defined as the difference between the final value of an investment and the total contributions.

Who will have the better financial gain after 15 years? Using the Table below* and appropriate formulas, justify your answer with suitable calculations. (4 marks)

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

Show Answers Only

`text(Josephine – see Worked Solutions)`

Show Worked Solution

♦ Mean mark 42%
COMMENT: Note that compound interest vs annuity comparisons are commonly tested.

`text(Josephine)`

`text(Investment)`	`= 50\ 000 (1 + 0.06)^15`
	`= 50\ 000 (1.06)^15`
	`= $119\ 827.91`

`text(Financial gain)`	`= 119\ 827.91-50\ 000`
	`= $69\ 827.91`

`text(Emma)`

`text{Monthly interest rate} = text(6%)-:12=text(0.5%)`

`text{# Monthly Payments}=12 xx 15=180`

`=>\ text{Annuity Factor = 290.8187 (from Table)}`

`text(Investment)`	`= 500 xx 290.8187`
	`= $145\ 409.35`

`text(Financial gain)`	`= 145\ 409.35\-text(total contributions)`
	`= 145\ 409.35\-(500 xx 12 xx 15)`
	`= 145\ 409.35\-90\ 000`
	`= $55\ 409.35`

`:.\ text(Josephine will have the better financial gain.)`

Statistics, STD2 S5 2011 HSC 27c

Two brands of light bulbs are being compared. For each brand, the life of the light bulbs is normally distributed.

One of the Brand B light bulbs has a life of 400 hours.

What is the `z`-score of the life of this light bulb? (1 mark)

--- 1 WORK AREA LINES (style=lined) ---
A light bulb is considered defective if it lasts less than 400 hours. The following claim is made:

‘Brand A light bulbs are more likely to be defective than Brand B light bulbs.’

Is this claim correct? Justify your answer, with reference to `z`-scores or standard deviations or the normal distribution. (2 marks)

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

Show Answers Only

`-2`
`text(The claim is incorrect.)`

Show Worked Solution

i. `z text{-score of Brand B bulb (400 hrs)}`

`= (x-mu)/sigma`

`= (400\-500)/50`

`= –2`

ii. `z text{-score of Brand A bulb (400 hours)}`

♦ Mean mark 42%
MARKER’S COMMENT: A number of students found drawing normal curves in their solution advantageous.

`=(400-450)/25`

`=–2`

`text(S)text(ince the)\ z text(-score for both brands is –2,)`

`text(they are equally likely to be defective.)`

`:.\ text(The claim is incorrect.)`

Measurement, 2UG 2011 HSC 27b

Pontianak has a longitude of 109°E, and Jarvis Island has a longitude of 160°W`.

Both places lie on the Equator.

Find the shortest distance between these two places, to the nearest kilometre. You may assume that the radius of the Earth is 6400 km. (2 marks)
The position of Rabaul is 4° to the south and 48° to the west of Jarvis Island. What is the latitude and longitude of Rabaul? (2 marks)

Show Answers Only

`10\ 165\ text(km)\ \ \ text{(nearest km)}`
`152^@ text(E)`

Show Worked Solution

♦♦ Mean mark 29%

(i)	`text(Longitude difference)`	`= 109 + 160`
		`= 269^@`

`=> text(Shortest distance)\ text{(by degree)}`	`= 360\-269`
	`= 91^@`

`:.\ text(Shortest distance)`	`= 91/360 xx 2 pi r`
	`= 91/360 xx 2 xx pi xx 6400`
	`= 10\ 164.79…`
	`=10\ 165\ text(km)\ text{(nearest km)}`

♦♦ Mean mark 33%

(ii)	`text(Latitude)`
	`4^@\ text(South of Jarvis Island)`
	`text(S)text(ince Jarvis Island is on equator)`
	`=> text(Latitude is)\ 4^@ text(S)`
	`text(Longitude)`
	`text(Jarvis Island is)\ 160^@ text(W)`
	`text(Rubail is)\ 48^@\ text(West of Jarvis Island, or 208° West)`
	`text(which is)\ 28^@\ text{past meridian (180°)}`

`=>\ text(Longitude)`	`= (180\ -28)^@ text(E)`
	`= 152^@ text(E)`

`:.text(Position is)\ (4^@text{S}, 152^@text{E})`

Statistics, STD2 S1 2011 HSC 27a

A company sells handbags in Paris, New York and Florence.

Use the data in the table to complete the area chart below. (2 marks)

Show Answers Only

Show Worked Solution

♦♦ Mean mark 27%
COMMENT: Most common error was to not realise Area charts show cumulative data.

Financial Maths, 2UG 2011 HSC 26c

Furniture priced at $20 000 is purchased. A deposit of 15% is paid.

The balance is borrowed using a flat-rate loan at 19% per annum interest, to be repaid in equal monthly instalments over five years.

What will be the amount of each monthly instalment? Justify your answer with suitable calculations. (4 marks)

Show Answers Only

`$552.50`

Show Worked Solution

`text(Purchase price) = $20\ 000`

♦ Mean mark 44%

`text(Deposit)`	`=\ text(15%)\ xx 20\ 000`
	`= $3000`

`:.\ text(Loan)`	`= 20\ 000\-3000`
	`= $17\ 000`

`text(Using)\ I`	`= Prn`
`I`	`= 17\ 000 xx\ text(19%)\ xx 5`
	`= 16\ 150`

`=>\ text(Total amount to repay)`	`= 17\ 000 + 16\ 150`
	`= $33\ 150`

`:.\ text(Monthly instalment)`	`= (33\ 150)/(12 xx 5)`
	`= (33\ 150)/60`
	`= $552.50`

Probability, 2UG 2011 HSC 26a

The two spinners shown are used in a game.

Each arrow is spun once. The score is the total of the two numbers shown by the arrows.
A table is drawn up to show all scores that can be obtained in this game.

What is the value of `X` in the table? (1 mark)
What is the probability of obtaining a score less than 4? (1 mark)
On Spinner `B`, a 2 is obtained. What is the probability of obtaining a score of 3? (1 mark)
Elise plays a game using the spinners with the following financial outcomes.

⇒ Win `$12` for a score of `4`

⇒ Win nothing for a score of less than `4`

⇒ Lose `$3` for a score of more than `4`

It costs `$5` to play this game. Will Elise expect a gain or a loss and how much will it be?

Justify your answer with suitable calculations. (3 marks)

Show Answers Only

`5`
`1/2`
`2/3`
`text(Loss of)\ $1.50`

Show Worked Solution

(i) `X=3+2=5`

(ii) `P(text{score}<4)=6/12=1/2`

(iii) `P(3)=2/3`

(iv) `P(4)=4/12=1/3`

♦ Mean mark 34%
MARKER’S COMMENT: Better responses remembered to deduct the $5 cost to play and recognised the negative result as a loss.

`P(text{score}<4)`	`=6/12=1/2`
`P(text{score}>4)`	`=2/12=1/6`

`text(Financial Expectation)`

`=(1/3xx12)+(1/2xx0)-(1/6xx3)-5`

`=4-0.5-5`

`=-1.50`

`:.\ text(Elise should expect a loss of $1.50) `

Financial Maths, STD2 F4 2010 HSC 25b

William wants to buy a car. He takes out a loan for $28 000 at 7% per annum interest for four years.

Monthly repayments for loans at different interest rates are shown in the spreadsheet.

How much interest does William pay over the term of this loan? (2 marks)

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Show Answers Only

`$4183.52`

Show Worked Solution

♦ Mean mark 42%
MARKER’S COMMENT: An incorrect table value used correctly in the following calculations received half-marks here. Show your working!

`text(Loan) = $28\ 000,\ \ \ \ r =\ text(7% p.a.)`

`text(Monthly repayment = $670.49`

`text(# Repayments) = 4 xx 12 = 48`

`text(Total repaid)`	`= 48 xx 670.49`
	`= $32\ 183.52`

`:.\ text(Interest paid)`	`=32\ 183.52\-28\ 000`
	`=$4183.52`

Statistics, STD2 S1 2011 HSC 25d

Data was collected from 30 students on the number of text messages they had sent in the previous 24 hours. The set of data collected is displayed.

What is the outlier for this set of data? (1 mark)

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What is the interquartile range of the data collected from the female students? (1 mark)

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Show Answers Only

`71`
`9`

Show Worked Solution

i. `text(Outlier is 71)`

♦♦ Mean mark 34%
COMMENT: Ensure you can quickly and accurately find quartile values using stem and leaf graphs!

ii. `text{Lower quartile = 9 (4th female data point)}`

`text{Upper quartile = 20 (11th female data point)}`

`:.\ text{Interquartile range (female)}=20-11=9`

Statistics, STD2 S1 2011 HSC 25a

A study on the mobile phone usage of NSW high school students is to be conducted.

Data is to be gathered using a questionnaire.

The questionnaire begins with the three questions shown.

Classify the type of data that will be collected in Q2 of the questionnaire. (1 mark)

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Write a suitable question for this questionnaire that would provide discrete ordinal data. (1 mark)

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An initial study is to be conducted using a stratified sample.

Describe a method that could be used to obtain a representative stratified sample. (1 mark)

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Who should be surveyed if it is decided to use a census for the study? (1 mark)

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Show Answers Only

`text(Categorical)`
`text(See Worked Solutions)`
`text(See Worked Solutions)`
`text(A census would involve all high school students in NSW.)`

Show Worked Solution

i. `text(Categorical)`

ii. `text(How many outgoing calls do you make per day?)`

`text{(Ensure it can be answered with a numerical score.)}`

iii. `text(The method could be to work out how many)`

♦♦♦ Mean mark 7%. Toughest mark to get in the 2011 exam!
COMMENT: Know and be able to describe random, systematic and stratified sampling!

`text{students are in each year and ask 10% of the}`

`text{students in each year. (Note the sample of}`

`text{students in each year must be proportional to}`

`text{their percentage in the population).}`

♦♦♦ Mean mark 18%.
MARKER’S COMMENT: A specific population needed (i.e. high school students).

iv. `text(A census would involve all high school)`

`text(students in NSW.)`

Financial Maths, STD2 F4 2011 HSC 23c

An amount of $5000 is invested at 10% per annum, compounded six-monthly.

Use the table to find the value of this investment at the end of three years. (2 marks)

Show Answers Only

`$6700`

Show Worked Solution

♦♦ Mean mark 28%
MARKER’S COMMENT: Remember that the number of periods is the number of “compounding periods” and when asked to use the table, use the table!

`text(Interest rate)= text(10% pa)= text(5% per 6 months)`

`text(Period)= 6\ \ \ \ text{(6 x 6 months in 3 years)}`

`=> text(Table value)=1.340`

`:.\ text(Value of investment)`	`=5000xx1.34`
	`=$6700`

Measurement, STD2 M1 2010 HSC 28b

Moivre’s manufacturing company produces cans of Magic Beans. The can has a diameter of 10 cm and a height of 10 cm.

Cans are packed in boxes that are rectangular prisms with dimensions 30 cm × 40 cm × 60 cm.

What is the maximum number of cans that can be packed into one of these boxes? (1 mark)

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The shaded label on the can shown wraps all the way around the can with no overlap. What area of paper is needed to make the labels for all the cans in this box when the box is full? (2 marks)

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The company is considering producing larger cans. Monica says if you double the diameter of the can this will double the volume.

Is Monica correct? Justify your answer with suitable calculations. (2 marks)

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The company wants to produce a can with a volume of 1570 cm³, using the least amount of metal. Monica is given the job of determining the dimensions of the can to be produced. She considers the following graphs.
What radius and height should Monica recommend that the company use to minimise the amount of metal required to produce these cans? Justify your choice of dimensions with reference to the graphs and/or suitable calculations. (2 marks)

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Show Answers Only

`72\ text(cans)`
`20\ 358\ text{cm² (nearest cm²)}`
`text(Monica is incorrect because the volume)`

`text(doesn’t double. It increases by a factor of 4.)`
`text(Radius 6.3 cm and height 12.6 cm.)`

Show Worked Solution

♦♦ Mean mark 27%

i.	`text(Maximum # Cans)`	`= 4 xx 3 xx 6`
		`= 72\ text(cans)`

♦ Mean mark 38%
MARKER’S COMMENT: Many students didn’t account for the clearance of 0.5 cm at the top and bottom of each can.

ii.	`text(Label Area)\ text{(1 can)}`	`= 2 pi rh`
		`= 2 xx pi xx 5 xx 9`
		`= 90 pi`
		`= 282.7433…\ text(cm²)`

`:.\ text(Label Area)\ text{(72 cans)}`

`= 72 xx 282.7433…`

`= 20\ 357.52…`

`= 20\ 358\ text(cm²)` `text{(nearest cm²)}`

♦ Mean mark 44%
MARKER’S COMMENT: Many students performed calculations in this part without concluding if Monica is correct or not. Read the question carefully.

iii.	`text(Original volume)`	`= pi r^2 h`
		`= pi xx 5^2 xx 10`
		`= 785.398…\ text(cm³)`

`text(If the diameter doubles, radius) = 10\ text(cm)`

`text(New volume)`	`= pi xx 10^2 xx 10`
	`= 3141.592…\ text(cm³)`

`:.\ text(Monica is incorrect because the volume)`

`text(doesn’t double. It increases by a factor of 4.)`

♦♦ Mean mark 26%

iv.

`text(Minimum metal used when surface area is a minimum.)`

`text(From graph, minimum surface area when)\ r = 6.3\ text(cm)`

`text(When)\ r = 6.3\ text(cm,)\ h = 12.6\ text(cm)\ \ \ text{(from graph)}`

`:.\ text(She should recommend radius 6.3 cm and height 12.6 cm)`

Financial Maths, STD2 F4 2010 HSC 28a

The table shows monthly home loan repayments with interest rate changes from February to October 2009.

What is the change in monthly repayments on a $250 000 loan from February 2009 to April 2009? (1 mark)

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Xiang wants to borrow $307 000 to buy a house.

Xiang’s bank approves loans for customers if their loan repayments are no more than 30% of their monthly gross salary.

Xiang’s monthly gross salary is $6500.

If she had applied for the loan in October 2009, would her bank have approved her loan?

Justify your answer with suitable calculations. (3 marks)

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Jack took out a loan at the same time and for the same amount as Xiang.

Graphs of their loan balances are shown.

Identify TWO differences between the graphs and provide a possible explanation for each difference, making reference to interest rates and/or loan repayments. (2 marks)

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Show Answers Only

`text(Monthly repayments decrease by $15)`
`text(S)text(ince repayments of 1987.29 > 1950, the loan)`
`text(would not have been approved.)`
`text(Differences)`
`text(Jack’s loan balance falls more sharply for first 12 years)`
`text(Jack’s loan balance falls less sharply between years 12-30.)`
`text(Explanation)`
`text(Jack made larger repayments for first 12 years, or)`
`text(Jack made the same repayments but had a lower interest rate)`
`text(for the first 12 years.)`
`text(Jack made smaller repayments in years 12-30.)`

Show Worked Solution

i.	`text(Repayment)\ text{(Feb 09)}`	`= 1588`
	`text(Repayment)\ text{(Apr 09)}`	`= 1573`

`text(Difference) = 1588\-1573 = 15`

`:.\ text(Monthly repayments decrease by $15)`

♦ Mean mark 39%
MARKER’S COMMENT: Borrowing $307,000 can be achieved by borrowing $300,000, and then `7` times the table repayment value for borrowing $1000.

ii.

`text(Loan) = $307\ 000`

`text{Repayments (Oct 09)}`	`= 1942 + (7 xx 6.47)`
	`= 1942 + 45.29`
	`= $1987.29\ text(per month)`

`text(30% Gross salary)`	`= 6500 xx\ text(30%)`
	`= $1950\ text(per month)`

`:.\ text(S)text(ince repayments of $1987.29 > $1950, the loan)`

`text(would not have been approved.)`

iii. `text(Differences)`

`text(Jack’s loan balance falls more sharply for first 12 years)`

`text(Jack’s loan balance falls less sharply between years 12-30.)`

`text{Explanation(s)}`

♦ Mean mark 36%
MARKER’S COMMENT: Explanations were generally poor and many failed to refer directly to the graphs shown, or reference Xiang or Jack directly.

`text(Jack made larger repayments for 1st 12 years, or)`

`text(Jack made the same repayments but had a)`

`text(lower interest rate for the first 12 years.)`

`text(Jack made smaller repayments in years 12-30.)`

Algebra, STD2 A2 2010 HSC 27c

The graph shows tax payable against taxable income, in thousands of dollars.

2010 27c

Use the graph to find the tax payable on a taxable income of $21 000. (1 mark)

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Use suitable points from the graph to show that the gradient of the section of the graph marked `A` is `1/3`. (1 mark)

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How much of each dollar earned between $21 000 and $39 000 is payable in tax? (1 mark)

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Write an equation that could be used to calculate the tax payable, `T`, in terms of the taxable income, `I`, for taxable incomes between $21 000 and $39 000. (2 marks)

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Show Answers Only

`$3000\ \ \ text{(from graph)}`
`1/3`
`33 1/3\ text(cents per dollar earned)`
`text(Tax payable on)\ I = 1/3 I\-4000`

Show Worked Solution

`text(Income on)\ $21\ 000=$3000\ \ \ text{(from graph)}`

ii. `text(Using the points)\ (21,3)\ text(and)\ (39,9)`

♦♦ Mean mark 25%

`text(Gradient at)\ A`	`= (y_2\-y_1)/(x_2\ -x_1)`
	`= (9000-3000)/(39\ 000 -21\ 000)`
	`= 6000/(18\ 000)`
	`= 1/3\ \ \ \ \ text(… as required)`

iii. `text(The gradient represents the tax applicable to each dollar)`

♦♦♦ Mean mark 12%!
MARKER’S COMMENT: Interpreting gradients is an examiner favourite, so make sure you are confident in this area.

`text(Tax)`	` = 1/3\ text(of each dollar earned)`
	` = 33 1/3\ text(cents per dollar earned)`

iv. `text( Tax payable up to $21 000 = $3000)`

`text(Tax payable on income between $21 000 and $39 000)`

♦♦♦ Mean mark 15%.
STRATEGY: The earlier parts of this question direct students to the most efficient way to solve this question. Make sure earlier parts of a question are front and centre of your mind when devising strategy.

` = 1/3 (I\-21\ 000)`

`:.\ text(Tax payable on)\ \ I`	`= 3000 + 1/3 (I\-21\ 000)`
	`= 3000 + 1/3 I\-7000`
	`= 1/3 I\-4000`

Statistics, STD2 S1 2010 HSC 27b

The graphs show the distribution of the ages of children in Numbertown in 2000 and 2010.

In 2000 there were 1750 children aged 0–18 years.

How many children were aged 12–18 years in 2000? (1 mark)

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The number of children aged 12–18 years is the same in both 2000 and 2010.

How many children aged 0–18 years are there in 2010? (1 mark)

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Identify TWO changes in the distribution of ages between 2000 and 2010. In your answer, refer to measures of location or spread or the shape of the distributions. (2 marks)

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What would be ONE possible implication for government planning, as a consequence of this change in the distribution of ages? (1 mark)

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Show Answers Only

i. `875`

ii. `3500`

iii. `text{Changes in distribution (include 2 of the following):}`

`text(the lower quartile age is lower in 2010)`
`text(the median is lower in 2010)`
`text(the upper quartile age is lower in 2010)`
`text(the interquartile range is greater in 2010)`
`text(2010 is positively skewed while 2000 is negatively)`

iv. `text(Implication for government planning:)`

`text(Since the children are getting younger in 2010,)`

`text(Approve and build more childcare facilities)`
`text(Build more school and public playgrounds)`

Show Worked Solution

i. `text{Since the median = 12 years}`

♦ Mean mark (i) 45%

`=>\ text{50% of children are aged 12–18 years}`

`:.\ text{Children aged 12–18}\ = 50\text{%}\ xx 1750 = 875`

♦♦ Mean mark (ii) 25%

ii. `text{Upper quartile (2010) = 12 years}`

`text{Children in upper quartile = 875 (from part (i))}`

`:.\ text{Children aged 0–18}\ =4 xx 875= 3500`

iii. `text{Changes in distribution (include 2 of the following):}`

♦ Mean mark (iii) 35%
MARKER’S COMMENT: A number of students incorrectly identified “positive” skew as “negative” skew here.

`text(the lower quartile age is lower in 2010)`
`text(the median is lower in 2010)`
`text(the upper quartile age is lower in 2010)`
`text(the interquartile range is greater in 2010)`
`text(2010 is positively skewed while 2000 is negatively)`

iv. `text(Implication for government planning:)`

♦ Mean mark (iv) 46%
MARKER’S COMMENT: Answers should reflect the 1 mark allocation.

`text(Since the children are getting younger in 2010,)`

`text(Approve and build more childcare facilities)`
`text(Build more school and public playgrounds)`

Algebra, 2UG 2010 HSC 27a

Fully simplify `(4x^2)/(3y) -: (xy)/5`. (3 marks)

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Show Answers Only

`(20x)/(3y^2)`

Show Worked Solution

♦♦ Mean mark 33%.
MARKER’S COMMENT: Remember to “invert and multiply” when dividing a fraction by a fraction (see Worked Solution).

`(4x^2)/(3y) -: (xy)/5`	`= (4x^2)/(3y) xx 5/(xy)`
	`= (20x^2)/(3xy^2)`
	`= (20x)/(3y^2)`

Probability, 2UG 2010 HSC 26c

Tai plays a game of chance with the following outcomes.

• `1/5` chance of winning `$10`

• `1/2` chance of winning `$3`

• `3/10` chance of losing `$8`

The game has a `$2` entry fee.

What is his financial expectation from this game? (2 marks)

Show Answers Only

`text(A loss of $0.90.)`

Show Worked Solution

♦♦ Mean mark 31%
MARKER’S COMMENT: A common error was not to deduct the chance of losing or the entry fee. BE CAREFUL!

`text(Financial Expectation)`

`= (1/5 xx 10) + (1/2 xx 3)\-(3/10 xx 8)\-2`

`= 2 + 1.5\-2.4\-2`

`=-0.9`

`:.\ text(The financial expectation is a loss of $0.90.)`

Statistics, STD2 S1 2010 HSC 26b

A new shopping centre has opened near a primary school. A survey is conducted to determine the number of motor vehicles that pass the school each afternoon between 2.30 pm and 4.00 pm.

The results for 60 days have been recorded in the table and are displayed in the cumulative frequency histogram.

Find the value of Χ in the table. (1 mark)

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On the cumulative frequency histogram above, draw a cumulative frequency polygon (ogive) for this data. (1 mark)
Use your graph to determine the median. Show, by drawing lines on your graph, how you arrived at your answer. (1 mark)
Prior to the opening of the new shopping centre, the median number of motor vehicles passing the school between 2.30 pm and 4.00 pm was 57 vehicles per day.

What problem could arise from the change in the median number of motor vehicles passing the school before and after the opening of the new shopping centre?

Briefly recommend a solution to this problem. (2 marks)

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Show Answers Only

`15`
`text(Problems)`
`text(- increased traffic delays)`
`text(- increased danger to students leaving school)`
`text(Solutions)`
`text(- signpost alternative routes around school)`
`text(- decrease the speed limit in the area)`

Show Worked Solution

i.	`X`	`= 25\ -10`
		`= 15`

♦♦♦ Mean mark 18%
MARKER’S COMMENT: The ogive was poorly drawn with many students incorrectly joining the middle of each column rather than from corner to corner.

ii.

♦♦ Mean mark 25%
MARKER’S COMMENT: Many students did not “show by drawing lines on the graph” as the question asked.

iii. `text(Median)\ ~~155`

♦ Mean mark 47%
MARKER’S COMMENT: Short answers were often the best. Be concise when you can.

iv.	`text(Problems)`
	`text(- increased traffic delays)`
	`text(- increased danger to students leaving school)`

	`text(Solutions)`
	`text(- signpost alternative routes around school)`
	`text(- decrease the speed limit in the area)`

Probability, STD2 S2 2010 HSC 26a

A design of numberplates has a two-digit number, two letters and then another two-digit number, for example

How many different numberplates are possible using this design? (1 mark)

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Jo can order a numberplate with ‘JO’ in the middle but will have to have randomly selected numbers on either side.

Jo’s birthday is 30 December 1992, so she would like a numberplate with either

What is the probability that Jo is issued with one of the numberplates she would like? (2 marks)

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Show Answers Only

`6\ 760\ 000`
`P = 1/5000`

Show Worked Solution

♦ Mean mark 41%

i.	`text(# Combinations)`	`=10 xx 10 xx 26 xx 26 xx 10 xx 10`
		`=6\ 760\ 000`

♦♦ Mean mark 30%
IMPORTANT: Since the middle letters of “JO” can be guaranteed, the focus becomes purely on the 4 surrounding digits.

ii. `text(# Possible numberplates)`

`=10 xx 10 xx 10 xx 10`

`=10\ 000`

`:.P (30\ text(JO)\ 12) + P (19\ text(JO)\ 92)`

`= 1/(10\ 000) + 1/(10\ 000)`

`= 1/5000`

Measurement, STD2 M7 2013 HSC 30c

Joel mixes petrol and oil in the ratio 40 : 1 to make fuel for his leaf blower.

Joel pours 5 litres of petrol into an empty container to make fuel for his leaf blower.

How much oil should he add to the petrol to ensure that the fuel is in the correct ratio? (1 mark)

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Joel has 4.1 litres of fuel left in his container after filling his leaf blower.

He wishes to use this fuel in his lawnmower. However, his lawnmower requires the petrol and oil to be mixed in the ratio 25 : 1.

How much oil should he add to the container so that the fuel is in the correct ratio for his lawnmower? (3 marks)

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Show Answers Only

`text(0.125 L)`
`60\ text(mL)`

Show Worked Solution

♦ Mean mark 50%

i.	`text(Petrol : Oil) = 40\ :\ 1`
	`5\ text(Litres :)\ X = 40\ :\ 1`

`:. 5/X`	`= 40/1`
`40X`	`=5`
`X`	`= 5/40 = 0.125\ text(L)`

`:.\ text(Joel should add 0.125 L of oil)`

ii. `text(4.1 L)\ = 4100\ text(mL)`

`text(In ratio 40:1, there is)`

♦♦♦ Mean mark 8%
STRATEGY: The critical information required is how much petrol is in the container. Once known, the oil required to satisfy a 25:1 ration can be easily found.

`=>\ text(4000 mL Petrol and 100 mL Oil)`

`text(Lawnmower ratio) = 25:1`

`text(Let Oil required for 4000 mL Petrol)\ = X\ text(mL)`

`X/4000`	`=1/25`
`25X`	`=4000`
`X`	`=4000/25`
	`=160\ text(mL)`

`text(4 L of petrol requires 160 mL of oil)`

`text(Container already has 100 mL of oil)`

`:.\ text(Oil to add)\ `	`=160\ -100`
	`=60\ text(mL)`

Probability, STD2 S2 2013 HSC 30b

In a class there are 15 girls (G) and 7 boys (B). Two students are chosen at random to be class representatives.

Complete the tree diagram below. (2 marks)

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What is the probability that the two students chosen are of the same gender? (2 marks)

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Show Answers Only

`6/11`

Show Worked Solution

ii.	`Ptext{(same gender)}`	`=P(G,G) + P(B,B)`
		`=(15/22 xx 14/21) + (7/22 xx 6/21)`
		`=210/462 + 42/462`
		`=252/462`
		`=6/11`

♦ Mean mark (ii) 40%.

Algebra, STD2 A4 2013 HSC 30a

Wind turbines, such as those shown, are used to generate power.

In theory, the power that could be generated by a wind turbine is modelled using the equation

`T = 20\ 000w^3`

where	`T` is the theoretical power generated, in watts
	`w` is the speed of the wind, in metres per second.

Using this equation, what is the theoretical power generated by a wind turbine if the wind speed is 7.3 m/s ? (1 mark)

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In practice, the actual power generated by a wind turbine is only 40% of the theoretical power.

If `A` is the actual power generated, in watts, write an equation for `A` in terms of `w`. (1 mark)

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The graph shows both the theoretical power generated and the actual power generated by a particular wind turbine.

Using the graph, or otherwise, find the difference between the theoretical power and the actual power generated when the wind speed is 9 m/s. (1 mark)

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A particular farm requires at least 4.4 million watts of actual power in order to be self-sufficient.

What is the minimum wind speed required for the farm to be self-sufficient? (1 mark)

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A more accurate formula to calculate the power (`P`) generated by a wind turbine is

`P = 0.61 xx pi xx r^2 × w^3`

where	`r` is the length of each blade, in metres
	`w` is the speed of the wind, in metres per second.

Each blade of a particular wind turbine has a length of 43 metres.The turbine operates at a wind speed of 8 m/s.

Using the formula above, if the wind speed increased by 10%, what would be the percentage increase in the power generated by this wind turbine? (3 marks)

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Show Answers Only

`7\ 780\ 340\ text(watts)`
`8000w^3`
`8.8\ text(million watts, or 8 748 000 watts)`
`w = 8.2\ text(m/s)\ \ \ text{(1 d.p.)}`
`text(33%)`

Show Worked Solution

i.	`T=20\ 000w^3`
	`text(If)\ \ w = 7.3`

`T`	`=20\ 000 xx (7.3)^3`
	`= 7\ 780\ 340\ text(watts)`

♦ Mean mark 34%

ii.

`text(We know)\ A = 40% xx T`

`=> A`	`=0.4 xx 20\ 000 xx w^3`
	`=8000w^3`

♦ Mean mark 38%

iii.	`text(Solution 1)`
	`text(At)\ w=9`
	`A = text(5.8 million watts)\ \ \ text{(from graph)}`
	`T = text(14.6 million watts)\ \ \ text{(from graph)}`

`text(Difference)`	`= text(14.6 million)\-text(5.8 million)`
	`= text(8.8 million watts)`

`text(Alternative Solution)`

`text(At)\ w=9`

`T`	`= 20\ 000 xx 9^3`
	`= 14\ 580\ 000\ text(watts)`

`A`	`= 8000 xx 9^3`
	`= 5\ 832\ 000\ text(watts)`

`text(Difference)`	`=14\ 580\ 000\-5\ 832\ 000`
	`=8\ 748\ 000\ \ text(watts)`

♦♦ Mean mark 25%
COMMENT: Students need to be comfortable in finding the cube roots of values – a calculation that can be required in a number of topic areas and is regularly examined.

iv.

`text(Find)\ w\ text(if)\ A=4.4\ text(million)`

`8000w^3`	`= 4\ 400\ 000`
`w^3`	`= (4\ 400\ 000)/8000`
	`= 550`

`:. w`	`= root(3)(550)`
	`=8.1932…`
	`=8.2\ text(m/s)\ \ \ text{(1 d.p.)}`

`:.\ text(The minimum wind speed required is 8.2 m/s)`

♦ Mean mark 41%
MARKER’S COMMENT: Students are reminded that a % increase requires them to find the difference in power generated at different wind speeds and divide this result by the original power output, as shown in the Worked Solution.

v.	`text(Find)\ P\ text(when)\ w=8\ text(and)\ r=43`

`P`	`= 0.61 xx pi xx r^2 xx w^3`
	`= 0.61 xx pi xx 43^2 xx 8^3`
	`= 1\ 814\ 205.92\ text(watts)`

`text(When speed of wind)\ uarr10%`

`w′ = 8 xx 110text(%) = 8.8\ text(m/s)`

`text(Find)\ P\ text(when)\ w′ = 8.8`

`P`	`=0.61 xx pi xx 43^2 xx 8.8^3`
	`= 2\ 414\ 708.08\ text(watts)`

`text(Increase in Power)`	`=2\ 414\ 708.08\-1\ 814\ 205.92`
	`= 600\ 502.16`

`:.\ text(% Power increase)`	`= (600\ 502.16)/(1\ 814\ 205.92)`
	`= 0.331`
	`= text(33%)\ \ \ text{(nearest %)}`

Measurement, STD2 M6 2010 HSC 26d

Find the area of triangle `ABC`, correct to the nearest square metre. (3 marks)

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

Show Answers Only

`717\ text(m²)` `text{(nearest m²)}`

Show Worked Solution

♦♦ Mean mark 32%.
TIP: The allocation of 3 marks to this question should flag the need for more than 1 step.

`cos/_C`	`=(AC^2 + CB^2-AB^2)/(2 xx AC xx CB)`
	`=(50^2 + 40^2-83^2)/(2 xx 50 xx 40)`
	`= -0.69725…`
`/_C`	`=134.2067…^@`

`text(Using Area) = 1/2 ab\ sinC :`

`text(Area)\ Delta ABC`	`=1/2 xx 50 xx 40 xx sin134.2067…^@`
	`=716.828…`
	`=717\ text(m²)\ \ \ \ text{(nearest m²)}`

Probability, 2UG 2013 HSC 29d

Jane plays a game which involves two coins being tossed. The amounts to be won for the different possible outcomes are shown in the table.

It costs `$4` to play one game. Will Jane expect a gain or a loss, and how much will it be? Justify your answer with suitable calculations. (3 marks)

Show Answers Only

`text(Loss of $1.50.)`

Show Worked Solution

`P (H,H) = 1/2 xx 1/2 = 1/4`

♦♦ Mean mark 31%
MARKER’S COMMENT: Students must be comfortable with how to set out and calculate such financial expectation examples, as well as interpreting their result.

`P (H\ text{and}\ T)`	`= P (H,T) + P (T,H)`
	`= (1/2 xx 1/2) + (1/2 xx 1/2)`
	`= 1/2`

`P (T,T) = 1/2 xx 1/2 = 1/4`

`text(Financial Expectation)`

`=(1/4 xx 6) + (1/2 xx 1) + (1/4 xx 2)-4`

`=1.50 + 0.5 + 0.5 -4`

`=-1.50`

`:.\ text(Jane should expect a loss of $1.50.)`

Probability, STD2 S2 2011 HSC 24b

A die was rolled 72 times. The results for this experiment are shown in the table.

\begin{array} {|c|c|}
\hline
\rule{0pt}{2.5ex} \textit{Number obtained} \rule[-1ex]{0pt}{0pt} & \textit{Frequency} \\
\hline
\rule{0pt}{2.5ex} \ 1 \rule[-1ex]{0pt}{0pt} & 16 \\
\hline
\rule{0pt}{2.5ex} \ 2 \rule[-1ex]{0pt}{0pt} & 11 \\
\hline
\rule{0pt}{2.5ex} \ 3 \rule[-1ex]{0pt}{0pt} & \textbf{A} \\
\hline
\rule{0pt}{2.5ex} \ 4 \rule[-1ex]{0pt}{0pt} & 8 \\
\hline
\rule{0pt}{2.5ex} \ 5 \rule[-1ex]{0pt}{0pt} & 12 \\
\hline
\rule{0pt}{2.5ex} \ 6 \rule[-1ex]{0pt}{0pt} & 15 \\
\hline
\end{array}

Find the value of `A`. (1 mark)

--- 1 WORK AREA LINES (style=lined) ---
What was the relative frequency of obtaining a 4. (1 mark)

--- 1 WORK AREA LINES (style=lined) ---
If the die was unbiased, which number was obtained the expected number of times? (1 mark)

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

Show Answers Only

$10$
$\dfrac{1}{9}$
$5$

Show Worked Solution

i. $\text{Since die rolled 72 times}$

$\therefore\ A$	$=72-(16+11+8+12+15)$
	$=72-62$
	$=10$

♦ Mean mark 38%
IMPORTANT: Many students confused ‘relative frequency’ with ‘frequency’ and incorrectly answered 8.

ii. $\text{Relative frequency of 4}$	$=\dfrac{8}{72}$
	$=\dfrac{1}{9}$

iii. $\text{Expected frequency of any number}$

$=\dfrac{1}{6}\times 72$

$=12$

$\therefore\ \text{5 was obtained the expected number of times.}$

Algebra, STD2 A2 2011 HSC 23b

Sticks were used to create the following pattern.

The number of sticks used is recorded in the table.

\begin{array} {|l|c|}
\hline
\rule{0pt}{2.5ex} \text{Shape $(S)$} \rule[-1ex]{0pt}{0pt} & \;\;\; 1 \;\;\; & \;\;\; 2 \;\;\; & \;\;\; 3 \;\;\; \\
\hline
\rule{0pt}{2.5ex} \text{Number of sticks $(N)$}\; \rule[-1ex]{0pt}{0pt} & \;\;\; 5 \;\;\; & \;\;\; 8 \;\;\; & \;\;\; 11 \;\;\; \\
\hline
\end{array}

Draw Shape 4 of this pattern. (1 mark)

--- 4 WORK AREA LINES (style=lined) ---
How many sticks would be required for Shape 100? (1 mark)

--- 3 WORK AREA LINES (style=lined) ---
Is it possible to create a shape in this pattern using exactly 543 sticks?

Show suitable calculations to support your answer. (2 marks)

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

Show Answers Only

`text(See Worked Solutions.)`
`302`
`text(No)`

Show Worked Solution

i. `text(Shape 4 is shown below:)`

ii. `text(S)text(ince)\ \ N=2+3S`

♦ Mean mark 48%.
MARKER’S COMMENT: Students should attempt to find a “rule” in such questions, and use this formula to solve the question, as per the Worked Solution.

`text(If)\ \ S`	`=100`,
`N`	`=2+(3xx100)`
	`=302`

iii.	`543`	`=2+3S`
	`3S`	`=541`
	`S`	`=180 1/3`

`text(S)text(ince S is not a whole number, 543 sticks)`

`text(will not create a figure.)`

Financial Maths, STD2 F4 2011 HSC 22 MC

Ying borrowed $250 000 to buy a house. The interest rate and monthly repayment for her loan are shown in the spreadsheet.

What is the total interest charged for the first four months of this loan?

$6364.32
$6366.11
$6369.67
$6376.25

Show Answers Only

`A`

Show Worked Solution

`text(Month 3)`

`P+I-R`	`=251\ 032.04-1871.94`
	`=249\ 160.10`

`text(Month 4)`

`P`	`=249\ 160.10`
`:.I`	`=249\ 160.10xx0.0765/12`
	`=1588.40`

`:.\ text(Total interest)`	`=1593.75+1591.98+1590.19+1588.40`
	`=6364.32`

`=> A`

Algebra, STD2 A4 2011 HSC 20 MC

A function centre hosts events for up to 500 people. The cost `C`, in dollars, for the centre
to host an event, where `x` people attend, is given by:

`C = 10\ 000 + 50x`

The centre charges $100 per person. Its income `I`, in dollars, is given by:

`I = 100x`

How much greater is the income of the function centre when 500 people attend an event, than its income at the breakeven point?

`$15\ 000`
`$20\ 000`
`$30\ 000`
`$40\ 000`

Show Answers Only

`C`

Show Worked Solution

♦ Mean mark 50%
COMMENT: Students can read the income levels directly off the graph to save time and then check with the equations given.

`text(When)\ x=500,\ I=100xx500=$50\ 000`

`text(Breakeven when)\ \ x=200\ \ \ text{(from graph)}`

`text(When)\ \ x=200,\ I=100xx200=$20\ 000`

`text(Difference)`	`=50\ 000-20\ 000`
	`=$30\ 000`

`=> C`

Financial Maths, STD2 F1 2011 HSC 19 MC

Simon is a mechanic who receives a normal rate of pay of $22.35 per hour for a 40-hour
week.

When he is needed for emergency call-outs he is paid a special allowance of $150 for that
week. Additionally, every time he is called out to an emergency he is paid for a minimum
of 4 hours at double time.

In the week beginning 2 February, 2011 Simon worked 40 hours normal time and was
needed for emergency call-outs. His emergency call-out log book for the week is shown.

What was Simon’s total pay for that week?

$1189.28
$1296.30
$1334.55
$1446.30

Show Answers Only

`D`

Show Worked Solution

♦ Mean mark 41%
COMMENT: To reduce errors, calculate each element of pay separately in your working and then add up the total, as per the Worked Solution.

`text(Normal pay)`	`=40xx22.35`
	`=894.00`

`text(Special Allowance)= 150.00`

`text(Emergency Calls)`	`= (5xx2xx22.35)+(4xx2xx22.35)`
	`=223.50+178.80`
	`=402.30`

`text(Total Weekly Pay)`	`=894.00+150.00+402.30`
	`=1446.30`

`=> D`

Measurement, 2UG 2011 HSC 24c

A ship sails 6 km from `A` to `B` on a bearing of 121°. It then sails 9 km to `C`. The
size of angle `ABC` is 114°.

Copy the diagram into your writing booklet and show all the information on it.

What is the bearing of `C` from `B`? (1 mark)
Find the distance `AC`. Give your answer correct to the nearest kilometre. (2 marks)
What is the bearing of `A` from `C`? Give your answer correct to the nearest degree. (3 marks)

Show Answers Only

`055^@`
`13\ text(km)`
`261^@`

Show Worked Solution

♦♦ Mean mark 24%
STRATEGY: This deserves repeating again: Draw North-South parallel lines through major points to make the angle calculations easier.

(i)

`text(Let point)\ D\ text(be due North of point)\ B`

`/_ABD`	`=180-121\ text{(cointerior with}\ \ /_A text{)}`
	`=59^@`
`/_DBC`	`=114-59`
	`=55^@`

`:. text(Bearing of)\ \ C\ \ text(from)\ \ B\ \ text(is)\ 055^@`

(ii) `text(Using cosine rule:)`

♦ Mean mark 39%

`AC^2`	`=AB^2+BC^2-2xxABxxBCxxcos/_ABC`
	`=6^2+9^2-2xx6xx9xxcos114^@`
	`=160.9275…`
`:.AC`	`=12.685…\ \ \ text{(Noting}\ AC>0 text{)}`
	`=13\ text(km)\ text{(nearest km)}`

(iii) `text(Need to find)\ /_ACB\ \ \ text{(see diagram)}`

♦♦♦ Mean mark 15%
MARKER’S COMMENT: The best responses clearly showed what steps were taken with working on the diagram. Note that all North/South lines are parallel.

`cos/_ACB`	`=(AC^2+BC^2-AB^2)/(2xxACxxBC)`
	`=((12.685…)^2+9^2-6^2)/(2xx(12.685..)xx9)`
	`=0.9018…`
`/_ACB`	`=25.6^@\ text{(to 1 d.p.)}`

`text(From diagram,)`

`/_BCE=55^@\ text{(alternate angle,}\ DB\ text(||)\ CE text{)}`

`:.\ text(Bearing of)\ A\ text(from)\ C`

	`=180+55+25.6`
	`=260.6`
	`=261^@\ text{(nearest degree)}`

Financial Maths, STD2 F5 2009 HSC 27a

The table shows the future value of a $1 annuity at different interest rates over different numbers of time periods.

What would be the future value of a $5000 per year annuity at 3% per annum for 6 years, with interest compounding yearly? (1 mark)

--- 1 WORK AREA LINES (style=lined) ---
What is the value of an annuity that would provide a future value of $407100 after 7 years at 5% per annum compound interest? (1 mark)

--- 1 WORK AREA LINES (style=lined) ---
An annuity of $1000 per quarter is invested at 4% per annum, compounded quarterly for 2 years. What will be the amount of interest earned? (3 marks)

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

Show Answers Only

`$32\ 342`
`$50\ 000`
`$285.70`

Show Worked Solution

i. `text(Table factor when)\ \ n = 6,\ \ \ r =\ 3text(%) \ => \ 6.4684`

`:.\ FV`	`= 5000 xx 6.4684`
	`= $32\ 342`

ii. `text(Table factor when)\ \ n = 7,\ \ \ r =\ text(5%)`

♦ Mean mark 45%
MARKER’S COMMENT: A common error was to multiply $407 100 by 8.1420 rather than divide.

`=> 8.1420`

`text(Let)\ \ A = text(annuity)`

`FV`	`= A xx 8.1420`
`A`	`= (FV)/8.1420`
	`= (407\ 100)/8.1420`
	`= $50\ 000`

iii. `n=8\ \ \ (text(8 quarters in 2 years) )`

♦♦ Mean mark 31%
MARKER’S COMMENT: When questions asked for the interest paid on annuities, remember to subtract the total principal amounts contributed.

`r = text(4%)/4 =\ text{1% per quarter}`

`:.\ text(Table factor) => 8.2857`

`FV`	`=1000 xx 8.2857`
	`=8285.70`

`text(Interest)`	`= FV (text(annuity) )\ – text(Principal)`
	`= 8285.70\ – (8 xx 1000)`
	`= 285.70`

`:.\ text(Interest earned is $285.70)`

Algebra, STD2 A4 2009 HSC 28c

The height above the ground, in metres, of a person’s eyes varies directly with the square of the distance, in kilometres, that the person can see to the horizon.

A person whose eyes are 1.6 m above the ground can see 4.5 km out to sea.

How high above the ground, in metres, would a person’s eyes need to be to see an island that is 15 km out to sea? Give your answer correct to one decimal place. (3 marks)

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

Show Answers Only

`17.8\ text(m)\ \ text{(to 1 d.p.)}`

Show Worked Solution

♦♦ Mean mark 22%
CRITICAL STEP: Reading the first line of the question carefully and establishing the relationship `h=k d^2` is the key part of solving this question.

`h prop d^2`

`h=kd^2`

`text(When)\ h = 1.6,\ d = 4.5`

`1.6`	`= k xx 4.5^2`
`:. k`	`= 1.6/4.5^2`
	`= 0.07901` `…`

`text(Find)\ h\ text(when)\ d = 15`

`h`	`= 0.07901… xx 15^2`
	`= 17.777…`
	`= 17.8\ text(m)\ \ \ text{(to 1 d.p.)}`

Statistics, STD2 S4 2009 HSC 28b

The height and mass of a child are measured and recorded over its first two years.

\begin{array} {|l|c|c|}
\hline \rule{0pt}{2.5ex} \text{Height (cm), } H \rule[-1ex]{0pt}{0pt} & \text{45} & \text{50} & \text{55} & \text{60} & \text{65} & \text{70} & \text{75} & \text{80} \\
\hline \rule{0pt}{2.5ex} \text{Mass (kg), } M \rule[-1ex]{0pt}{0pt} & \text{2.3} & \text{3.8} & \text{4.7} & \text{6.2} & \text{7.1} & \text{7.8} & \text{8.8} & \text{10.2} \\
\hline
\end{array}

This information is displayed in a scatter graph.

Describe the correlation between the height and mass of this child, as shown in the graph. (1 mark)

--- 2 WORK AREA LINES (style=lined) ---
A line of best fit has been drawn on the graph.

Find the equation of this line. (2 marks)

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

Show Answers Only

`text(The correlation between height and)`

`text(mass is positive and strong.)`
`M = 0.23H-8`

Show Worked Solution

i. `text(The correlation between height and)`

♦ Mean mark 48%.

`text(mass is positive and strong.)`

ii. `text(Using)\ \ P_1(40, 1.2)\ \ text(and)\ \ P_2(80, 10.4)`

♦♦♦ Mean mark 18%.
MARKER’S COMMENT: Many students had difficulty due to the fact the horizontal axis started at `H= text(40cm)` and not the origin.

`text(Gradient)`	`= (y_2-y_1)/(x_2-x_1)`
	`= (10.4-1.2)/(80-40)`
	`= 9.2/40`
	`= 0.23`

`text(Line passes through)\ \ P_1(40, 1.2)`

`text(Using)\ \ \ y-y_1`	`= m(x-x_1)`
`y-1.2`	`= 0.23(x-40)`
`y-1.2`	`= 0.23x-9.2`
`y`	`= 0.23x-8`

`:. text(Equation of the line is)\ \ M = 0.23H-8`

Algebra, STD2 A4 2009 HSC 28a

Anjali is investigating stopping distances for a car travelling at different speeds. To model this she uses the equation

`d = 0.01s^2+ 0.7s`,

where `d` is the stopping distance in metres and `s` is the car’s speed in km/h.

The graph of this equation is drawn below.

Anjali knows that only part of this curve applies to her model for stopping distances.

In your writing booklet, using a set of axes, sketch the part of this curve that applies for stopping distances. (1 mark)
What is the difference between the stopping distances in a school zone when travelling at a speed of 40 km/h and when travelling at a speed of 70 km/h? (2 marks)

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

Show Answers Only

`54\ text(metres)`

Show Worked Solution

(i)

(ii) `text(When)\ \ s = 40`

♦ Mean mark 41%
COMMENT: Students could easily have used the graph for calculating `d` at `text(40 km/h)`, although the formula was required when the speed increased to `text(70 km/h)`.

`d`	`= 0.01(40^2)+ 0.7 (40)`
	`= 16+28`
	`= 44\ text(m)`

`text(When)\ \ s = 70`

`d`	`= 0.01 (70^2) +0.7(70)`
	`= 49 +49`
	`= 98\ text(m)`

`:.\ text(Difference)`	`= 98\ -44`
	`= 54\ text(metres)`