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Logarithm, SMB-018

Evaluate  `log_a 6`  given  `log_a 2=0.62`  and  `log_a 24=2.67`.  (2 marks)

Show Answers Only

`1.43`

Show Worked Solution
`log_a 6` `=log_a (24/4)`  
  `=log_a 24-log_b 2^2`  
  `=log_a 24-2log_a 2`  
  `=2.67-2 xx 0.62`  
  `=1.43`  

Logarithms, SMB-010

Solve the equation  `2 log_2(x + 5)-log_2(x + 9) = 1`.  (3 marks)

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`x = text{−1}`

Show Worked Solution
`2 log_2(x + 5)-log_2(x + 9)` `= 1`
`log_2(x + 5)^2-log_2(x + 9)` `= 1`
`log_2(((x + 5)^2)/(x + 9))` `= 1`
`((x + 5)^2)/(x + 9)` `= 2`
`x^2 + 10x + 25` `= 2x + 18`
`x^2 + 8x + 7` `= 0`
`(x + 7)(x + 1)` `= 0`

 
`:. x = -1\ \ \ \ (x != text{−7}\ \ text(as)\ \ x > text{−5})`

L&E, 2ADV E1 2008 HSC 7a

Solve  `log_2 x-3/log_2 x=2`   (3 marks)

Show Answers Only

`x=8\ \ text(or)\ \ 1/2`

Show Worked Solution
 
IMPORTANT: Students should recognise this equation as a quadratic, and the best responses substituted `log_2 x` with a variable such as `X`.
`log_2 x-3/(log_2 x)` `=2`
`(log_2 x)^2-3` `=2log_2 x`
`(log_2 x)^2-2log_2 x-3` `=0`
   
`text(Let)\  X=log_2 x`  
`:.\ X^2-2X-3` `=0`
`(X-3)(X+1)` `=0`
MARKER’S COMMENT: Many responses incorrectly stated that there is no solution to `log_2 x=-1` or could not find `x` given `log_2 x=3`.
`X` `=3` `\ \ \ \ \ \ \ \ \ \ ` `X` `=-1`
`log_2 x` `=3` `\ \ \ \ \ \ \ \ \ \ ` `log_2 x` `=-1`
`x` `=2^3=8` `\ \ \ \ \ \ \ \ \ \ ` `x` `=2^{-1}=1/2`

 

`:.x=8\ \ text(or)\ \ 1/2`

Logarithms, SMB-006 MC

The expression

`log_c(a) + log_a(b) + log_b(c)`

is equal to

  1. `1/(log_c(a)) + 1/(log_a(b)) + 1/(log_b(c))`
  2. `1/(log_a(c)) + 1/(log_b(a)) + 1/(log_c(b))`
  3. `-1/(log_a(b))-1/(log_b(c))-1/(log_c(a))`
  4. `1/(log_a(a)) + 1/(log_b(b)) + 1/(log_c(c))`
Show Answers Only

`B`

Show Worked Solution

`text(Solution 1)`

`text(Using Change of Base:)`

`log_c(a) + log_a(b) + log_b(c)`

`=(log_a(a))/(log_a(c)) + (log_b(b))/(log_b(a)) + (log_c(c))/(log_c(b))`

`=1/(log_a(c)) + 1/(log_b(a)) + 1/(log_c(b))`

 
`=> B`

Logarithms, SMB-005

Solve  `log_3(t)-log_3(t^2-4) = -1`  for  `t`.  (3 marks)

Show Answers Only

`4 `

Show Worked Solution
`log_3(t)-log_3(t^2-4)` `= -1`
`log_3 ({t}/{t^2-4})` `= -1`
`(t)/(t^2-4)` `= (1)/(3)`
`t^2-4` `= 3t`
`t^2-3t – 4` `= 0`
`(t-4)(t+ 1)` `= 0`

 
`:. t=4 \ \ \ (t > 0, \ t!= –1)`

Logarithms, SMB-001 MC

It is given that  `log_10 a = log_10 b-log_10 c`, where  `a, b, c > 0.`

Which statement is true?

  1. `a = b-c`
  2. `a = b/c`
  3. `log_10 a = b/c`
  4. `log_10 a = (log_10 b)/(log_10 c)`
Show Answers Only

`B`

Show Worked Solution
Mean mark 51%.
COMMENT: Use of log laws here proved difficult for many students.
`log_10 a` `= log_10 b-log_10 c`
`log_10 a` `= log_10 (b/c)`
`:. a` `= b/c`

 
`=>  B`

Functions, 2ADV F1 2022 HSC 12

A student believes that the time it takes for an ice cube to melt (`M` minutes) varies inversely with the room temperature `(T^@ text{C})`. The student observes that at a room temperature of `15^@text{C}` it takes 12 minutes for an ice cube to melt.

  1. Find the equation relating `M` and `T`.    (2 marks)

  2. By first completing this table of values, graph the relationship between temperature and time from `T=5^@C` to `T=30^@ text{C}`.   (2 marks)
     

\begin{array} {|c|c|c|c|}
\hline  \ \ T\ \  & \ \ 5\ \  & \ 15\  & \ 30\  \\
\hline M &  &  &  \\
\hline \end{array}

 
                   

Show Answers Only

a.    `M=180/T`

 b.    

\begin{array} {|c|c|c|c|}
\hline  \ \ T\ \  & \ \ 5\ \  & \ 15\  & \ 30\  \\
\hline M & 36 & 12 & 6 \\
\hline \end{array}       

 

Show Worked Solution
a.    `M` `prop 1/T`
  `M` `=k/T`
  `12` `=k/15`
  `k` `=15 xx 12`
    `=180`

 
`:.M=180/T`
 


♦ Mean mark (a) 49%.

b.   

\begin{array} {|c|c|c|c|}
\hline  \ \ T\ \  & \ \ 5\ \  & \ 15\  & \ 30\  \\
\hline M & 36 & 12 & 6 \\
\hline \end{array}

Functions, 2ADV F1 2021 HSC 8 MC

The graph of  `y = f(x)`  is shown.

Which of the following could be the equation of this graph?

  1. `y = (1 - x)(2 + x)^3`
  2. `y = (x + 1)(x - 2)^3`
  3. `y = (x + 1)(2 - x)^3`
  4. `y = (x - 1)(2 + x)^3`
Show Answers Only

`C`

Show Worked Solution

`text(By elimination:)`

`text(A single negative root occurs when)\ \ x =–1`

`->\ text(Eliminate A and D)`

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

`->\ text(Eliminate B)`

`=> C`

Functions, EXT1 F2 2020 HSC 11a

Let  `P(x) = x^3 + 3x^2-13x + 6`.

  1. Show that  `P(2) = 0`.  (1 mark)

  2. Hence, factor the polynomial  `P(x)`  as  `A(x)B(x)`, where  `B(x)`  is a quadratic polynomial.  (2 marks)

Show Answers Only
  1. `text(See Worked Solutions)`
  2. `P(x) = (x-2)(x^2 + 5x – 3)`
Show Worked Solution
i.    `P(2)` `= 8 + 12-26 + 6`
    `= 0`

 

ii.   

`:. P(x) = (x-2)(x^2 + 5x – 3)`

Functions, EXT1 F2 2019 HSC 11d

Find the polynomial  `Q(x)`  that satisfies  `x^3 + 2x^2-3x-7 = (x-2) Q(x) + 3`.  (2 marks)

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`Q(x ) = x^2 + 4x + 5`

Show Worked Solution
`(x-2) ⋅ Q(x) + 3` `= x^3 + 2x^2-3x-7`
`(x-2) ⋅ Q(x)` `= x^3 + 2x^2-3x-10`

 

`:. Q(x ) = x^2 + 4x + 5`

Algebra, STD2 A2 2019 HSC 34

The relationship between British pounds `(p)` and Australian dollars `(d)` on a particular day is shown in the graph.
 

  1. Write the direct variation equation relating British pounds to Australian dollars in the form  `p = md`. Leave `m` as a fraction.  (1 mark)

  2. The relationship between Japanese yen `(y)` and Australian dollars `(d)` on the same day is given by the equation  `y = 76d`.

     

    Convert 93 100 Japanese yen to British pounds.  (2 marks)

Show Answers Only
  1. `p = 4/7 d`
  2. `93\ 100\ text(Yen = 700 pounds)`
Show Worked Solution

a.   `m = text(rise)/text(run) = 4/7`

♦ Mean mark 42%.

`p = 4/7 d`

 

b.   `text(Yen to Australian dollars:)`

`y` `=76d`
`93\ 100` `= 76d`
`d` `= (93\ 100)/76`
  `= 1225`

 
`text(Aust dollars to pounds:)`

`p` `= 4/7 xx 1225`
  `= 700\ text(pounds)`

 
`:. 93\ 100\ text(Yen = 700 pounds)`

Algebra, STD2 A4 2019 HSC 33

The time taken for a car to travel between two towns at a constant speed varies inversely with its speed.

It takes 1.5 hours for the car to travel between the two towns at a constant speed of 80 km/h.

  1. Calculate the distance between the two towns.  (1 mark)

  2. By first plotting four points, draw the curve that shows the time taken to travel between the two towns at different constant speeds.  (3 marks)

Show Answers Only
  1. `120\ text(km)`
  2.  
Show Worked Solution
a.    `D` `= S xx T`
    `= 80 xx 1.5`
    `= 120\ text(km)`

 
b.   

\begin{array} {|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} \ \ s\ \  \rule[-1ex]{0pt}{0pt} & 20 & 40 & 60 & 80 \\
\hline
\rule{0pt}{2.5ex} t \rule[-1ex]{0pt}{0pt} & 6 & 3 & 2 & 1.5 \\
\hline
\end{array}

Plane Geometry, EXT1 2018 HSC 11d

Two secants from the point `C` intersect a circle as shown in the diagram.
 

 
What is the value of `x`?  (2 marks)

Show Answers Only

`4`

Show Worked Solution

`text(Using formula for intercepts of intersecting secants:)`

`x (x + 2)` `= 3 (3 + 5)`
`x^2 + 2x` `= 24`
`x^2 + 2x – 24` `= 0`
`(x + 6) (x – 4)` `= 0`
`:. x` `= 4 \ \ \ (x > 0)`

Functions, EXT1 F2 2018 HSC 11a

Consider the polynomial  `P(x) = x^3-2x^2-5x + 6`.

  1. Show that  `x = 1`  is a zero of  `P(x)`.  (1 mark)

  2. Find the other zeros.  (2 marks)

Show Answers Only
  1. `text(See Worked Solution)`
  2. `x = -2 and x = 3`
Show Worked Solution

i.   `P(1) = 1-2-5 + 6 = 0`

`:. x=1\ \ text(is a zero)`

 

ii.   `text{Using part (i)} \ => (x-1)\ text{is a factor of}\ P(x)`

`P(x) = (x-1)*Q(x)`
 

`text(By long division:)`

`P(x)` `= (x-1) (x^2-x-6)`
  `= (x-1) (x-3) (x + 2)`

 
`:.\ text(Other zeroes are:)`

`x = -2 and x = 3`

Functions, 2ADV F1 2017 HSC 11h

Find the domain of the function  `f(x) = sqrt (3-x)`.  (2 marks)

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`x <= 3 or (-oo,3].`

Show Worked Solution

`text(Domain of)\ \ f(x) = sqrt (3-x)`

`3-x` `>= 0`
`x` `<= 3`

 

`text(Note domain can also be expressed as:)\ \ (-oo,3]`

Functions, EXT1* F1 2017 HSC 8 MC

The region enclosed by  `y = 4 - x,\ \ y = x`  and  `y = 2x + 1`  is shaded in the diagram below.
 

Which of the following defines the shaded region?

A.   `y <= 2x + 1, qquad` `y <= 4-x, qquad` `y >= x`
B.   `y >= 2x + 1, qquad` `y <= 4-x, qquad` `y >= x`
C.   `y <= 2x + 1, qquad` `y >= 4-x, qquad` `y >= x`
D.   `y >= 2x + 1, qquad` `y >= 4-x, qquad` `y >= x`
Show Answers Only

`A`

Show Worked Solution

`text(Consider)\ \ y = 2x + 1,`

`text(Shading is below graph)`

`=> y <= 2x + 1`

`text(Consider)\ \ y = 4-x,`

`text(Shading is below graph)`

`=> y <= 4-x`

`=>  A`

L&E, 2ADV E1 2016 HSC 14e

Write  `log 2 + log 4 + log 8 + … + log 512`  in the form  `a log b`  where `a` and `b` are integers greater than `1.`  (2 marks)

Show Answers Only

`45 log 2`

Show Worked Solution

`log 2 + log 4 + log 8 + … + log 512`

`= log 2^1 + log 2^2 + log2^3 + … + log 2^9`

`= log 2 + 2 log 2 + 3 log 2 + … + 9 log 2`

`= 45 log 2`

♦ Mean mark 40%.
 
TIP: Note that `log 2 = log_10 2`

Functions, EXT1 F2 2007 HSC 2c

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

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

Show Answers Only

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

Show Worked Solution

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

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

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

 
`P(-1) = 18,`

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

 
`text(Subtract)\ \ (1)-(2),`

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

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

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

 
`:.a = -7\ \ text(and)\ \ b = 10`

Functions, EXT1 F2 2015 HSC 11f

Consider the polynomials  `P(x) = x^3-kx^2 + 5x + 12`  and  `A(x) = x - 3`.

  1. Given that  `P(x)`  is divisible by  `A(x)`, show that  `k = 6`.  (1 mark)

  2. Find all the zeros of  `P(x)`  when  `k = 6`.  (2 marks)

Show Answers Only
  1. `text(See Worked Solutions)`
  2. `3, 4, −1`
Show Worked Solution
i.    `P(x)` `= x^3-kx^2 + 5x + 12`
  `A(x)` `= x-3`

 
`text(If)\ P(x)\ text(is divisible by)\ A(x)\ \ =>\ \ P(3) = 0`

`3^3-k(3^2) + 5 xx 3 + 12` `= 0`
`27-9k + 15 + 12` `= 0`
`9k` `= 54`
`:.k` `= 6\ \ …\ text(as required)`

 

ii.  `text(Find all roots of)\ P(x)`

`P(x)=(x-3)*Q(x)`

`text{Using long division to find}\ Q(x):`
 

`:.P(x)` `= x^3-6x^2 + 5x + 12`
  `= (x-3)(x^2-3x − 4)`
  `= (x-3)(x-4)(x + 1)`

 
`:.\ text(Zeros at)\ \ \ x = -1, 3, 4`

Plane Geometry, EXT1 2015 HSC 3 MC

Two secants from the point `P` intersect a circle as shown in the diagram.
  

What is the value of `x`?

  1. `2`
  2. `5`
  3. `7`
  4. `8`
Show Answers Only

`B`

Show Worked Solution

`text{Property: products of intercepts of secants from external point are equal}`

`x(x + 3)` `= 4(4 + 6)`
`x^2 + 3x` `= 40`
`x^2 + 3x-40` `= 0`
`(x-5)(x + 8)` `= 0`

 
`:.x = 5,\ \  (x>0)`

`=>B`

Algebra, STD2 A2 2007 HSC 24c

Sandy travels to Europe via the USA. She uses this graph to calculate her currency conversions.
  
  
 

  1. After leaving the USA she has US$150 to add to the A$600 that she plans to spend in Europe.

     

    She converts all of her money to euros.How many euros does she have to spend in Europe?    (3 marks)

  2. If the value of the euro falls in comparison to the Australian dollar, what will be the effect on the gradient of the line used to convert Australian dollars to euros?   (1 mark)

Show Answers Only
  1. `480\ €`
  2. `text(If the value of the euro falls against the)`

  3.  

    `text(A$, then 1 A$ will buy more euros than)`

  4.  

    `text(before and the gradient used to convert)`

  5.  

    `text{the currencies will steepen (increase).}`

Show Worked Solution

i.   `text(From graph:)`

`75\ text(US$)` `=\ text(100 A$)`
`=> 150\ text(US$)` `=\ text(200 A$)`

 
`:.\ text(Sandy has a total of 800 A$)`
 

`text(Converting A$ to €:)`

`text(100 A$)` `= 60\ €`
`:.\ text(800 A$)` `= 8 xx 60`
  `= 480\ €`

 

ii.   `text(If the value of the euro falls against the)`

`text(A$, then 1 A$ will buy more euros than)`

`text(before and the gradient used to convert)`

`text{the currencies will steepen (increase).}`

Algebra, STD2 A4 2007 HSC 15 MC

If pressure (`p`) varies inversely with volume (`V`), which formula correctly expresses  `p`  in terms of  `V`  and  `k`, where  `k`  is a constant?

  1. `p = k/V`
  2. `p = V/k`
  3. `p = kV`
  4. `p = k + V`
Show Answers Only

`A`

Show Worked Solution

`p prop 1/V`

`p = k/V`

`=>  A`

Functions, EXT1 F2 2008 HSC 1a

The polynomial  `x^3`  is divided by  `x + 3`. Calculate the remainder.   (2 marks)

Show Answers Only

`-27`

Show Worked Solution
`P(-3)` `= (-3)^3`
  `= -27`

 
`:.\ text(Remainder when)\ x^3 -: (x + 3) = -27`

MARKER’S COMMENT: “Grave concern” that many who found `P(-3)=-27` stated the remainder was 27.

Functions, EXT1 F2 2014 HSC 9 MC

The remainder when the polynomial  `P(x) = x^4-8x^3-7x^2 + 3`  is divided by  `x^2 + x`  is  `ax + 3`.

What is the value of  `a`?

  1. `-14`
  2. `-11`
  3. `-2`
  4. `5`
Show Answers Only

`C`

Show Worked Solution

`P(x) = x^4-8x^3-7x^2 + 3`

`text(Given)\ \ P(x)` `= (x^2 + x) *Q(x) + ax + 3`
  `= x (x + 1) Q(x) + ax + 3`

 
`P(-1) = 1 + 8-7 + 3 = 5`

`:. -a + 3` `= 5`
`a` `= -2`

 
`=>  C`

Plane Geometry, EXT1 2014 HSC 1 MC

The points \(A\), \(B\) and \(C\) lie on a circle with centre \(O\), as shown in the diagram.

The size of \(\angle ACB\) is 40°.

 What is the size of \(\angle AOB\)?

  1. \(20^{\circ}\)
  2. \(40^{\circ}\)
  3. \(70^{\circ}\)
  4. \(80^{\circ}\)
Show Answers Only

\(D\)

Show Worked Solution

\(\angle AOB = 2 \times 40 = 80^{\circ}\)

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

\(\Rightarrow D\)

Functions, EXT1 F2 2009 HSC 2a

The polynomial  `p(x) = x^3-ax + b`  has a remainder of  `2`  when divided by  `(x-1)`  and a remainder of  `5`  when divided by  `(x + 2)`.  

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

Show Answers Only
`a` `= 4`
`b` `= 5`
Show Worked Solution
`p(x)` `= x^3-ax + b`
`P(1)` `= 2`
`1-a + b` `= 2`
`b` `= a+1\ \ \ …\ text{(1)}`
`P (-2)` `= 5`
`-8 + 2a + b` `= 5`
`2a + b` `= 13\ \ \ …\ text{(2)}`

 

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

`2a + a+1` `= 13`
`3a` `= 12`
`:. a` `= 4`
`:. b` `= 5`

Algebra, STD2 A2 2014 HSC 26f

The weight of an object on the moon varies directly with its weight on Earth.  An astronaut who weighs 84 kg on Earth weighs only 14 kg on the moon.

A lunar landing craft weighs 2449 kg when on the moon. Calculate the weight of this landing craft when on Earth.   (2 marks)

Show Answers Only

 `14\ 694\ text(kg)`

Show Worked Solution

`W_text(moon) prop W_text(earth)`

`=> W_text(m) = k xx W_text(e)`

`text(Find)\ k\ text{given}\  W_text(e) = 84\ text{when}\ W_text(m) = 14`

`14` `= k xx 84`
`k` `= 14/84 = 1/6`

 

`text(If)\ W_text(m) = 2449\ text(kg),\ text(find)\ W_text(e):`

`2449` `= 1/6  xx W_text(e)`
`W_text(e)` `= 14\ 694\ text(kg)`

 

`:.\ text(Landing craft weighs)\ 14\ 694\ text(kg on earth)`

Functions, EXT1 F2 2013 HSC 1 MC

The polynomial  `P(x) = x^3-4x^2-6x + k`  has a factor  `x-2`.

What is the value of  `k`? 

  1. `2` 
  2. `12`
  3. `20` 
  4. `36`  
Show Answers Only

`C`

Show Worked Solution

`P(x) = x^3-4x^2-6x  + k`

`text(S)text(ince)\ \ (x-2)\ \ text(is a factor,)\ \ P(2) = 0`

`2^3-4*2^2-6*2 + k` `= 0`
`8-16-12 + k` `= 0`
`k` `= 20`

 
`=>  C`

Functions, EXT1 F2 2010 HSC 2c

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

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

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

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

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

  2. Find the remainder when  `P(x)`  is divided by  `(x + 1)(x-3)`.     (1 mark)

Show Answers Only
  1. `a = -2,\ b = 6`
  2. ` -2x + 6`
Show Worked Solution

i.  `P(x) = (x+1)(x-3)Q(x) + ax + b`

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

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

 
`P(x) ÷ (x+1)=8\ \ \ text{(given)}`

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

 
`text{Subtract  (1) – (2)}`

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

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

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

 
`:. a= – 2, \ b=6` 
 

ii.  `P(x) -: (x + 1)(x-3)`

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

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

 
`:.\ text(Remainder is)\ \ -2x + 6`

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

Functions, EXT1 F2 2011 HSC 2a

Let  `P(x) = x^3-ax^2 + x`  be a polynomial, where  `a`  is a real number.

When  `P(x)`  is divided by  `x-3`  the remainder is  `12`.

Find the remainder when  `P(x)`  is divided by  `x + 1`.    (3 marks)

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`-4`

Show Worked Solution

`P(x) = x^3 – ax^2 + x`

`text(S)text(ince)\ \ P(x) -: (x – 3)\ \ text(has remainder 12,)`

`P(3) = 3^3-a xx 3^2 + 3` `=12`
`27-9a + 3` `= 12`
`9a` `= 18`
`a` `=2`

 
`:.\ P(x) = x^3-2x^2 + x`

 

`text(Remainder)\ \ P(x) -: (x + 1)\ \ text(is)\ \ P(–1)`

`P(-1)` `= (-1)^3-2(-1)^2-1`
  `= – 4`

Plane Geometry, EXT1 2012 HSC 10 MC

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

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

  1. `theta/2`  
  2. `90^@-theta/2`
  3. `theta` 
  4. `180^@-theta` 
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`B`

Show Worked Solution

`/_ BOA= 2 xx /_ APB`

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

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

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

 
`=>  B`

Functions, EXT1 F2 2012 HSC 8 MC

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

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

  1. `1` 
  2. `7` 
  3. `9` 
  4. `13` 
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`D`

Show Worked Solution

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

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

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

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

 
`=>  D`

Functions, 2ADV F1 2010 HSC 1g

Let  `f(x) = sqrt(x-8)`.  What is the domain of  `f(x)`?   (1 mark)

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 `x >= 8`

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♦ Mean mark 49%.
MARKER’S COMMENT: `x>8` was a common incorrect answer.

`f(x) = sqrt(x-8)`

`text(Domain exists for:)`

`(x-8)` `>= 0`
`x` `>= 8`

Functions, 2ADV F1 2013 HSC 3 MC

Which inequality defines the domain of the function  `f(x) = 1/sqrt(x+3)` ?

  1. `x > -3`  
  2. `x >= -3`  
  3. `x < -3`  
  4. `x <= -3` 
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`A`

Show Worked Solution

`text(Given)\ f(x) = 1/sqrt(x+3)`

`(x + 3)` `> 0`
`x` `> -3`

 

`:.\ text(The domain of)\ f(x)\ text(is)\ \ \ f(x)> -3`

`=>  A`

Algebra, STD2 A4 2011 HSC 28a

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

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

  2. 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)

  3. 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)


Show Answers Only
  1. `P = a/V`
  2. `P = 1 1/2`
  3.  
     
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.

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)

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 `17.8\ text(m)\ \ text{(to 1 d.p.)}`

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♦♦ 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.)}`

Algebra, STD2 A1 2009 HSC 16 MC

The time for a car to travel a certain distance varies inversely with its speed.

Which of the following graphs shows this relationship?
 

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`A`

Show Worked Solution
`T` `prop 1/S`
`T` `= k/S`

 
`text{By elimination:}`

`text(As   S) uarr text(, T) darr => text(cannot be B or D)`

♦ Mean mark 38%

`text(C  is incorrect because it graphs a linear relationship)`

`=>  A`

Algebra, STD2 A4 2010 HSC 13 MC

The number of hours that it takes for a block of ice to melt varies inversely with the temperature. At 30°C it takes 8 hours for a block of ice to melt.

How long will it take the same size block of ice to melt at 12°C?  

  1. 3.2 hours
  2. 20 hours
  3. 26  hours
  4. 45 hours
Show Answers Only

`B`

Show Worked Solution
 
♦ Mean mark 50% 

`text{Time to melt}\ (T) prop1/text(Temp) \ => \ T=k/text(Temp)`

`text(When) \ T=8, text(Temp = 30)`

`8` `=k/30`
`k` `=240`

  

`text{Find}\ T\ text{when  Temp = 12:}`

`T` `=240/12`
  `=20\ text(hours)`

 
`=>  B`