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Linear Functions, EXT1 2017 HSC 11a

The point `P` divides the interval from `Atext{(−4, −4)}` to `B(1,6)` internally in the ratio 2:3.

Find the `x`-coordinate of `P`.  (1 mark)

Show Answers Only

`-2`

Show Worked Solution

`text(Find)\ xtext(-coordinate:)`

`Atext{(−4, −4)}, B(1,6),\ text{Ratio 2:3 (internal)}`

`x` `= (nx_1 + mx_2)/(m + n)`
  `= (3 xx −4 + 2 xx 1)/(2 + 3)`
  `= -2`

Calculus, EXT1* C1 2017 HSC 15c

Two particles move along the `x`-axis.

When  `t = 0`, particle `P_1` is at the origin and moving with velocity 3.

For  `t >= 0`, particle `P_1` has acceleration given by  `a_1 = 6t + e^(-t)`.

  1. Show that the velocity of particle `P_1` is given by  `v_1 = 3t^2 + 4-e^(-t)`  (2 marks)

When  `t = 0`, particle `P_2` is also at the origin.

For  `t >= 0`, particle `P_2` has velocity given by  `v_2 = 6t + 1-e^(-t)`.

  1. When do the two particles have the same velocity?  (2 marks)

  2. Show that the two particles do not meet for  `t > 0`.  (3 marks)

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `t = 1`
  3. `text(Proof)\ \ text{(See Worked Solutions)}`
Show Worked Solution
i.   `a_1` `= 6t + e^(-t)`
  `v_1` `= int a_1\ dt`
    `= int 6t + e^(-t)\ dt`
    `= 3t^2-e^(-t) + c`

 

`text(When)\ t = 0,\ v_1 = 3`

`3` `= 0-1 + c`
`c` `= 4`
`:. v_1` `= 3t^2 + 4-e^(-t) …\ text(as required)`

 

ii.  `v_2 = 6t + 1-e^(-t)`

`text(Find)\ \ t\ \ text(when)\ \ v_1 = v_2`

`3t^2 + 4-e^(-t)` `= 6t + 1-e^(-t)`
`3t^2-6t + 3` `= 0`
`t^2-2t + 1` `= 0`
`(t-1)^2` `= 0`
`:. t` `=1`

 

iii.   `x_1` `= int v_1\ dt`
    `= int 3t^2 + 4-e^(-t)\ dt`
    `= t^3 + 4t + e^(-t) + c`

 

`text(When)\ \ t = 0,\ \ x_1 = 0`

♦ Mean mark (iii) 39%.
`0` `= 0 + 0 + 1 + c`
`c` `= -1`
`:. x_1` `= t^3 + 4t + e^(-t)-1`

 

`x_2` `= int 6t + 1-e^(-t)\ dt`
  `= 3t^2 + t + e^(-t) + c`

 
`text(When)\ \ t = 0,\ \ x_2 = 0`

`0` `= 0 + 0 + 1 + c`
`c` `= -1`
`:. x_2` `= 3t^2 + t + e^(-t)-1`

 

`text(Find)\ \ t\ \ text(when)\ \ x_1 = x_2`

`t^3 + 4t + e^(-t)-1` `= 3t^2 + t + e^(-t)-1`
`t^3-3t^2 + 3t` `= 0`
`t(t^2-3t + 3)` `= 0`

 

`text(S)text(ince)\ \ Delta < 0\ \ text(for)\ \ t^2-3t + 3`

`=>\ text(No real solution)`

 

`:.\ text(The particles do not meet)`

`(x_1 != x_2)\ \ text(for)\ \ t > 0.`

Calculus, EXT1* C1 2017 HSC 14c

Carbon-14 is a radioactive substance that decays over time. The amount of carbon-14 present in a kangaroo bone is given by

`C(t) = Ae^(kt),`

where `A` and `k` are constants, and `t` is the number of years since the kangaroo died.

  1. Show that `C(t)` satisfies  `(dC)/(dt) = kC`.  (1 mark)

  2. After 5730 years, half of the original amount of carbon-14 is present.

     

    Show that the value of `k`, correct to 2 significant figures, is – 0.00012.  (2 marks)

  3. The amount of carbon-14 now present in a kangaroo bone is 90% of the original amount.

     

    Find the number of years since the kangaroo died. Give your answer correct to 2 significant  figures.  (2 marks)

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `text(Proof)\ \ text{(See Worked Solutions)}`
  3. `870\ text{years (2 sig. fig.)}`
Show Worked Solution
i.   `C` `= Ae^(kt)`
   `(dC)/(dt)` `= k * Ae^(kt)`
    `= kC …\ text(as required)`

 

ii.  `text(When)\ \ t = 5730, qquad A = 0.5 A_0`

`0.5 A_0` `= A_0 * e^(5730 k)`
`e^(5730 k)` `= 0.5`
`text(ln)\ e^(5730 k)` `= text(ln)\ 0.5`
`5730 k` `= text(ln)\ 0.5`
`k` `= {text(ln)\ 0.5}/5730`
  `= -0.0001209…`
  `= -0.00012\ text{(2 sig fig) … as required}`

 

iii.  `text(Find)\ t\ text(when)\ A = 0.9 A_0`

`0.9 A_0` `= A_0 e^(kt)`
`e^(kt)` `= 0.9`
`kt` `= text(ln)\ 0.9`
`t` `= (text(ln)\ 0.9)/k`
  `= (5730 xx text(ln)\ 0.9)/(text(ln)\ 0.5)`
  `= 870.97…`
  `= 870\ text{years (2 sig.fig.)}`

Calculus, EXT1 C1 2017 HSC 8 MC

A stone drops into a pond, creating a circular ripple. The radius of the ripple increases from 0 cm, at a constant rate of `5\ text(cm s)^(−1)`.

At what rate is the area enclosed within the ripple increasing when the radius is 15 cm?

A.     `25pi\ text(cm)^2\ text(s)^(−1)`

B.     `30pi\ text(cm)^2\ text(s)^(−1)`

C.     `150pi\ text(cm)^2\ text(s)^(−1)`

D.     `225pi\ text(cm)^2\ text(s)^(−1)`

Show Answers Only

`C`

Show Worked Solution

`(dr)/(dt) = 5\ text(cm)^2\ text(s)^(−1)`

`A` `=pi r^2`
`(dA)/(dr)` `= 2pir`

 

`(dA)/(dt)` `= (dA)/(dr) · (dr)/(dt)`
  `= 2pi r · 5`
  `= 10pir`

 
`text(When)\ r = 15`

`(dA)/(dr)` `= 10pi · 15`
   `= 150pi\ text(cm)^2\ text(s)^(−1)`

 
`=>C`

Calculus, 2ADV C4 2017 HSC 14b

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

  3. Using Simpson’s rule with one application, find an approximation to the integral
  4.  
    `qquad int_0^(pi/3) cos x\ dx,`
     
  5. leaving your answer in terms of `pi` and `sqrt 3`.  (2 marks)
     

  6. Using parts (i) and (ii), show that
     
  7. `qquad pi ~~ (18 sqrt 3)/(3 + 4 sqrt 3)`.  (1 mark)

 

 

Show Answers Only
(i)   `sqrt 3/2`
(ii)   `((4 sqrt 3 + 3)pi)/36`
(iii)   `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)  
    `x`     `0`    `overset(pi) underset(6) _`     `overset(pi) underset(3) _`  
    `y`     `1`     `overset(sqrt 3) underset(2) _`     `overset(1) underset(2) _`  
      `y_0`     `y_1`     `y_2`  
`int_0^(pi/3) cos x\ dx` `~~ h/3 [y_0 + 4y_1 + y_2]`
  `~~ pi/6 ⋅ 1/3 [1 + 4 ⋅ sqrt 3/2 + 1/2]`
  `~~ pi/18 ((4 sqrt 3 + 3)/2)`
  `~~ ((4 sqrt 3 + 3) pi)/36`

 

(iii)  `text{Using parts (i) and (ii)}`

♦ Mean mark 49%.
`((4 sqrt 3 + 3) pi)/36` `~~ sqrt 3/2`
`:. pi` `~~ (36 sqrt 3)/(2(3 + 4 sqrt 3))`
  `~~ (18 sqrt 3)/(3 + 4 sqrt 3) … text( as required)`

Geometry and Calculus, EXT1 2017 HSC 5 MC

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

A. B.
       
C. D.
Show Answers Only

`D`

Show Worked Solution
`y` `= (2x^2)/((1 – x^2))`
  `= −((2 – 2x^2 – 2))/((1 – x^2))`
  `= −(2(1 – x^2))/((1 – x^2)) – 2/((1 – x^2))`
  `= −2 – 2/((1 – x^2))`

 

`text(As)\ \ x -> oo,\ \ y -> −2`

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

`⇒D`

Plane Geometry, EXT1 2017 HSC 3 MC

The points `A`, `B`, `C` and `D` lie on a circle and the tangents at `A` and `B` meet at `T`, as shown in the diagram.The angles `BDA` and `BCD` are 65° and 110° respectively.

What is the value of  `angleTAD`?

A.     `130°`

B.     `135°`

C.     `155°`

D.     `175°`

Show Answers Only

`B`

Show Worked Solution
`angleBAD` `= 180 – 110` `(text(opposite angles of)`
  `= 70^@`    `text(cyclic quad))`

 

`angleTAB = 65^@\ \ (text(angle in alternate segment))`

`:. angleTAD` `= 70 + 65`
  `= 135^@`

`⇒ B`

L&E, EXT1 2017 HSC 2 MC

It is given that  `log_a8 = 1.893`, correct to 3 decimal places.

What is the value of  `log_a4`, correct to 2 decimal places?

A.     `0.95`

B.     `1.26`

C.     `1.53`

D.     `2.84`

Show Answers Only

`B`

Show Worked Solution
`log_a8` `= log_a4^(3/2)`
  `= 3/2 xx log_a4`
`:. log_a4` `= 2/3 xx log_a8`
  `= 2/3 xx 1.893`
  `= 1.263`
  `= 1.26`

`⇒ B`

Calculus, 2ADV C4 2017 HSC 13d

The rate at which water flows into a tank is given by

`(dV)/(dt) = (2t)/(1 + t^2)`,

where `V` is the volume of water in the tank in litres and `t` is the time in seconds.

Initially the tank is empty.

Find the exact amount of water in the tank after 10 seconds.  (3 marks)

Show Answers Only

`text(ln)\ 101`

Show Worked Solution
`(dV)/(dt)` `= (2t)/(1 + t^2)`
`V` `= int (2t)/(1 + t^2)\ dt`
  `= text(ln)\ (1 + t^2) + c`

 
`text(When)\ \ t = 0,\ \ V = 0`

`0` `= text(ln)\ 1 + c`
`:. c` `= 0`

 
`text(Find)\ V\ text(when)\ t = 10:`

`V` `= text(ln)\ (1 + 10^2)`
  `= text(ln)\ 101`

Calculus, 2ADV C3 2017 HSC 13b

Consider the curve  `y = 2x^3 + 3x^2 - 12x + 7`.

  1. Find the stationary points of the curve and determine their nature.  (4 marks)

  2. Sketch the curve, labelling the stationary points.  (2 marks)

  3. Hence, or otherwise, find the values of `x` for which `(dy)/(dx)` is positive.  (1 mark)

Show Answers Only
  1. `text(maximum at)\ (-2, 27)`

     

    `text(minimum at)\ (1, 0)`

  2.    
  3. `x < -2 and x > 1`
Show Worked Solution
i.   `y` `= 2x^3 + 3x^2 – 12x + 7`
  `(dy)/(dx)` `= 6x^2 + 6x – 12`
  `(d^2y)/(dx^2)` `= 12x + 6`

 

`text(S.P. when)\ (dy)/(dx)` `= 0`
`6x^2 + 6x – 12` `= 0`
`x^2 + x – 2` `= 0`
`(x + 2) (x – 1)` `= 0`

 
`x = -2 or 1`
 

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

`:.\ text(MAX at)\ (–2, 27)`
 

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

`:.\ text(MIN at)\ (1, 0)`

 

ii.  

 

iii.  `text(Solution 1)`

`text(From graph, gradient is positive for)`

`x < –2 and x > 1`

`:. (dy)/(dx) > 0\ \ text(for)\ \ x < –2 and x > 1`

 

`text(Solution 2)`

`(dy)/(dx) > 0`

`6x^2 + 6x – 12` `> 0`
`(x + 2) (x – 1)` `> 0`

 
 
`:. x < –2 and x > 1`

Statistics, 2ADV 2017 HSC 12e

A spinner is marked with the numbers 1, 2, 3, 4 and 5. When it is spun, each of the five numbers is equally likely to occur.
 

 
The spinner is spun three times.

  1. What is the probability that an even number occurs on the first spin?  (1 mark)
  2. What is the probability that an even number occurs on at least one of the three spins?  (1 mark)
  3. What is the probability that an even number occurs on the first spin and odd numbers occur on the second and third spins?  (1 mark)
  4. What is the probability that an even number occurs on exactly one of the three spins?  (1 mark)
Show Answers Only
  1. `2/5`
  2. `98/125`
  3. `18/125`
  4. `54/125`
Show Worked Solution

i.   `Ptext{(even)} = 2/5`
 

ii.  `Ptext{(at least 1 even)}`

`= 1 – Ptext{(no evens)}`

`= 1 – 3/5 ⋅ 3/5 ⋅ 3/5`

`= 1 – 27/125`

`= 98/125`
 

iii.  `Ptext{(even, odd, odd)}`

`= 2/5 ⋅ 3/5 ⋅ 3/5`

`= 18/125`
 

iv.  `Ptext{(even occurs exactly once)}`

`= Ptext{(e, o, o)} + P text{(o, e, o)} + P text{(o, o, e)}`

`= 2/5 ⋅ 3/5 ⋅ 3/5 + 3/5 ⋅ 2/5 ⋅ 3/5 + 3/5 ⋅ 3/5 ⋅ 2/5`

`= 54/125`

Linear Functions, 2UA 2017 HSC 12d

The points  `A(–4, 0)`  and  `B(1, 5)`  lie on the line  `y = x + 4`.

The length of  `AB`  is  `5 sqrt 2`.

The points  `C(0, –2)`  and  `D(3, 1)`  lie on the line  `x - y - 2 = 0`.

The points `A, B, D, C` form a trapezium as shown.
 


 

  1. Find the perpendicular distance from point `A(–4, 0)` to the line  `x - y - 2 = 0`.  (1 mark)
  2. Calculate the area of the trapezium.  (2 marks)
Show Answers Only

(i)   `3sqrt2\ text(units)`

(ii)  `24\ text(u²)`

Show Worked Solution

(i)   `A(-4, 0), qquad qquad x – y – 2 = 0`

`_|_\ text(dist)` `= |ax_1 + by_1 + c|/sqrt(a^2 + b^2)`
  `= |-4 + 0 – 2|/sqrt (1 + 1)`
  `= 6/sqrt 2 xx sqrt 2/sqrt 2`
  `= 3 sqrt 2\ text(units)`

 

(ii)  `text(Area) = 1/2 ⋅ h ⋅ (AB + CD)`

`AB` `= 5 sqrt 2\ text{(given)}`
`CD` `= sqrt((3 – 0)^2 + (1 + 2)^2)`
  `= sqrt 18`
  `= 3 sqrt 2`

 

`:.\ text(Area)` `= 1/2 ⋅ 3 sqrt 2\ \ (5 sqrt 2 + 3 sqrt 2)`
  `= 1/2 ⋅ 3 sqrt 2 ⋅ 8 sqrt 2`
  `= 24\ text(u)²`

Calculus, EXT1* C3 2017 HSC 12b

The diagram shows the region bounded by  `y = sqrt (16 - 4x^2)`  and the `x`-axis.
 


 

The region is rotated about the `x`-axis to form a solid.

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

Show Answers Only

`(128 pi)/3\ text(u³)`

Show Worked Solution
`y` `= sqrt (16 – 4x^2)`
`V` `= pi int_(-2)^2 y^2\ dx`
  `= 2 pi int_0^2 16 – 4x^2\ dx`
  `= 2 pi [16x – 4/3 x^3]_0^2`
  `= 2 pi [(16 ⋅ 2 – 4/3 2^3)-0]`
  `= 2 pi (64/3)`
  `= (128 pi)/3\ text(u³)`

Calculus, 2ADV C1 2017 HSC 12a

Find the equation of the tangent to the curve  `y = x^2 + 4x - 7`  at the point  `(1, -2)`.  (2 marks)

Show Answers Only

`y = 6x – 8`

Show Worked Solution
`y` `= x^2 + 4x – 7`
`(dy)/(dx)` `= 2x + 4`

 
`text(When)\ x = 1,\ \ (dy)/(dx) = 6`

`text(Equation of tangent through)\ (1, -2)`

`y + 2` `= 6 (x – 1)`
`y` `= 6x – 8`

Trigonometry, 2ADV T1 2017 HSC 11e

In the diagram, `OAB` is a sector of the circle with centre `O` and radius 6 cm, where `/_ AOB = 30^@`.

  1.  Find the exact value of the area of the triangle `OAB`.   (1 mark)

  2. Find the exact value of the area of the shaded segment.   (1 mark)

Show Answers Only
  1.  `9\ text(cm)^2`
  2. `3 pi-9\ text(cm)^2`
Show Worked Solution
i.   `text(Area)\ Delta OAB` `= 1/2 ab sin C`
    `= 1/2 xx 6^2 xx sin 30^@`
    `= 9\ text(cm)^2`

 

ii.  `text(Area segment)` `= text(Area sector)-text(Area)\ Delta OAB`
    `= 30/360 xx pi xx 6^2-9`
    `= 3 pi-9\ \ text(cm)^2`

Functions, 2ADV F1 2017 HSC 2 MC

Which expression is equal to  `3x^2-x-2`?

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

`D`

Show Worked Solution

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

`=>  D`

Functions, 2ADV F1 2017 HSC 1 MC

What is the gradient of the line  `2x + 3y + 4 = 0`?

  1. `-2/3`
  2. `2/3`
  3. `-3/2`
  4. `3/2`
Show Answers Only

`A`

Show Worked Solution
`2x + 3y + 4` `= 0`
`3y` `= -2x-4`
`y` `= -2/3 x-4/3`

 
`=>  A`

Statistics, STD2 S5 2017 HSC 29d

All the students in a class of 30 did a test.

The marks, out of 10, are shown in the dot plot.
 


 

  1. Find the median test mark.  (1 mark)

  2. The mean test mark is 5.4. The standard deviation of the test marks is 4.22.

     

    Using the dot plot, calculate the percentage of the marks which lie within one standard deviation of the mean.  (2 marks)

  3. A student states that for any data set, 68% of the scores should lie within one standard deviation of the mean. With reference to the dot plot, explain why the student’s statement is NOT relevant in this context.  (1 mark)

Show Answers Only
  1. `6`
  2. `text(43%)`
  3. `text(The statement assumes the data is normally)`
    `text(distributed which is not the case here.)`
Show Worked Solution
♦ Mean mark 50%.
i.    `text(Median)` `= text(15th + 16th score)/2`
    `= (4 + 8)/2`
    `= 6`

 

ii.   `text(Lower limit) = 5.4 – 4.22 = 1.18`

♦♦ Mean mark 34%.

`text(Upper limit) = 5.4 + 4.22 = 9.62`

`:.\ text(Percentage in between)`

`= 13/30 xx 100`

`= 43.33…`

`= 43text{%  (nearest %)}`

 

iii.   `text(The statement assumes the data is normally)`

♦♦♦ Mean mark 13%.

`text(distributed which is not the case here.)`

Algebra, 2UG 2017 HSC 28a

Temperature can be measured in degrees Celsius (`C`) or degrees Fahrenheit (`F`).

The two temperature scales are related by the equation  `F = (9C)/5 + 32`.

  1. Calculate the temperature in degrees Fahrenheit when it is  −20 degrees Celsius.  (1 mark)
  2. Solve the following equations simultaneously, using either the substitution method or the elimination method.  (2 marks)
    `qquadF = (9C)/5 + 32`

    `qquadF = C`

     

  3. The graphs of  `F = (9C)/5 + 32`  and  `F = C`  are shown below.
  4.  


  5. What does the result from part (ii) mean in the context of the graph?  (1 mark)     

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Show Answers Only
  1. `−4^@F`
  2. `C = −40, F = −40`
  3. `text(It means the two graphs intersect)`
    `text(at)\ (−40,−40).`

 

 

Show Worked Solution
(i)   `F` `= (9(−20))/5 + 32`
    `= −4^@F`

 

(ii)   `F` `= (9C)/5 + 32` `…\ (1)`
  `F` `= C` `…\ (2)`

 

♦♦ Mean mark 31%.
MARKER’S COMMENT: An area that requires attention.

`text(Substitute)\ \ F = C\ \ text{from (2) into (1)}`

`C` `= (9C)/5 + 32`
`(9C)/5 – C` `= −32`
`(4C)/5 – C` `= −32`
`C` `= −32 xx 5/4 = −40`

 

`text{From (2),}`

`F = −40`

`text{(i.e. when}\ C = −40, F = −40)`

♦♦♦ Mean mark 20%.

 

(iii)   `text(It means the two graphs intersect)`

`text{at (−40,−40).}`

Statistics, STD2 S1 2017 HSC 27a

Jamal surveyed eight households in his street. He asked them how many kilolitres (kL) of water they used in the last year. Here are the results.

`220, 105, 101, 450, 37, 338, 151, 205`

  1. Calculate the mean of this set of data.   (1 mark)

  2. What is the standard deviation of this set of data, correct to one decimal place?   (1 mark)

Show Answers Only

a.    `200.875`

b.    `127.4\ \ text{(1 d.p.)}`

Show Worked Solution
a.   `text(Mean)` `= (220 + 105 + 101 + 450 + 37 + 338 + 151 + 205) ÷ 8`
    `= 200.875`
♦ Mean mark part (b) 47%.
IMPORTANT: The population standard deviation is required here.
  

b.   `text(Std Dev)` `= 127.357…\ \ text{(by calc)}`
    `= 127.4\ \ text{(1 d.p.)}`

Algebra, STD2 A2 2017 HSC 14 MC

Kate is comparing two different models of car. Car A uses fuel at the rate of 9 L/100 km. Car B uses 3.5 L/100 km.

Suppose Kate plans on driving 8000 km in the next year.

How much less fuel will she use driving car B instead of car A?

  1. 280 L
  2. 440 L
  3. 720 L
  4. 1000 L
Show Answers Only

`B`

Show Worked Solution

`text(Fuel car)\ A= 8000/100 xx 9= 720\ text(L)`

`text(Fuel car)\ B= 8000/100 xx 3.5= 280\ text(L)`

`:.\ text(Fuel saved car)\ B= 720-280= 440\ text(L)`

`=>B`

Statistics, STD2 S5 2017 HSC 13 MC

The heights of Year 12 girls are normally distributed with a mean of 165 cm and a standard deviation of 5.5 cm.

What is the `z`-score for a height of 154 cm?

A.     `−2`

B.    `−0.5`

C.     `0.5`

D.     `2`

Show Answers Only

`text(A)`

Show Worked Solution
`ztext(-score)` `= (x – mu)/sigma`
  `= (154 – 165)/5.5`
  `= −2`

 
`=>A`

Statistics, STD2 S4 2017 HSC 12 MC

Which of the data sets graphed below has the largest positive correlation coefficient value?
 

A.      B.     
C.      D.     
Show Answers Only

\(C\)

Show Worked Solution

\(\text{Largest positive correlation occurs when both variables move}\)

\(\text{in tandem. The tighter the linear relationship, the higher the}\)

\(\text{correlation.}\)

\(\Rightarrow C\)

\(\text{(Note that B is negatively correlated)}\)

Algebra, STD2 A1 2017 HSC 9 MC

What is the value of  `x`  in the equation  `(5-x)/3 = 6`?

  1. `-13`
  2. `-3`
  3. `3`
  4. `13`
Show Answers Only

`A`

Show Worked Solution
`(5-x)/3` `= 6`
`5-x` `= 18`
`-x` `=13`
`x` `=-13`

`=>A`

Financial Maths, STD2 F1 2017 HSC 6 MC

Tom earns a weekly wage of $1025. He also receives an additional allowance of $87.50 per day when handling toxic substances.

What is Tom’s income in a fortnight in which he handles toxic substances on 5 separate days?

  1. $1112.50
  2. $1462.50
  3. $2225.00
  4. $2487.50
Show Answers Only

`D`

Show Worked Solution

`text(Fortnightly wage)= 2 xx 1025= $2050`

`text(Allowances)= 5 xx 87.50= $437.50`

`:.\ text(Income)= 2050 + 437.50= $2487.50`

   
`=>D`

Probability, STD2 S2 2017 HSC 5 MC

In a survey of 200 randomly selected Year 12 students it was found that 180 use social media.

Based on this survey, approximately how many of 75 000 Year 12 students would be expected to use social media?

A.     60 000

B.     67 500

C.     74 980

D.     75 000

Show Answers Only

`B`

Show Worked Solution
`text(Expected number)` `= 180/200 xx 75\ 000`
  `= 67\ 500`

`=> B`

Algebra, STD2 A2 2017 HSC 3 MC

The graph shows the relationship between infant mortality rate (deaths per 1000 live births) and life expectancy at birth (in years) for different countries.
 

What is the life expectancy at birth in a country which has an infant mortality rate of 60?

  1. 68 years
  2. 69 years
  3. 86 years
  4. 88 years
Show Answers Only

\(A\)

Show Worked Solution

\(\text{When infant mortality rate is 60, life expectancy}\)

\(\text{at birth is 68 years (see below).}\)
 

\(\Rightarrow A\)

Algebra, STD2 A1 2017 HSC 2 MC

A car is travelling at 95 km/h.

How far will it travel in 2 hours and 30 minutes?

  1. 38 km
  2. 41.3 km
  3. 218.5 km
  4. 237.5 km
Show Answers Only

`D`

Show Worked Solution

`text(Distance)= 95 xx 2.5= 237.5\ text(km)`

`=>D`

CHEMISTRY, M4 2009 HSC 20

  1. Calculate the mass of ethanol \(\ce{C2H6O}\) that must be burnt to increase the temperature of 210 g of water by 65°C, if exactly half of the heat released by this combustion is lost to the surroundings.
  2. The heat of combustion of ethanol is 1367 kJ mol −1.  (3 marks)

  3. What are TWO ways to limit heat loss from the apparatus when performing a first-hand investigation to determine and compare heat of combustion of different liquid alkanols?  (1 mark)

Show Answers Only

a.    3.85 grams

b.    Answers could include two of the following:

  • Use of an insulated vessel (Styrofoam cup)
  • Place the vessel as close as safely possible to the Bunsen’s flame.
  • Use a lid for the beaker
Show Worked Solution

a.    \(q=mC \Delta T = 210 \times 4.18 \times 65 = 57\ 057\ \text{J} = 57.057\ \text{kJ}\)

\(\ce{n(C2H5OH) = \dfrac{57.057}{1367} = 0.04174\ \text{mol}}\)

\(\ce{m(C2H5OH) = n \times MM = 0.04174 \times 46.068 = 1.923\ \text{g}}\)

Since half of the heat is lost to environmental surroundings.

\(\ce{m(C2H5OH)_{\text{init}} = 2 \times 1.923= 3.85\ \text{g}}\)
 

b.    Answers could include two of the following:

  • Use of an insulated vessel (Styrofoam cup)
  • Place the vessel as close as safely possible to the Bunsen’s flame.
  • Use a lid for the beaker

GEOMETRY, FUR1 SM-Bank 11 MC

In this diagram of the Earth, `O` represents the centre and `B` lies on both the Equator and the Greenwich Meridan.

 2010 15 MC

What is the latitude and longitude of point `A`?

A.   `text(30°N  110°E)`

B.   `text(70°N  30°W)`

C.   `text(110°N  60°E)`

D.   `text(30°N  110°W)`

E.    `text(60°N  110°W)`

Show Answers Only

`A`

Show Worked Solution

 `text(S)text(ince A is)  30^circ\  text(North of the Equator)`

   `→ text(Latitude is)  30^circ text(N)`

  `text(S)text(ince A is)  110^circ\  text(East of Greenwich)`

    `→ text(Longitude is)  110^circ text(E)`

`:. A\ text(coordinates are:) \  30^circ text(N)  110^circ text(E)`

`=>  A`

Calculus, MET1 2016 VCAA 1a

Let `y = (cos(x))/(x^2 + 2)`.

Find  `(dy)/(dx)`.   (2 marks)

Show Answers Only

`(-x^2sin(x)-2sin(x)-2xcos(x))/((x^2 + 2)^2)`

Show Worked Solution

`text(Using Quotient Rule:)`

`(h/g)^{prime}` `= (h^{prime}g-hg^{prime})/(g^2)`
`(dy)/(dx)` `= (-sin(x)(x^2 + 2)-cos(x)(2x))/((x^2 + 2)^2)`
  `= (-x^2sin(x)-2sin(x)-2xcos(x))/((x^2 + 2)^2)`

Probability, MET2 2007 VCAA 19 MC

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

VCAA 2007 19mc

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):}`
  5. `{:(a = 0­­­, and b = 0.3):}`
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}\ \ text(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`

Algebra, MET2 2007 VCAA 11 MC

The solution set of the equation  `e^(4x) - 5e^(2x) + 4 = 0`  over `R` is

A.   `{1, 4}`

B.   `{– 4, – 1})`

C.   `{– 2, – 1, 1, 2})`

D.   `{– log_e(2), 0, log_e(2)}`

E.   `{0, log_e(2)}`

Show Answers Only

`E`

Show Worked Solution

`text(Solve for)\ \ x\ \ text(on CAS:)`

`x = 0 \ or \ x = log_e(2)`

`=>   E`

Probability, MET2 2007 VCAA 7 MC

The random variable `X` has a normal distribution with mean 11 and standard deviation 0.25.

If the random variable `Z` has the standard normal distribution, then the probability that `X` is less than 10.5 is equal to

  1. `text(Pr) (Z > 2)`
  2. `text(Pr) (Z < – 1.5)`
  3. `text(Pr) (Z < 1)`
  4. `text(Pr) (Z >= 1.5)`
  5. `text(Pr) (Z < – 4)`
Show Answers Only

`A`

Show Worked Solution
`text(Pr) (X < 10.5)` `= text(Pr) (Z < (10.5 – 11)/0.25)`
  `= text(Pr) (Z < – 2)`
  `= text(Pr) (Z > 2)`

 
`=> A`

Algebra, MET2 2007 VCAA 3 MC

If  `y = log_a (7x - b) + 3`, then `x` is equal to

  1. `1/7 a^(y - 3) + b`
  2. `1/7 (a^y - 3) + b`
  3. `1/7 (a^(y - 3) + b)`
  4. `a^(y - 3) - b/7`
  5. `(y - 3)/(log_a(7 - b))`
Show Answers Only

`C`

Show Worked Solution
`y – 3` `= log_a (7x – b)`
`a^(y – 3)` `= 7x – b`
`a^(y – 3) + b` `= 7x`
`:. x` `= 1/7 (a^(y – 3) + b)`

 
`=>   C`

Algebra, MET2 2007 VCAA 2 MC

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

Then  `f(g(x))`  is given by 

  1. `e^(4x + 6) + 2 e^(2x + 3) - 3`
  2. `2x^2 + 4x - 6`
  3. `e^(2x^2 + 4x + 9)`
  4. `e^(2x^2 + 4x - 3)`
  5. `e^(2x^2 + 4x - 6)`
Show Answers Only

`D`

Show Worked Solution

`text(Solution 1)`

`text(Define)\ \ f(x) and g(x)\ \ text(on CAS)`

`f(g(x)) = e^(2x^2 + 4x – 3)`

`=>   D`
 

`text(Solution 2)`

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

`=>D`

Algebra, MET2 SM-Bank 8 MC

For the polynomial  `P(x) = x^3 − ax + 4,\ \ P( – 3) = – 5.`

The value of  `a`  is

A.   `− 12`

B.    `− 5`

C.    `– 3`

D.       `3`

E.       `6`

Show Answers Only

`E`

Show Worked Solution
`(-3)^3 -a(-3)+4` `= -5`
`-27+3a+4`  `= -5`
`3a`  `=18`
`a`  `= 6`

 
`⇒ E`

Algebra, MET2 SM-Bank 7 MC

If  `x-2`  is a factor of  `2x^3 - 10x^2 + 6x + a`  where  `a in R text{\}{0},`  then the value of `a` is

A.   `-68`

B.   `-20`

C.     `-2`

D.        `2`

E.      `12`

Show Answers Only

`E`

Show Worked Solution

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

`=> P(2)=0`

`P(2)` `= 2 · 2^3 – 10 · 2^2 + 6 · 2 + a`
`0`  `= 16-40+12+a`
`a` `=12`

 

`=>  E`

Graphs, MET2 2008 VCAA 8 MC

The graph of the function  `f: D -> R,\ f(x) = (x - 3)/(2 - x),` where `D` is the maximal domain has asymptotes

  1. `x = 3,\ \ \ \ \ \ \ \ \ \ y = 2`
  2. `x = -2,\ \ \ \ \ y = 1`
  3. `x = 1,\ \ \ \ \ \ \ \ \ \ y = -1`
  4. `x = 2,\ \ \ \ \ \ \ \ \ \ y = -1`
  5. `x = 2,\ \ \ \ \ \ \ \ \ \ y = 1`
Show Answers Only

`D`

Show Worked Solution

`text(Use proper fraction tool on CAS:)`

`[text(CAS: propFrac) ((x – 3)/(2 – x))]`

`f(x) = -1 – 1/(x – 2)`

`:.\ text(Asymptotes:)\ \ x = 2, y = – 1`

`=>   D`

Algebra, MET2 2008 VCAA 7 MC

The inverse of the function  `f: R^+ -> R,\ f(x) = 1/sqrt x - 3`  is

  1. `{:f^-1: R^+ -> R, qquad qquad qquad qquad f^-1(x) = (x + 3)^2:}`
  2. `{:f^-1: R^+ -> R, qquad qquad qquad qquad f^-1(x) = 1/x^2 + 3:}`
  3. `{:f^-1: (3, oo) -> R, qquad qquad qquad f^-1 (x) = (-1)/(x - 3)^2:}`
  4. `{:f^-1: text{(−3, ∞)} -> R, qquad qquad f^-1 (x) = 1/(x + 3)^2:}`
  5. `{:f^-1: text{(−3, ∞)} -> R, qquad qquad f^-1 (x) = -1/x^2 - 3:}`
Show Answers Only

`D`

Show Worked Solution

`text(Let)\ \ y = f(x)`

`text(Inverse:  swap)\ \ x harr y`

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

 

`text(Domain)\ (f^-1(x))` `= text(Range)\ (f)`
  `= (– 3, oo)`

`=>   D`

Graphs, MET1 SM-Bank 27

The graph shown is  `y = A sin bx`.

  1. Write down the value of  `A`.   (1 mark)

  2. Find the value of  `b`.   (1 mark)

  3. Copy or trace the graph into your writing booklet.

     

    On the same set of axes, draw the graph  `y = 3 sin x + 1`  for  `0 <= x <= pi`.   (2 marks)

Show Answers Only
  1. `A = 4`
  2. `b = 2`
  3. `text(See Worked Solutions for sketch)`
Show Worked Solution

a.   `A = 4`

b.  `text(S)text(ince the graph passes through)\ \ (pi/4, 4)`

`text(Substituting into)\ \ y = 4 sin bx`

`4 sin (b xx pi/4)` `=4`
`sin (b xx pi/4)` `= 1`
`b xx pi/4` `= pi/2`
`:. b` `= 2`

  

 MARKER’S COMMENT: Graphs are consistently drawn too small by many students. Aim to make your diagrams 1/3 to 1/2 of a page. 
c.

Calculus, MET1 2006 ADV 2ai

Differentiate with respect to `x`:

Let  `f(x)=x tan x`.  Find  `f^{prime}(x)`.   (2 marks)

 

Show Answers Only

`f^{prime}(x) = x  sec^2 x + tan x `

Show Worked Solution

`y = x tan x`

`text(Using product rule)`

`f^{prime} (uv)` `= u^{prime}v + uv ^{prime}`
`:.f^{prime}(x)` `= tan x + x xx sec^2 x`
  `= x sec^2 x + tan x`

Graphs, MET1 SM-Bank 20

The rule for  `f` is  `f(x) = x-1/2 x^2`  for  `x <= 1`.  This function has an inverse,  `f^(-1) (x)`.

  1. Sketch the graphs of  `y = f(x)`  and  `y = f^(-1) (x)`  on the same set of axes. (Use the same scale on both axes.)   (2 marks)

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

  3. Evaluate  `f^(-1) (3/8)`.    (1 mark)

Show Answers Only
  1.  
    Inverse Functions, EXT1 2008 HSC 5a Answer
  2. `y = 1-sqrt(1-2x)`
  3. `1/2`
Show Worked Solution
a. 

Inverse Functions, EXT1 2008 HSC 5a Answer
b.   `y = x-1/2 x^2,\ \ \ x <= 1`

 

`text(For the inverse function, swap)\ \ x↔y,`

`x` `= y-1/2 y^2,\ \ \ y <= 1`
`2x` `= 2y-y^2`
`y^2-2y + 2x` `= 0`

 

`text(Using quadratic formula,)`

`y` `= (2 +- sqrt( (-2)^2-4 * 1 * 2x) )/2`
  `= (2 +- sqrt(4-8x))/2`
  `= (2 +- 2 sqrt(1-2x))/2`
  `= 1 +- sqrt (1-2x)`

 

`:. y = 1-sqrt(1-2x), \ \ (y <= 1)`

 

c.    `f^(-1) (3/8)` `= 1-sqrt(1-2(3/8))`
    `= 1-sqrt(1-6/8)`
    `= 1-sqrt(1/4)`
    `= 1-1/2`
    `= 1/2`

Calculus, MET1 2008 ADV 3b

  1. Differentiate  `log_e(cos x)` with respect to `x`.   (2 marks)

  2. Hence, or otherwise, evaluate  `int_0^(pi/4) tan x\ dx`.   (2 marks)

Show Answers Only
  1. `-tan x`
  2. `-log_e(1/sqrt2)\  text{or  0.35  (2 d.p.)}`
Show Worked Solution
a.   `y` `= log_e(cos x)`
  `(dy)/(dx)` `= (-sin x)/(cos x)`
    `=-tan x`

 

b.   `int_0^(pi/4) tan x\ dx`

`= -[log_e(cos x)]_0^(pi/4)`

`= -[log_e(cos(pi/4))-log_e(cos 0)]`

`= -[log_e(1/sqrt2)-log_e 1]`

`= -[log_e(1/sqrt2)-0]`

`= -log_e(1/sqrt2)`

`= 0.346…`

`= 0.35\ \ (text(2 d.p.))`