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Calculus, EXT1 C1 EQ-Bank 12

A tank is initially full. It is drained so that at time `t` seconds the volume of water, `V`, in litres, is given by

`V = 50(1 - t/80)^2`  for  `0 <= t <= 100`

  1.  How much water was initially in the tank?  (1 mark)

  2.  After how many seconds was the tank one-quarter full?  (1 mark)

  3.  At what rate was the water draining out the tank when it was one-quarter full?  (2 marks)

Show Answers Only
  1.  `50\ text(L)`
  2.  `40\ text(seconds)`
  3.  `5/8\ text(litres per second)`
Show Worked Solution

i.   `text(Initially,)\ \ t = 0.`

`V` `= 50(1 – 0)^2`
  `= 50\ text(L)`

 

ii.   `text(Find)\ t\ text(when tank is)\ \ 1/4\ \ text(full:)`

`50/4` `= 50(1 – t/80)^2`
`(1 – 1/80)^2` `= 1/4`
`1 – t/80` `= 1/2`
`t/80` `= 1/2`
`t` `= 40\ text(seconds)`

 

iii.   `(dv)/(dt)` `= 2 · 50(1 – t/80)(−1/80)`
  `= −5/4(1 – t/80)`

 
`text(When)\ \ t = 40,`

`(dv)/(dt)` `= −5/4(1 – 40/80)`
  `= −5/8`

 
`:. text(Water is draining out at)\ \ 5/8\ \ text(litres per second.)`

Trigonometry, EXT1 T2 SM-Bank 4

If  `costheta = −3/4`  and  `0 < theta < pi`,

determine the exact value of  `tantheta`.  (2 marks)

Show Answers Only

`tan theta = – sqrt7/3`

Show Worked Solution

`text(Consider the angle graphically:)`

 
`text(S)text(ince)\ \ costheta\ \ text(is negative  ⇒  2nd quadrant.)`

`text(Using Pythagoras,)`

`x^2` `= 4^2 – 3^2`
`x` `= sqrt7`

 
`:. tan theta = – sqrt7/3`

Trigonometry, EXT1 T2 EQ-Bank 1

Find  `a`  and  `b`  such that

`tan75^@ = a + bsqrt3`   (2 marks)

Show Answers Only

`a = 2, b = 1`

Show Worked Solution
`tan75^@` `= tan(45 + 30)^@`
  `= (tan45^@ + tan30^@)/(1 – tan45^@tan30^@)`
  `= (1 + 1/sqrt3)/(1 – 1 · 1/sqrt3) xx sqrt3/sqrt3`
  `= (sqrt3 + 1)/(sqrt3 – 1) xx (sqrt3 + 1)/(sqrt3 + 1)`
  `= (3 + 2sqrt3 + 1)/((sqrt3)^2 – 1^2)`
  `= 2 + sqrt3`

 
`:. a = 2, \ b = 1`

Functions, EXT1 F1 EQ-Bank 7

The parametric equations of a graph are

`x = 1 - 1/t`

`y = 1 + 1/t`
 

Sketch the graph.  (2 marks)

Show Answers Only

Show Worked Solution
`x` `= 1 – 1/t\ \ …\ (1)`
`tx` `= t – 1`
`t(1 – x)` `= 1`
`t` `= 1/(1 – x)\ \ …\ (1′)`
`y` `= 1 + 1/t\ \ …\ (2)`

 
`text(Substitute)\ \ t = 1/(1 – x)\ \ text{from (1′) into (2)}:`

`y` `= 1 + 1/(1/(1 – x))`
`y` `= 1 + 1 – x`
`y` `= 2 – x`

 

Functions, EXT1 F1 EQ-Bank 6

The parametric equations of a graph are

`x = t^2`

`y = 1/t`  for  `t > 0`
 

  1. Find the Cartesian equation for the graph.  (1 mark)

  2. Sketch the graph.  (1 mark)

Show Answers Only

i.   `y = sqrt(1/x)`

ii.

Show Worked Solution

i.     `x = t^2\ …\ (1)`

`y = 1/t\ …\ (2)`
 

`text(Substitute)\ \ t = 1/y\ \ text{from (1) into (1)}`

`x` `= (1/y)^2`
`y^2` `= 1/x`
`y` `= sqrt(1/x)\ \ \ (y > 0\ \ text(as)\ \ t > 0)`

 

ii.

Functions, EXT1 F1 SM-Bank 4

Sketch the curve described by these two parametric equations

`x = 3cost + 2`

`y = 3sint - 3`   for   `0 <= t < 2pi`.  (3 marks)

Show Answers Only

Show Worked Solution

`(x – 2) = 3cost`

`(y + 3) = 3sint`

`(x – 2)^2 + (y + 3)^2` `= (3cost)^2 + (3sint)^2`
  `= 9cos^2t + 9sin^2t`
  `= 9(cos^2t + sin^2t)`
  `= 9`

 
`text(Sketch:) \ (x – 2)^2 + (y + 3)^2 = 3^2`

Functions, EXT1 F1 SM-Bank 2

Solve  `3/(|\ x - 3\ |) < 3`.  (3 marks)

Show Answers Only

`x < 2\ ∪\ x > 4`

Show Worked Solution

`text(Solution 1)`

`3/(|\ x – 3\ |) < 3`

`|\ x – 3\ |` `> 1`
`(x^2 – 6x + 9)` `> 1^2`
`x^2 – 6x + 8` `> 0`
`(x – 4)(x – 2)` `> 0`

 

 

`:. {x: \ x < 2\ ∪\ x > 4}`

 

`text(Solution 2)`

`|\ x – 3\ | > 1`

`text(If)\ \ (x – 3)` `> 0,\ text(i.e.)\ x >3`
`x – 3` `> 1`
`x` `> 4`

 
`=> x > 4\ (text(satisfies both))`
 

`text(If)\ \ (x – 3)` `< 0,\ text(i.e.)\ x <3`
`−(x – 3)` `> 1`
`−x + 3` `> 1`
`x` `< 2`

 
`=> x < 2\ (text(satisfies both))`

`:. {x: \ x < 2\ ∪\ x > 4}`

Functions, EXT1 F1 SM-Bank 3

A circle has centre `(5,3)` and radius 3.

  1.  Describe, with inequalities, the region that consists of the interior of the circle and more than 2 units above the `x`-axis.  (2 marks)

  2.  Sketch the region.  (1 mark)

Show Answers Only
  1. `(x-5)^2 + (y-3)^2 < 9\ ∩\ y > 2`

  2.  

Show Worked Solution

i.   `text(Equation of circle:)`

`(x-5)^2 + (y-3)^2 = 3^2`
 

`:.\ text(Region is:)`

`(x-5)^2 + (y- 3)^2 < 9\ ∩\ y > 2`

COMMENT: The broken line on the graph represents an excluded boundary.

 

ii.   

Functions, EXT1 F1 SM-Bank 1

Find the values of `x` that satisfy the equation

  `x^2 + 8x + 3 <= 0`.  (3 marks)

Show Answers Only

`{x:\  -4-sqrt{13} <=x <= -4 + sqrt{13}}`

Show Worked Solution
`x` `= (-8 ± sqrt(8^2-4 · 1 · 3))/2`
  `= (-8 ± sqrt(52))/2`
  `= -4 ± sqrt13`

 


 

`{x:\  -4-sqrt13 <=x <= -4 + sqrt13}`

Complex Numbers, SPEC2 2017 VCAA 5 MC

 On an Argand diagram, a point that lies on the path defined by  `|\ z - 2 + i\ | = |\ z - 4\ |`  is

  1.  `(3, −1/2)`
  2.  `(−3, −1/2)`
  3.  `(−3, 3/2)`
  4.  `(3, 1/2)`
  5.  `(3, −3/2)`
Show Answers Only

`A`

Show Worked Solution

`text(Equidistant from)\ \ z = 4\ \ text(and)\ \ z= 2 + i`

`text(Midpoint:)\ ((2 + 4)/2, (−1 + 0)/2)`

`= (3, −1/2)`

`=>A`

Functions, EXT1 F1 2010 HSC 3b*

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

 Inverse Functions, EXT1 2010 HSC 3b

  1. The graph has two points of inflection.  

     

    Find the  `x`  coordinates of these points.   (3 marks)

  2. Explain why the domain of  `f(x)`  must be restricted if  `f(x)`  is to have an inverse function.    (1 mark)

  3. Find a formula for  `f^(-1) (x)`  if the domain of  `f(x)`  is restricted to  `x ≥ 0`.   (2 marks)

  4. State the domain of  `f^(-1) (x)`.    (1 mark)

  5. Sketch the curve  `y = f^(-1) (x)`.    (1 mark)

Show Answers Only
  1. `x = +- 1/sqrt2` 
  2. `text(There can only be 1 value of)\ y\ text(for each value of)\ x.`
  3. `f^(-1)x = sqrt(ln(1/x))`
  4. `0 <= x <= 1`
  5.  
  6. Inverse Functions, EXT1 2010 HSC 3b Answer
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. 

Inverse Functions, EXT1 2010 HSC 3b Answer

Functions, EXT1 F1 2004 HSC 5b*

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

Inverse Functions, EXT1 2004 HSC 5b

  1. On the same set of axes, sketch the graph of the inverse function,  `y = f^(−1)(x)`.  (1 mark)
  2. State the domain of  `f^(−1)(x)`.  (1 mark)

  3. Find an expression for  `y = f^(−1)(x)`  in terms of  `x`.  (2 marks)

  4. The graphs of  `y = f(x)`  and  `y = f^(−1)(x)`  meet at exactly one point  `P`.

     

    Let  `α`  be the `x`-coordinate of  `P`. Explain why  `α`  is a root of the equation  `x^3 + x − 1 = 0`.  (1 mark)

Show Answers Only
  1. `text(See Worked Solutions)`
  2. `0 < x ≤ 1`
  3. `y = sqrt((1 − x)/x), \ y > 0`
  4. `text(See Worked Solutions)`
Show Worked Solution
i.  

Inverse Functions, EXT1 2004 HSC 5b Answer

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

`0 < x ≤ 1`
 

iii.  `f(x) = 1/(1 + x^2)`

`text(Inverse: swap)\ \ x↔y`

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

 

`:.y = sqrt((1 − x)/x), \ \ y >= 0`

 

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

`text(i.e. where)`

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

 

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

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

Functions, 2ADV F1 SM-Bank 11

Given  `f(x) = sqrt (x^2 - 9)`  and  `g(x) = x + 5`

  1.  Find integers `c` and `d` such that  `f(g(x)) = sqrt {(x + c) (x + d)}`   (2 marks)

  2.  State the domain for which  `f(g(x))`  is defined.   (2 marks)

Show Answers Only
  1. `c = 2, d = 8 or c = 8, d = 2`
  2. `x in (– oo, – 8] uu [– 2, oo)`
Show Worked Solution
a.   `f(g(x))` `= sqrt {(x + 5)^2 – 9}`
    `= sqrt (x^2 + 10x + 16)`
    `= sqrt {(x + 2) (x + 8)}`

 
 `:. c = 2, d = 8 or c = 8, d = 2`

 

b.   `text(Find)\ x\ text(such that:)`

♦♦♦ Mean mark 13%.
MARKER’S COMMENT: “Very poorly answered” with a common response of  `-3 <=x<=3`  that ignored the information from part (a).

`(x+2)(x+8) >= 0`
 

 vcaa-2011-meth-4ii

`(x + 2) (x + 8) >= 0\ \ text(when)`

`x <= -8 or x >= -2`
 

`:.\ text(Domain:)\ \ x<=-8\ \ and\ \  x>=-2`

Functions, 2ADV F1 SM-Bank 9 MC

If  `f(x) = 1/2e^(3x)  and  g(x) = log_e(2x) + 3`  then  `g (f(x))` is equal to
 

A.   `3(x + 1)`

B.   `e^(3x) + 3`

C.   `e^(8x + 9)`

D.   `log_e (3x) + 3`

Show Answers Only

`A`

Show Worked Solution
`g(f(x))` `= log_e(2 xx 1/2e^(3x)) + 3`
  `=log_e e^(3x) + 3`
  `=3x + 3`
  `= 3 (x + 1)`

 
`=> A`

Functions, 2ADV F1 SM-Bank 5 MC

Let  `g(x) = x^2 + 2x - 3`  and  `f(x) = e^(2x + 3).`

Then  `f(g(x))`  is given by
 

A.   `e^(4x + 6) + 2 e^(2x + 3) - 3`

B.   `2x^2 + 4x - 6`

C.   `e^(2x^2 + 4x - 3)`

D.   `e^(2x^2 + 4x - 6)`

Show Answers Only

`C`

Show Worked Solution

`text(By trial and error,)`

`text(Consider:)\ \ f(x) = e^(2x^2 + 4x – 3)`

`f(g(x))` `=e^(2 (x^2 + 2x – 3)+3)`
  `= e^(2x^2 + 4x – 3)`

 
`=> C`

Calculus, 2ADV C4 2013* HSC 15a

The diagram shows the front of a tent supported by three vertical poles. The poles are 1.2 m apart. The height of each outer pole is 1.5 m, and the height of the middle pole is 1.8 m. The roof hangs between the poles.

2013 15a

The front of the tent has area  `A\ text(m²)`. 

  1. Use the trapezoidal rule to estimate  `A`.    (1 mark)

  2. Does the Trapezoidal rule give a higher or lower estimate of the actual area? Justify your answer.  (1 mark)

Show Answers Only
  1. `3.96\ text(m²)`
  2. `text(See Worked Solutions)`
Show Worked Solution
i.    `A` `~~ h/2 [y_0 + 2y_1 + y_2]`
    `~~ 1.2/2 [1.5 + (2 xx 1.8) + 1.5]`
    `~~ 0.6 [6.6]`
    `~~ 3.96\ text(m²)`

 

ii.       

`text(S)text(ince the tent roof is concave up, the)`

`text(Trapezoidal rule uses straight lines and)`

`text(will estimate a higher area.)`

Probability, 2ADV S1 2015 MET1 8

For events `A` and `B` from a sample space, `P(A | B) = 3/4`  and  `P(B) = 1/3`.

  1.  Calculate  `P(A ∩ B)`.  (1 mark)

  2.  Calculate  `P(A′ ∩ B)`, where `A′` denotes the complement of `A`.  (1 mark)

  3.  If events `A` and `B` are independent, calculate  `P(A ∪ B)`.  (1 mark)

Show Answers Only
  1. `1/4`
  2. `1/12`
  3. `5/6`
Show Worked Solution

i.   `text(Using Conditional Probability:)`

`P(A | B)` `= (P(A ∩ B))/(P(B))`
`3/4` `= (P(A ∩ B))/(1/3)`
`:. P(A ∩ B)` `= 1/4`

 

ii.    met1-2015-vcaa-q8-answer
`P(A′ ∩ B)` `= P(B) – P(A ∩B)`
  `= 1/3 – 1/4`
  `= 1/12`

 

iii.   `text(If)\ A, B\ text(independent)`

♦♦ Mean mark 28%.
MARKER’S COMMENT: A lack of understanding of independent events was clearly evident.
`P(A ∩ B)` `= P(A) xx   P(B)`
`1/4` `= P(A) xx 1/3`
`:. P(A)` `= 3/4`

 

`P(A ∪ B)` `= P(A) + P(B) – P(A ∩ B)`
  `= 3/4 + 1/3 – 1/4`
`:. P(A ∪ B)` `= 5/6`

Probability, 2ADV S1 2009 MET1 5

Four identical balls are numbered 1, 2, 3 and 4 and put into a box. A ball is randomly drawn from the box, and not returned to the box. A second ball is then randomly drawn from the box.

  1. What is the probability that the first ball drawn is numbered 4 and the second ball drawn is numbered 1?  (1 mark)

  2. What is the probability that the sum of the numbers on the two balls is 5?  (1 mark)

  3. Given that the sum of the numbers on the two balls is 5, what is the probability that the second ball drawn is numbered 1?  (2 marks)

Show Answers Only
  1. `1/12`
  2. `1/3`
  3. `1/4`
Show Worked Solution
i.    `P (4, 1)` `= 1/4 xx 1/3`
    `= 1/12`

 

ii.   `P (text(Sum) = 5)`

`= P (1, 4) + P (2, 3) + P (3, 2) + P (4, 1)`

`= 4 xx (1/4 xx 1/3)`

`= 1/3`

 

iii.   `text(Conditional Probability)`

♦ Mean mark 46%.

`P (2^{text(nd)} = 1\ |\ text(Sum) = 5)`

`= (P (4, 1))/(P (text(Sum) = 5))`

`= (1/12)/(1/3)`

`= 1/4`

Probability, 2ADV S1 2015 MET2 14 MC

Consider the following discrete probability distribution for the random variable `X.`

\begin{array} {|c|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} x \rule[-1ex]{0pt}{0pt} & \ \ \ 1\ \ \  & \ \ \ 2\ \ \  & \ \ \ 3\ \ \ &\ \ \ 4\ \ \ &\ \ \ 5\ \ \ \\
\hline
\rule{0pt}{2.5ex} P(X=x) \rule[-1ex]{0pt}{0pt} & p & 2p & 3p & 4p & 5p \\
\hline
\end{array} 

The mean of this distribution is

  1. `2`
  2. `3`
  3. `7/2`
  4. `11/3`
Show Answers Only

`D`

Show Worked Solution

`text(Find)\ p:`

`p + 2p + 3p + 4p + 5p` `= 1`
`:. p` `= 1/15`

 

`E(X)` `= 1 xx p + 2(2p) + 3(3p) + 4(4p) + 5(5p)`
  `= 55p`
  `= 55 xx (1/15)`
  `= 11/3`

`=>   D`

Probability, 2ADV S1 2012 MET1 4

On any given day, the number `X` of telephone calls that Daniel receives is a random variable with probability distribution given by

\begin{array} {|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} x \rule[-1ex]{0pt}{0pt} & \ \ \ 0\ \ \  & \ \ \ 1\ \ \  & \ \ \ 2\ \ \  & \ \ \ 3\ \ \  \\
\hline
\rule{0pt}{2.5ex} P(X=x) \rule[-1ex]{0pt}{0pt} & 0.2 & 0.2 & 0.5 & 0.1 \\
\hline
\end{array}

  1. Find the mean of  `X`.   (2 marks)

  2. What is the probability that Daniel receives only one telephone call on each of three consecutive days?   (1 mark)

  3. Daniel receives telephone calls on both Monday and Tuesday.

     

    What is the probability that Daniel receives a total of four calls over these two days?   (3 marks)

Show Answers Only
  1. `1.5`
  2. `0.008`
  3. `29/64`
Show Worked Solution
i.    `E(X)` `= 0 xx 0.2 + 1 xx 0.2 + 2 xx 0.5 + 3 xx 0.1`
    `= 0 + .2 + 1 + 0.3`
    `= 1.5`

 

ii.   `P(1, 1, 1)` `= 0.2 xx 0.2 xx 0.2`
    `= 0.008`

 

iii.   `text(Conditional Probability:)`

♦ Mean mark 36%.

`P(x = 4 | x >= 1\ text{both days})`

`= (P(1, 3) + P(2, 2) + P(3, 1))/(P(x>=1\ text{both days}))`

`= (0.2 xx 0.1 + 0.5 xx 0.5 + 0.1 xx 0.2)/(0.8 xx 0.8)`

`= (0.02 + 0.25 + 0.02)/0.64`

`= 0.29/0.64`

`= 29/64`

Probability, 2ADV S1 2016 MET2 7 MC

The number of pets, `X`, owned by each student in a large school is a random variable with the following discrete probability distribution.

\begin{array} {|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} x \rule[-1ex]{0pt}{0pt} & \ \ \ \ 0\ \ \ \  & \ \ \ \ 1\ \ \ \  & \ \ \ \ 2\ \ \ \  & \ \ \ \ 3\ \ \ \  \\
\hline
\rule{0pt}{2.5ex} P(X=x) \rule[-1ex]{0pt}{0pt} & 0.5 & 0.25 & 0.2 & 0.05 \\
\hline
\end{array}

If two students are selected at random, the probability that they own the same number of pets is

  1. `0.3`
  2. `0.305`
  3. `0.355`
  4. `0.405`
Show Answers Only

`C`

Show Worked Solution

`P (0, 0)+ P (1, 1)+ P (2, 2)+ P (3, 3)`

`= 0.5^2+0.25^2+0.2^2+0.05^2`

`=0.355`

 
`=>   C`

Probability, 2ADV S1 2010 MET2 15 MC

The discrete random variable `X` has the following probability distribution.

\begin{array} {|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} x \rule[-1ex]{0pt}{0pt} & \ \ \ 0\ \ \  & \ \ \ 1\ \ \  & \ \ \ 2\ \ \ \\
\hline
\rule{0pt}{2.5ex} P(X=x) \rule[-1ex]{0pt}{0pt} & a & b & 0.4 \\
\hline
\end{array}

If the mean of `X` is 1 then

  1. `a = 0.3 and b = 0.1`
  2. `a = 0.2 and b = 0.2`
  3. `a = 0.4 and b = 0.2`
  4. `a = 0.1 and b = 0.3`
Show Answers Only

`C`

Show Worked Solution

`E(X) = 1:`

`1 xx b + 2 xx 0.4` `=1`
`b` `=0.2`

 
`text(Sum of probabilities) = 1:`

`a + 0.2 + 0.4` `= 1`
`a` `=0.4`

`=>   C`

Probability, 2ADV S1 2007 MET2 19 MC

The discrete random variable `X` has probability distribution as given in the table. The mean of `X` is 5.

\begin{array} {|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} x \rule[-1ex]{0pt}{0pt} & \ \ \ 0\ \ \  & \ \ \ 2\ \ \  & \ \ \ 4\ \ \  & \ \ \ 6\ \ \ &\ \ \ 8\ \ \ \\
\hline
\rule{0pt}{2.5ex} P(X=x) \rule[-1ex]{0pt}{0pt} & a & 0.2 & 0.2 & 0.3 & b \\
\hline
\end{array}

The values of `a` and `b` are

  1. `{:(a = 0.05,\ and b = 0.25):}`
  2. `{:(a = 0.1­,\ and b = 0.29):}`
  3. `{:(a = 0.2­,\ and b = 0.9):}`
  4. `{:(a = 0.3­,\ and b = 0):}`
Show Answers Only

`A`

Show Worked Solution

`text(Sum of probabilities) = 1`

`a + 0.2 + 0.2 + 0.3 + b = 1`
  

`text(S)text{ince}\ E(X) = 5,`

`5` `=(0 xx a) + (2 xx 0.2) + (4 xx 0.2) + (6 xx 0.3) + 8b`
`8b` `=2`
`:. b` `=0.25`

 
`:. a = 0.05,\ \  b = 0.25`

`=>   A`

Calculus, 2ADV C4 2017* HSC 14b

  1. Find the exact value of  `int_0^(pi/3) cos x\ dx`.  (1 mark)

  2. Using the Trapezoidal rule with three function values, find an approximation to the integral  `int_0^(pi/3) cos x\ dx,` leaving your answer in terms of `pi` and `sqrt 3`.  (2 marks)

  3. Using parts (i) and (ii), show that  `pi ~~ (12 sqrt 3)/(3 + 2 sqrt 3)`.  (1 mark)

Show Answers Only
  1. `sqrt 3/2`
  2. `((2sqrt3 + 3)pi)/24`
  3. `text{Proof (See Worked Solutions)}`
Show Worked Solution
i.   `int_0^(pi/3) cos x\ dx` `= [sin x]_0^(pi/3)`
    `= sin\ pi/3-0`
    `= sqrt 3/2`

 

ii.  

\begin{array} {|l|c|c|c|}\hline
x & \ \ \ 0\ \ \  & \ \ \ \dfrac{\pi}{6}\ \ \  & \ \ \ \dfrac{\pi}{3}\ \ \ \\ \hline
\text{height} & 1 & \dfrac{\sqrt{3}}{2} & \dfrac{1}{2} \\ \hline
\text{weight} & 1 & 2 & 1  \\ \hline \end{array}

`int_0^(pi/3) cos x\ dx` `~~ 1/2 xx pi/6[1 + 2(sqrt3/2) + 1/2]`
  `~~ pi/12((3 + 2sqrt3)/2)`
  `~~ ((3+2sqrt3)pi)/24`

 

♦ Mean mark part (iii) 49%.

(iii)    `((3+2sqrt3)pi)/24` `~~ sqrt3/2`
  `:. pi` `~~ (24sqrt3)/(2(3+2sqrt3))`
    `~~ (12sqrt3)/(3 + 2sqrt3)`

Calculus, 2ADV C4 2012* HSC 12d

At a certain location a river is 12 metres wide. At this location the depth of the river, in metres, has been measured at 3 metre intervals. The cross-section is shown below.
 

2012 12d
  

  1. Use the Trapezoidal rule with the five depth measurements to calculate the approximate area of the cross-section   (3 marks)

  2. The river flows at 0.4 metres per second.

     

    Calculate the approximate volume of water flowing through the cross-section in 10 seconds.   (1 mark)

Show Answers Only
  1. `30.9 \ text(m²)`
  2. `123.6 \ text(m³)`  
Show Worked Solution
i.   
`A` `~~ 3/2[0.5 + 2(2.3 + 2.9 + 3.8) + 2.1]`
  `~~ 3/2(20.6)`
  `~~ 30.9\  text(m²)`

 

♦ Mean mark 49%.

ii.  `text(Distance water flows)`  `= 0.4 xx 10`
  `= 4 \ text(metres)`

 

`text(Volume flow in 10 seconds)` `~~ 4 xx 30.9`
  `~~ 123.6  text(m³)`

Calculus, 2ADV C1 2008 HSC 6b

The graph shows the velocity of a particle,  `v`  metres per second, as a function of time,  `t`  seconds.

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

  2. When is the velocity of the particle equal to zero?    (1 mark)

  3. When is the acceleration of the particle equal to zero?    (1 mark)

Show Answers Only
  1. `20\ text(m/s)`
  2. `t=10\ text(seconds)`
  3. `t=6\ text(seconds)`
Show Worked Solution

i.    `text(Find)\   v  \ text(when)  t=0`

`v=20\ \ text(m/s)`
 

ii.    `text(Particle comes to rest at)\  t=10\ text{seconds  (from graph)}`

 

iii.  `text(Acceleration is zero when)\ t=6\ text{seconds  (from graph)}`

Calculus, 2ADV C4 2016* HSC 14a

The diagram shows the cross-section of a tunnel and a proposed enlargement.
 

hsc-2016-14a
 

The heights, in metres, of the existing section at 1 metre intervals are shown in Table `A.`
 

hsc-2016-14ai
 

The heights, in metres, of the proposed enlargement are shown in Table `B.`
  

hsc-2016-14aii
 

Use the Trapezoidal rule with the measurements given to calculate the approximate increase in area.  (3 marks)

Show Answers Only

`1.3\ text(m²)`

Show Worked Solution

`text(Consider the shaded area distances:)`

`A` `~~ 1/2[0 + 2(0.4 + 0.5 + 0.4) + 0]`
  `~~ 1/2(2.6)`
  `~~ 1.3\ text(m²)`

Trigonometry, 2ADV’ T1 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°.
 

5a
 

  1. Show that  `BP = (sqrt3 tan 3°)/(tan20°)`.   (3 marks) 

  2. Find the distance `AB`. Give your answer to 1 decimal place.   (1 mark)

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `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.)`

Trigonometry, 2ADV’ T1 2008 HSC 6a

From a point  `A`  due south of a tower, the angle of elevation of the top of the tower  `T`, is 23°. From another point  `B`, on a bearing of 120° from the tower, the angle of elevation of  `T`  is 32°. The distance  `AB`  is 200 metres.
 

Trig Ratios, EXT1 2008 HSC 6a 
 

  1. Copy or trace the diagram into your writing booklet, adding the given information to your diagram.  (1 mark)

  2. Hence find the height of the tower. Give your answer to the nearest metre.  (3 marks)

Show Answers Only
  1.  
    Trig Ratios, EXT1 2008 HSC 6a Answer

  2. `96\ text(m)`
Show Worked Solution

i.

Trig Ratios, EXT1 2008 HSC 6a Answer 

 

ii.  `text(Find)\ \ OT = h`

`text(Using the cosine rule in)\ Delta AOB :`

`200^2 = OA^2 + OB^2 – 2 * OA * OB * cos 60\ …\ text{(*)}`

 `text(In)\ Delta OAT,\tan 23^@= h/(OA)`

`=> OA= h/(tan 23^@)\  …\ (1)`

 `text(In)\ Delta OBT,\ tan 32^@= h/(OB)`

`=> OB= h/(tan 32^@)\ \ \ …\ (2)`
 

`text(Substitute)\ (1)\ text(and)\ (2)\ text(into)\ text{(*):}`

`200^2` `= (h^2)/(tan^2 23^@) + (h^2)/(tan^2 32^@) – 2 * h/(tan 23^@) * h/(tan 32^@) * 1/2`
  `= h^2 (1/(tan^2 23^@) + 1/(tan^2 32^@) + 1/(tan23^@ * tan32^@) )`
  `= h^2 (4.340…)`
`h^2` `= (40\ 000)/(4.340…)`
  `= 9214.55…`
`:. h` `= 95.99…`
  `= 96\ text(m)\ \ \ text{(to nearest m)}`

Trigonometry, 2ADV’ T1 2015 HSC 12c

A person walks 2000 metres due north along a road from point `A` to point `B`. The point `A` is due east of a mountain `OM`, where `M` is the top of the mountain. The point `O` is directly below point `M` and is on the same horizontal plane as the road. The height of the mountain above point `O` is `h` metres.

From point `A`, the angle of elevation to the top of the mountain is 15°.

From point `B`, the angle of elevation to the top of the mountain is 13°.
 

Trig Ratios, EXT1 2015 HSC 12c
 

  1. Show that  `OA = h\ cot\ 15°`.  (1 mark)

  2. Hence, find the value of  `h`.  (2 marks)

Show Answers Only
  1. `text(See Worked Solutions)`
  2. `910\ text{m  (nearest metre)}`
Show Worked Solution

i.   `text(Show)\ \ OA = h\ cot\ 15^@` 

Trig Ratios, EXT1 2015 HSC 12c Answer1

`text(In)\ \ Delta MOA,`

`tan\ 15^@` `= h/(OA)`
`OA` `= h/(tan\ 15^@)`
  `= h\ cot\ 15^@\ \ …text(as required)`

 

ii.   `text(In)\ \ ΔMOB`

`tan\ 13^@` `= h/(OB)`
`OB` `= h/(tan\ 13^@)`
  `= h\ cot\ 13^@`

 

Trig Ratios, EXT1 2015 HSC 12c Answer2 
 

`text(In)\ \ ΔAOB:`

`OA^2 + AB^2` `= OB^2`
`OB^2 − OA^2` `= AB^2`
`(h\ cot\ 13^@)^2 − (h\ cot\ 15^@)^2` `= 2000^2`
`h^2[(cot^2\ 13^@ − cot^2\ 15^@)]` `= 2000^2`
`h^2` `= (2000^2)/(cot^2\ 13^@ − cot^2\ 15^@)`
`:. h` `= sqrt((2000^2)/(cot^2\ 13^@ − cot^2\ 15^@))`
  `= 909.704…`
  `= 910\ text{m  (nearest metre)}`

Measurement, STD2 M1 SM-Bank 16 MC

The diagram represents a field.
 

What is the area of the field, using four applications of the Trapezoidal’s rule?

  1. 105 m²
  2. 136 m²
  3. 210 m²
  4. 420 m²
Show Answers Only

`A`

Show Worked Solution

`text(Solution 1)`

`text(Area)` `~~ 3/2(6 + 7) + 3/2(7 + 12) + 3/2(12 + 8) + 3/2(8 + 10)`
  `~~ 3/2(13 + 19 + 20 + 18)`
  `~~ 105\ text(m²)`

 

`text(Solution 2)`

\begin{array} {|l|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} x \rule[-1ex]{0pt}{0pt} & 0 & 3 & 6 & 9 & 12 \\
\hline
\rule{0pt}{2.5ex} \text{height} \rule[-1ex]{0pt}{0pt} & \ \ \ 6\ \ \  & \ \ \ 7\ \ \  & \ \ 12\ \  & \ \ \ 8\ \ \  & \ \ 10\ \  \\
\hline
\rule{0pt}{2.5ex} \text{weight} \rule[-1ex]{0pt}{0pt} & 1 & 2 & 2 & 2 & 1 \\
\hline
\end{array}

`text(Area)` `~~ 3/2(6 + 2 xx 7 + 2 xx 12 + 2 xx 8 + 10)`
  `~~ 3/2(70)`
  `~~ 105\ text(m²)`

 
`=> A`

Measurement, STD2 M1 SM-Bank 15 MC

The shaded region represents a block of land bounded on one side by a road.
 

2UG-2005-12MC

 
What is the approximate area of the block of land, using the Trapeziodal rule?

  1.  720 m²
  2.  880 m²
  3.  1140 m²
  4.  1440 m²
Show Answers Only

`A`

Show Worked Solution

`text(Area)` `≈ 20/2(23 + 15) + 20/2(15 + 19)`
  `≈ 10(38) + 10(34)`
  `≈ 720\ text(m²)`

 
`=> A`

Complex Numbers, EXT2 N2 2018 HSC 15b

  1. Use De Moivre's theorem and the expansion of `(costheta + isintheta)^8` to show that
     
    `sin8theta = ((8),(1)) cos^7thetasintheta - ((8),(3)) cos^5thetasin^3theta`
     
                        `+ ((8),(5)) cos^3thetasin^5theta - ((8),(7)) costhetasin^7theta`  (2 marks)

  2. Hence, show that
     
    `(sin8theta)/(sin2theta) = 4(1 - 10sin^2theta + 24sin^4theta - 16sin^6theta)`.  (3 marks)

Show Answers Only
  1. `text(See Worked Solutions)`
  2. `text(See Worked Solutions)`
Show Worked Solution

i.   `text(By De Moivre)`

`costheta + isintheta^8 = cos8theta + isin8theta\ \ …\ (text{*})`
 

`text(Using Binomial Expansion)`

`(costheta + isintheta)^8`

`= cos^8theta + ((8),(1))cos^7theta * isintheta + ((8),(2)) cos^6theta *i^2sin^2theta`

`+ ((8),(3)) cos^5theta *i^3sin^3theta + ((8),(4)) cos^4theta *i^4sin^4theta + ((8),(5)) cos^3theta *i^5sin^5theta`

`+ ((8),(6)) cos^2theta *i^6sin^6theta + ((8),(7)) costheta *i^7sin^7theta + i^8sin^8theta`

 
`text(Equating imaginary parts of the expansion equation (*)):`

`isin8theta = ((8),(1)) cos^7theta* isintheta + ((8),(3)) icos^5theta* i^3sin^3theta`

`+ ((8),(5)) cos^3theta* i ^5sintheta + ((8),(7)) costheta *i^7sin^7theta`

`:. sin8theta = ((8),(1)) cos^7theta sintheta – ((8),(3)) cos^5theta sin^3theta`

`+ ((8),(5)) cos^3theta sin^5theta – ((8),(7)) costheta sin^7theta`
 

ii.    `sin8theta` `= 8cos^7theta sintheta – 56cos^5 sin^3theta + 56cos^3theta sin^5theta – 8costheta sin^7theta`
    `= 2sinthetacostheta (4cos^6theta – 28cos^4theta sin^2theta + 28cos^2theta sin^4theta – 4sin^6theta)`

 
`:. (sin8theta)/(sin2theta)`

`= 4cos^6theta – 28cos^4theta sin^2theta + 28cos^2theta sin^4theta – 4sin^6theta`

`= 4(1 – sin^2theta)^3 – 28(1 – sin^2theta)^2 sin^2theta + 28(1 – sin^2theta) sin^4theta – 4sin^6theta`

`= 4(1 – 3sin^2theta + 3sin^4theta + sin^6theta) – 28sin^2theta (1 – 2sin^2theta + sin^4theta)`

`+ 28sin^4theta (1 – sin^2theta) – 4sin^6theta`

`= 4 – 40sin^2theta + 96sin^4theta – 56sin^6theta`

`= 4(1 – 10sin^2theta + 24sin^4theta – 16sin^6theta)`

Harder Ext1 Topics, EXT2 2018 HSC 14d

Three people, `A`, `B` and `C`, play a series of n games, where  `n ≥ 2`. In each of the games there is one winner and each of the players is equally likely to win.

  1.  What is the probability that player `A` wins every game?  (1 mark)
  2.  Show that the probability that `A` and `B` win at least one game each but `C` never wins, is
     
         `(2/3)^n - 2(1/3)^n`.  (1 mark)
     
  3.  Show that the probability that each player wins at least one game is 
     
         `(3^(n - 1) - 2^n + 1)/(3^(n - 1))`.  (2 marks)
Show Answers Only
  1. `(1/3)^n`
  2. `text(See Worked Solutions)`
  3. `text(See Worked Solutions)`
Show Worked Solution

i.   `text{Pr (A wins every game)} = (1/3)^n`

 

ii.   `text{Pr (C never wins)} = (2/3)^n`

♦ Mean mark 36%.

`text(If C never wins, only 2 scenarios occur where A or B)`

`text(don’t win at least 1 game)`

`→\ text(A wins all or B wins all.)`
 

`:.\ text{Pr (No C, A and B win at least 1 game)}`

`=\ text{Pr (No C) – Pr (A wins all) – Pr (B wins all)}`

`= (2/3)^n – 2(1/3)^n`

 

iii.   `=>\ text{A, B or C cannot win all games (part(i)).}`

♦♦♦ Mean mark 19%.

`=>\ text(A cannot lose all games, with B and C winning)`

`text{at least 1 each (part (ii)). Similarly for each player.}`
 

`:.\ text{Pr (each player wins at least 1 game)}`

`= 1 – 3(1/3)^n – 3[(2/3)^n – 2(1/3)^n]`

`= (3^n – 3 – 3 · 2^n + 6)/(3^n)`

`= (3^n – 3 · 2^n + 3)/(3^n)`

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

Calculus, EXT2 C1 2018 HSC 14c

Let  `I_n = int_(−3)^0 x^n sqrt(x + 3)\ dx`  for  `n = 0, 1, 2…`

  1.  Show that, for  `n >= 1`,
     
         `I_n = (−6n)/(3 + 2n) I_(n - 1)`.   (3 marks)

  2.  Find the value of  `I_2`.  (2 marks)

Show Answers Only
  1. `text(See Worked Solutions)`
  2. `(144sqrt3)/35`
Show Worked Solution

i.   `I_n = int_(−3)^0 x^n sqrt(x + 3)\ dx\ \ text(for)\ \ n = 0, 1, 2…`

`u` `= x^n` `vprime` `= (x + 3)^(1/2)`
`uprime` `= nx^(n – 1)` `v` `= 2/3 (x + 3)^(3/2)`

 

`I_n` `= [2/3 x^n(x + 3)^(3/2)]_(−3)^0 – 2/3int_(−3)^0  nx^(n – 1)(x + 3)sqrt(x + 3)\ dx`
`I_n` `= 0 – (2n)/3 int_(−3)^0 x^nsqrt(x + 3) + 3x^(n – 1)sqrt(x + 3)\ dx`
`I_n` `= −(2n)/3 (I_n + 3I_(n-1))`
`I_n + (2n)/3 I_n` `= −2nI_(n – 1)`
`I_n(1 + (2n)/3)` `= −2nI_(n – 1)`
`I_n((3 + 2n)/3)` `= −2nI_(n – 1)`
`:. I_n` `= (−6n)/(3 + 2n) I_(n – 1)\ \ \ text(… as required)`

 

ii.    `I_2` `= −12/7 I_1`
    `= −12/7 xx −6/5 I_0`
    `= 72/35 int_(−3)^0 (x + 3)^(1/2)dx`
    `= 72/35 xx 2/3 [(x + 3)^(3/2)]_(−3)^0`
    `= 48/35 (3^(3/2))`
    `= (144sqrt3)/35`

Mechanics, EXT2 M1 2018 HSC 14b

A falling particle experiences forces due to gravity and air resistance. The acceleration of the particle is  `g - kv^2`, where `g` and `k` are positive constants and `v` is the speed of the particle. (Do NOT prove this.)

Prove that, after falling from rest through a distance, `h`, the speed of the particle will be  `sqrt(g/k (1 - e^(−2kh)))`.  (3 marks)

Show Answers Only

`text(See Worked Solutions)`

Show Worked Solution

`a = v · (dv)/(dx) = g – kv^2`

`(dv)/(dx)` `= (g – kv^2)/v`
`(dx)/(dv)` `= v/(g – kv^2)`
`x` `= int v/(g – kv^2)\ dv`
  `= −1/(2k) log_e(g – kv^2) + c`

 

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

`=> c = 1/(2k) log_e (g)`

`x` `= −1/(2k)log_e(g – kv^2) + 1/(2k) log_e g`
  `= −1/(2k) log_e ((g – kv^2)/g)`

 

`text(Find)\ \ v\ \ text(when)\ \ x = h:`

`log_e ((g – kv^2)/g)` `= −2kh`
`(g – kv^2)/g` `= e^(−2kh)`
`kv^2` `= g – g e^(−2kh)`
`v^2` `= g/k (1 – e^(−2kh))`
`:. v` `= sqrt(g/k (1 – e^(−2kh)))`

Calculus, EXT2 C1 2018 HSC 14a

Using the substitution  `t = tan\ theta/2`  evaluate  `int_0^(pi/2) (d theta)/(2 - costheta)`.  (3 marks)

Show Answers Only

`(2sqrt3 pi)/9`

Show Worked Solution

`t = tan\ theta/2, \ costheta = (1 – t^2)/(1 + t^2), \ d theta = 2/(1 + t^2) dt`

`text(When)\ \ theta = pi/2, \ t = 1`

`text(When)\ \ theta = 0, \ t = 0`
 

`int_0^(pi/2) (d theta)/(2 – costheta)` `= int_0^1 1/(2 – (1 – t^2)/(1 + t^2)) · 2/(1 + t^2)\ dt`
  `= int_0^1 2/(2(1 + t^2) – (1 – t^2))\ dt`
  `= int_0^1 2/(1 + 3t^2)\ dt`
  `= [2/sqrt3 tan^(−1)(sqrt3 t)]_0^1`
  `= 2/sqrt3 tan^(−1) sqrt3 – 0`
  `= (2pi)/(3sqrt3)`
  `= (2sqrt3 pi)/9`

Harder Ext1 Topics, EXT2 2018 HSC 13d

The points  `P(cp, c/p)`  and  `Q(cp, c/q)`  lie on the rectangular hyperbola  `xy = c^2`.

The line `PQ` has equation  `x + pqy = c(p + q)`. (Do NOT prove this.)

The `x` and `y` intercepts of `PQ` are `R` and `S` respectively, as shown in the diagram.
 


  

  1. Show that  `PS = QR`.  (3 marks)

The point  `T(2at, at^2)`  lies on the parabola  `x^2 = 4ay`. The tangent to the parabola at `T` intersects the rectangular hyperbola  `xy = c^2`  at `A` and `B` and has equation  `y = tx - at^2`. (Do NOT prove this.) The point `M` is the midpoint of the interval `AB`. One such case is shown in the diagram.
 

  1. Using part (i), or otherwise, show that `M` lies on the parabola  `2x^2 = −ay`.  (2 marks)
Show Answers Only
  1. `text(See Worked Solutions)`
  2. `text(See Worked Solutions)`
Show Worked Solution

i.   `P(cp, c/p)\ text(and)\ Q(cq, c/q)`

`PQ:\ \  x + pqy = c(p + q)`

 
`text(When)\ \ y = 0, x = c(p + q)`

`=> R(c(p + q), 0)`

`text(When)\ x = 0, y = (c(p + q))/(pq)`

`=> S(0, (c(p + q))/(pq))`
 

`text(Using Pythagoras:)`

`(PS)^2` `= (0 – cp)^2 + ((c(p + q))/(pq) – c/p)^2`
  `= c^2p^2 + (c/q + c/p – c/p)^2`
  `= c^2p^2 + (c^2)/(q^2)`

 

`(QR)^2` `= (cq – c(p + q))^2 + (c^2)/(q^2)`
  `= (cq – cp – cq)^2 + (c^2)/(q^2)`
  `= c^2p^2 + (c^2)/(q^2)`

 
`:. PS = QR\ \ \ text(… as required)`

 

ii.    `y = tx – at^2`

♦ Mean mark part (ii) 50%.

`=> xtext(-intercept at)\ \ R(at, 0)`

`=> ytext(-intercept at)\ \ S(0,−at^2)`
 

`text(S)text(ince)\ \ AR = BS\ \ \ (text{from part(i)})`

`M_(AB)` `= M_(RS)= ((at)/2, (−at^2)/2)“

 
`text(Find locus of)\ M:`

`x = (at)/2 \ => \ t = (2x)/a`

`y` `= (−at^2)/2`
  `= −a/2 xx ((2x)/a)^2`
  `= (−2x^2)/a`
`:. 2x^2` `= −ay\ \ \ text(… as required)`

Complex Numbers, EXT2 N1 2018 HSC 13b

Let   `z = 1 - cos2theta + isin2theta`, where   `0 < theta <= pi`.

  1.  Show that  `|\ z\ | = 2sintheta`.  (2 marks)

  2.  Show that  `text(arg)(z) = pi/2 - theta`.  (2 marks)

Show Answers Only
  1. `text(See Worked Solutions)`
  2. `text(See Worked Solutions)`
Show Worked Solution

i.   `z = 1 – cos2theta + isin2theta`

`|\ z\ |` `= sqrt((1 – cos2theta)^2 + sin^2 2theta)`
  `= sqrt(1 – 2cos2theta + cos^2 2theta + sin^2 2theta)`
  `= sqrt(2 – 2cos2theta)`
  `= sqrt(2(1 – cos2theta))`
  `= sqrt(2(2sin^2theta))`
  `= 2sintheta\ \ \ text(… as required)`

 

ii.  `z= 1 – cos2theta + isin2theta`

`text(arg)(z)` `= tan^(−1)((sin2theta)/(1 – cos2theta))`
  `= tan^(−1)((2sinthetacostheta)/(2sin^2theta))`
  `= tan^(−1)(cottheta)`
  `= tan^(−1)(tan(pi/2 – theta))`
  `= pi/2 – theta`

 
`(text(S)text(ince)\ \ 0 < theta < pi => \ −pi/2 <= pi/2 – theta < pi/2)`

Functions, EXT1′ F1 2018 HSC 12d

The diagram shows the graph of the function  `f(x) = x/(x - 1)`.
  


 

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

  1.  `y = |\ f(x)\ |`  (1 mark)

  2.  `y = (f(x))^2`  (2 marks)

Show Answers Only

i.   

ii.

Show Worked Solution
i.   

 

ii.   

Graphs, EXT2 2018 HSC 12b

A curve has equation  `x^2 + xy + y^2 = 3`.

  1. Use implicit differentiation to show that  `dy/dx = −(2x + y)/(x + 2y)`.  (2 marks)
  2. Hence, or otherwise, find the coordinates of the points on the curve where  `dy/dx = 0`.  (2 marks)
Show Answers Only
  1. `text(See Worked Solutions)`
  2. `(1,−2)\ \ text(and)\ \ (−1,2).`
Show Worked Solution

i.   `x^2 + xy + y^2 = 3`

♦ Mean mark 98%!

`2x + y + x · dy/dx + 2y · dy/dx=0`

`dy/dx(x + 2y)` `= −2x – y`
`:. dy/dx` `= −(2x + y)/(x + 2y)\ \ \ text(… as required)`

 

ii.   `text(When)\ \ dy/dx = 0,`

`−(2x + y)/(x + 2y)` `= 0`
`-2x-y` `=0`
`y` `= −2x\ \ \ \ (text(*))`

 

`text(Substituting into)\ \ x^2 + xy + y^2 = 3`

`x^2 + x(−2x) + (−2x)^2` `= 3`
`x^2 – 2x^2 + 4x^2` `= 3`
`3x^2` `= 3`
`x` `= +-1`
`y` `= +-2\ \ \ \ (text(*))`

 
`:.\ text(Coordinates are)\ \ (1,−2)\ \ text(and)\ \ (−1,2).`

Complex Numbers, EXT2 N2 2018 HSC 11d

The points `A`, `B` and `C` on the Argand diagram represent the complex numbers `u`, `v` and `w` respectively.

The points `O`, `A`, `B` and `C` form a square as shown on the diagram.
 

 
It is given that  `u = 5 + 2i`.

  1.  Find  `w`.  (1 mark)

  2.  Find  `v`.  (1 mark)

  3.  Find  `text(arg)(w/v)`.  (1 mark)

Show Answers Only
  1. `−2 + 5i`
  2. `3 + 7i`
  3. `pi/4`
Show Worked Solution
i.    `w` `= iu`
    `= i(5 + 2i)`
    `= −2 + 5i`

 

ii.    `v` `= u + w`
    `= 5 + 2i + (−2 + 5i)`
    `= 3 + 7i`

 

iii.    `text(arg)(w/v)` `= text(arg)(w) – text(arg)(v)`
    `= pi/4\ \ (text(diagonal of square bisects corner))`

Calculus, EXT2 C1 2018 HSC 11c

By writing  `(x^2 - x - 6)/((x + 1)(x^2 - 3))`  in the form  `a/(x + 1) + (bx + c)/(x^2 - 3)`,

find  `int(x^2 - x - 6)/((x + 1)(x^2 - 3))\ dx`.  (4 marks)

Show Answers Only

`2 ln |\ x + 1\ | – 1/2 ln |\ x^2 – 3\ | + c`

Show Worked Solution
`(x^2 – x – 6)/((x + 1)(x^2 – 3))` `≡ a/(x + 1) + (bx + c)/(x^2 – 3)`
`x^2 – x – 6` `≡ a(x^2 – 3) + (bx + c)(x + 1)`

 
`text(When)\ x = −1`

`1 + 1 – 6 = −2a\ \ =>\ \ a = 2`
 

`text(Equating co-efficients of)\ x^2`

`1 = (a + b)x^2\ \ =>\ \ b = −1`
 

`text(Equating co-efficients of)\ x`

`−1 = b + c\ \ =>\ \ c = 0`
 

`:. int(x^2 – x – 6)/((x + 1)(x^2 – 3))\ dx` `= int2/(x + 1) – x/(x^2 – 3)\ dx`
  `= 2 ln |\ x + 1\ | – 1/2 ln |\ x^2 – 3\ | + c`

Functions, EXT1′ F2 2018 HSC 11b

The polynomial  `p(x) = x^3 + ax^2 + b`  has a zero at `r` and a double zero at 4.

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

Show Answers Only

`a =− 6, b = 32, r = – 2`

Show Worked Solution

`p(x) = x^3 + ax^2 + b`

`pprime(x) = 3x^2 + 2ax`
 

`text(Double root at 4:)`

`pprime(4)` `=0`
`3 xx 16 + 8a` `= 0`
`a` `= −6`

 

`p(4)` `=0`
`64 + 16a + b` `= 0`
`64 – 96 + b` `= 0`
`b` `= 32`

 
`text(Roots of)\ \ p(x)\ \ text(are)\ \ 4, 4, r`

` 4 + 4 +r` `= −a/1 = 6`
`:. r` `= – 2`

Trigonometry, 2ADV T3 2018 HSC 15a

The length of daylight, `L(t)`, is defined as the number of hours from sunrise to sunset, and can be modelled by the equation

`L(t) = 12 + 2 cos ((2 pi t)/366)`,

where `t` is the number of days after 21 December 2015, for  `0 ≤ t ≤ 366`.

  1. Find the length of daylight on 21 December 2015.   (1 mark)

  2. What is the shortest length of daylight?  (1 mark)

  3. What are the two values of  `t`  for which the length of daylight is 11?  (2 marks)

Show Answers Only
  1. `14\ text(hours)`
  2. `10\ text(hours)`
  3. `t = 122 or 244`
Show Worked Solution

i.   `L(t) = 12 + 2 cos ((2 pi t)/366)`

`text(On 21 Dec 2015) => t = 0`

`:. L(0)` `= 12 + 2 cos 0`
  `= 14\ text(hours)`

 

ii.   `text(Shortest length of daylight occurs when)`

♦ Mean mark 43%.

`cos ((2 pi t)/366) = -1`
 

`:.\ text(Shortest length)` `= 12 + 2 (-1)`
  `= 10\ text(hours)`

 

iii.   `text(Find)\ \ t\ \ text(such that)\ \ L(t) = 11:`

`11 = 12 + 2 cos ((2 pi t)/366)`

`cos ((2 pi t)/366) = -1/2`
 

`(2 pi t)/366` `= (2 pi)/3` `qquad\ \ text(or)`     `(2 pi t)/366` `= (4 pi)/3`
`t` `= 366/3`   `t` `= (366 xx 2)/3`
  `= 122`     `= 244`

 
`:. t = 122 or 244`

Functions, EXT1′ 2018 HSC 5 MC

The diagram shows the graph  `y = e^(3x)`  for  `0<= x <= 4`. The region bounded by  `y = −1`,  `y = e^(3x)`,  `x = 0`  and  `x = 4`  is rotated about the line  `y = −1`  to form a solid.
 

 
Which integral represents the volume of the solid formed?

  1. `pi int_0^4 (e^(3x) + 1)^2 dx`
  2. `2pi int_0^4 x(e^(3x) + 1)\ dx`
  3. `pi int_0^4 (e^(3x) - 1)^2 dx`
  4. `2pi int_0^4 x(e^(3x) - 1)\ dx`
Show Answers Only

`A`

Show Worked Solution

`text(Rotating)\ \ y = e^(3x)\ \ text(about)\ \ y = −1\ \ text(is equivalent)`

`text(to rotating)\ \ y = e^(3x) + 1\ \ text(about)\ xtext(-axis.)`
 

`V` `= pi int_0^4 y^2 dx`
  `= pi int_0^4 (e^(3x) + 1)^2 dx`

 
`=>A`

Functions, EXT1′ F1 2018 HSC 4 MC

Which graph best represents the curve  `y = 1/sqrt(x^2 - 4)`?
 

A.    B.   
C.    D.   
Show Answers Only

`=>\ text(C)`

Show Worked Solution

`text(S)text(ince)\ \ sqrt(x^2 – 4) > 0`

`=> y > 0\ \ (text(Eliminate B and D))`
 

`text(As)\ x -> 2^+, sqrt(x^2 – 4) -> 0, y -> ∞`

`text(As)\ x -> -2^-, sqrt(x^2 – 4) -> 0, y -> ∞`

`=>\ text(C)`

Functions, EXT1′ F2 2018 HSC 3 MC

The cubic equation  `x^3 + 2x^2 + 5x - 1 = 0`  has roots, `alpha`, `beta` and `gamma`.

Which cubic equation has roots  `(−1)/alpha, (−1)/beta, (−1)/gamma`?

  1. `x^3 - 5x^2 - 2x + 1 = 0`
  2. `x^3 - 5x^2 - 2x - 1 = 0`
  3. `x^3 + 5x^2 + 2x + 1 = 0`
  4. `x^3 + 5x^2 - 2x + 1 = 0`
Show Answers Only

`text(D)`

Show Worked Solution

`alpha beta gamma = 1`

`alpha beta + beta gamma + gamma alpha = 5`

`alpha + beta + gamma = −2`
 

`text(If roots are)\ \ (−1)/alpha, (−1)/beta, (−1)/gamma :`

`(−1)/alpha · (−1)/beta · (−1)/gamma = (−1)/(alphabetagamma) = −1`

`=> d = 1`
 

`1/(alphabeta) + 1/(betagamma) + 1/(gammaalpha) = ((alpha + beta + gamma))/(alphabetagamma) = −2`

`=>\ c = −2`
 

`(−1)/alpha – (−1)/beta – (−1)/gamma = (−(alphabeta + betagamma + gammaalpha))/(alphabetagamma) = −5`

`=> b = 5`
 

`:.\ text(Equation is)\ \ x^3 + 5x^2 – 2x + 1 = 0`

`=>D`

Conics, EXT2 2018 HSC 02 MC

What are the equations of the asymptotes of the hyperbola  `9x^2 - 4y^2 = 36`?

  1. `y = +- 9/4x`
  2. `y = +- 2/3x`
  3. `y = +- 3/2x`
  4. `y = +- 4/9x`
Show Answers Only

`text(C)`

Show Worked Solution
`9x^2 – 4y^2` `= 36`
`(x^2)/4 – (y^2)/9` `= 1`

 
`:.\ text(Asymptotes are:)\ \ y = +- 3/2x`

`=>\ text(C)`

Financial Maths, 2ADV M1 2018 HSC 14d

An artist posted a song online. Each day there were  `2^n + n`  downloads, where `n` is the number of days after the song was posted.

  1. Find the number of downloads on each of the first 3 days after the song was posted.  (1 mark)

  2. What is the total number of times the song was downloaded in the first 20 days after it was posted?  (2 marks)

Show Answers Only
  1. `text(Day 1) : 3`
    `text(Day 2) : 6`
    `text(Day 3) : 11`

  2. `2\ 097\ 360`
Show Worked Solution

i.   `text(Day 1:)\ \ 2^1 + 1 = 3`

`text(Day 2:)\ \ 2^2 + 2 = 6`

`text(Day 3:)\ \ 2^3 + 3 = 11`

 

ii.  `text{Total downloads (20 days)}`

♦ Mean mark 38%.

`= 2^1 + 1 + 2^2 + 2 + … + 2^20 + 20`

`= underbrace(2^1 + 2^2 + … + 2^20)_{text(GP),\ a = 2,\ r=2} + underbrace(1 + 2 + … + 20)_{text(AP),\ a = 1,\ d = 1}`

`= (2(2^20 – 1))/(2 – 1) + 20/2(1 + 20)`

`= 2\ 097\ 150 + 210`

`= 2\ 097\ 360`

Calculus, EXT1* C1 2018 HSC 13c

The population of a country grew exponentially between 1910 and 2010. This population can be modelled by the equation  `P(t) = 92e^(kt)`, where  `P(t)`  is the population of the country in millions, `t` is the time in years after 1910 and `k` is a positive constant. The population of the country in 1960 was 184 million.

  1. Show that the value of `k` is 0.0139, correct to 4 decimal places.  (2 marks)

  2. Assuming that this model continues to be valid after 2010, estimate the population of the country in 2020 to the nearest million.  (2 marks)

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `424\ text{million  (nearest million)}`
Show Worked Solution

i.   `P(t) = 92 e^(kt)`

`text(Find)\ \ k\ \ text(given)\ \ t = 50\ \ text(when)\ \ P = 184`

`184` `= 92 e^(50k)`
`e^(50 k)` `= 2`
`50k` `= ln 2`
`k` `= 1/50 xx ln 2`
  `= 0.01386…`
  `= 0.0139\ \ text{(to 4 d.p.)}`

 

ii.  `text(In 2020),\ t = 110`

`:.\ text(Estimated population)`

`= 92 e^(110 xx 0.0139)`

`= 92 e^1.529`

`= 424.44…`

`= 424\ text{million  (nearest million)}`