NCEA Level 2, 2024 Mathematics exams
on this page: algebra exam | calculus exam | conclusion
Mathematics and Statistics – Algebra
QUESTION ONE (d)
A cubic polynomial is given by `f(x) = 3x^3+ ax^2+ bx + c`, where `a`, `b`, and `c` are constants.
The graph of the polynomial has three roots, and cuts the x axis at `-2`, `1/3`, and `4`.
Using your knowledge of roots, find the value of the constants.
If we can factorise the cubic, each factor should correspond directly to one of the stated roots.
The coefficient `3` on the `x^3` term is a prime number, which is only factorisable one way (using whole numbers), so the coefficients on each x in the factors will be `3`, `1`, and `1`. These can be in any order, but if we keep them in the same order as the stated roots, the `3` will be in the middle factor.
The other factors are easy to get. Setting each factor equal to zero gives the three roots, and working backwards can be done just as easily. For example:
`x=4`
`implies x-4=0`
`implies (x-4)` is a factor
`(x + 2)(3x - 1)(x - 4)=0`
Expand to get the function with its coefficients.
`3x^3 - 7x^2 - 22x + 8 = 0`
Check each root to confirm the equation is correct. (Using the x variable memory on your calculator and your calculator history, you only need to enter the polynomial expression once.)
`a=-7`
`b=-22`
`c=8`
Comment: This doesn't seem hard enough to be an Excellence question.
QUESTION ONE (e)
Show that `((x^(3/2) + x^(1/2))(x^(1/2) - x^(-1/2)))/(x^(3/2) - x^(1/2))^2` can be simplified to `(x+1)/(x(x-1))`.
Use exponent rules to expand. The denominator is already a perfect square. The numerator becomes the difference of two squares, which can be factorised and simplified with the denominator.
`(x^2 - x + x - 1)/(x^3 - 2x^2 + x)`
`(x^2 - 1)/(x(x^2 - 2x + 1))`
`((x + 1)(x - 1))/(x(x-1)^2)`
`(x + 1)/(x(x-1))`
Comment: Intimidating fraction! But this is just a straight forward application of algebra. Students should be able to recognise the difference of two squares, and simple perfect squares.
QUESTION TWO (c)
Find the value of `k` for which the equation `(x^2 - 2x)/(4x - 1) = (k - 1)/(k + 1)` has roots numerically equal but opposite signs (for example, `2` and `-2`).
I think this equation will be easiest to work with without denominators and all terms on the left.
`(x^2 - 2x)(k + 1) = (k - 1)(4x - 1)`
`kx^2 + x^2 - 2kx - 2x = 4kx - k - 4x + 1`
`kx^2 + x^2 - 2kx - 4kx - 2x + 4x + k - 1 = 0`
`(k + 1)x^2 + (- 6k + 2)x + (k - 1) = 0`
Good so far. This is now in the typical exanded form of quadratic, a good starting point for whatever we need to do next.
`ax^2 + bx + c = 0`
For one root to be the negative of the other, the vertex needs to be at `x = 0`; we want a parabola with no horizontal movement from the origin. With an expanded form quadratic equation this means the `x` term must be zero, so `b` must be zero.
`ax^2 + c = 0`
`=> - 6k + 2 = 0`
`6k = 2`
`k = 1/3`
Comment: I imagine this will stump a few students, but a good understanding of how quadratics work and what we're trying to get to makes the problem quite easy to solve.
QUESTION TWO (d)
 The logo for an IT company, Doctor of Data, is shown in the diagram to the right.
The green background of the logo is the shape of a rectangle with length `x`. The letters D, O, D are formed using semi-circles for both Ds and a circle for the O. Each has the same radius.
Area of a circle = `pir^2`.
(i) Find an expression in terms of `x` for the area of the circle in the middle of the logo (i.e. the letter 'O').
radius of circle `= x/4`
area of circle `= pi(x/4)^2`
area of circle `= (pix^2)/16`
(ii) If the green background has an area of 10 cm2, find the length of the rectangle.
Make an equation in `x` (the thing we're asked to find) using the expression for the area of the circle, and solve for `x`. Sometimes I like to start with a word equation; it gives me a plan of attack.
area of 2 circles + green area = area of rectangle
`(2pix^2)/16 + 10 = x(x/2)`
`(pix^2)/8 + 10 = x^2/2`
`pix^2 + 80 = 4x^2`
`4x^2 - pix^2 = 80`
`(4 - pi)x^2 = 80`
`x^2 = 80/(4 - pi)`
`x = sqrt(80/(4 - pi))`
`x = 9.654` cm (4sf)
Comment: Not too hard.
QUESTION THREE (d)
Consider the equation `3x^2 - 4kx + k^2 = 0`, where `k` is a constant.
Using the quadratic formula, find the fully simplified solutions to the quadratic equation in terms of `k`.
I find it's helpful to write down the formula I'm going to substitute stuff into. It means I can more easily see what I'm going to put where. I also like to clearly identify a, b, and c by underlining the coefficients in the given equation – including any negative signs!
`x = (-b +- sqrt(b^2 - 4ac))/(2a)`
`x = (4k +- sqrt((-4k)^2 - 4(3)(k^2)))/(2(3))`
`x = (4k +- sqrt(16k^2 - 12k^2))/6`
`x = (4k +- sqrt(4k^2))/6`
`x = (4k +- 2k)/6`
`x = (2k +- k)/3`
`x = k/3 or x = k`
Comment: Straight forward.
QUESTION THREE (e)
The level of sound (intensity) is measured on a logarithmic scale using a unit called a decibel. The formula for the decibel level, `d`, is given by
`d = 10log_10(P/P_0)`
where `P` is the intensity of the sound and `P_0` is the weakest sound that the human ear can hear. A cooling fan has a decibel level of 38 decibels. A heat pump has a decibel level of 30 decibels.
Show that the cooling fan sound intensity is more that six times that of the heat pump sound intensity.
We need to show the ratio `P_"fan"/P_"pump" > 6`, which means rearranging the given formula. We don't need to separate the ratio `P/P_0` because the `P_0` will cancel out.
`d/10 = log_10(P/P_0)`
`10^(d/10) = P/P_0`
This is what we need. Now we can make the ratio we want.
`P/P_0 = 10^(d/10)`
`(P_"fan"/P_0)/(P_"pump"/P_0) = (10^(d_"fan"/10))/(10^(d_"pump"/10))`
`P_"fan"/P_"pump" = 10^((d_"fan"/10)-(d_"pump"/10))`
`P_"fan"/P_"pump" = 10^(38/10-30/10)`
`P_"fan"/P_"pump" = 10^(3.8-3.0)`
`P_"fan"/P_"pump" = 10^0.8`
`P_"fan"/P_"pump" = 6.309... > 6`
Comment: Again, not very hard for well-prepared students who are clear on what they need to do.
Mathematics and Statistics – Calculus
QUESTION ONE (c)
A curve is given by `y = 3x^3 - 9x^2 - 27x + 4`.
Using calculus methods:
(i) Find the x-coordinate of the local minimum.
(ii) Explain how you know this is a minimum point.
Differentiate to find stationary points, let it equal zero, etc.
`(dy)/(dx) = 9x^2 - 18x - 27 = 0` @ stationary points.
`x^2 - 2x - 3 = 0`
`(x + 1)(x - 3) = 0`
`x = -1 or x = 3`
`x = 3` is a local minimum.
There are two ways we know this, but the question specifically requests calculus methods be used for (i) and (ii).
We know the `x^3` coefficient is positive, meaning the first unique `x` value, `x = -1`, is a local maximum, while the second, `x = 3`, is a local minimum. (There could only be a point of inflexion stationary point if the two `x` values were the same.) But this explanation is not adequate, since calculus methods have been requested.
Using calculus methods, differentiate again to get the second derivative (the rate of change of the gradient) and test the two `x` values.
`(d^2y)/(dx^2) = 18x - 18`
`18(-1) - 18 < 0 =>` local maximum
`18(3) - 18 > 0 =>` local minimum
Comment: Pretty straight forward, but read the question carefully so an adequate reason is given for part (ii).
QUESTION ONE (d)
A drink manufacturer would like to start a new line of cylindrical cups that are designed to keep hot liquids warm far as long as possible. To do this, they wish to minimise the surface area of the new cup, which includes a lid.
They would like the cup to hold a volume of 500 mL [corrected from ml], and be cylindrical in shape.
Calculate the dimensions of the cylinder that would satisfy the above conditions.
Volume of a cylinder: `V = pir^2h`
Surface area of a cylinder: `SA = 2pir^2 + 2pirh`
How exciting. I'm betting `2r = h` (diameter = height) gives minimum surface area.
Start with an expression for h, substitute that into the expression for surface area, then rewrite to make it easier to differentiate.
`V = pir^2h = 500`
`h = 500/(pir^2)`
`SA = 2pir^2 + 2pir500/(pir^2)`
`SA = 2pir^2 + 1000/r`
`SA = 2pir^2 + 1000r^(-1)`
`(d(SA))/(dr) = 4pir - 1000r^(-2) = 0` at minimum surface area
`4pir = 1000/r^2`
`r^3 = 1000/(4pi)`
`r = root(3)(1000/(4pi))`
`r = 10/root(3)(4pi)`
`r = 4.301` (4sf)
`h = 500/(pir^2) = 8.603` (4sf)
Yep, `2r = h`.
Don't forget to reread the question after getting a result; we are asked for the cup's dimensions.
The cup is (internally) 8.6 cm in diameter and 8.6 cm high.
Comment: This seems a pretty standard question (quite predictable, even).
QUESTION TWO (d)
A trapezium is drawn within a parabola given by `y = 8x - x^2` and the `x`-axis, such that the two parallel sides are positioned at `x` and `2x`, as shown in the graph below, where `0 <= x <= 4`.
Given that the area of a trapezium is given by `A = 1/2 (a + b) h`, where `a` and `b` are the lengths of the parallel sides, and `h` is the perpendicular distance between them, use calculus to find the maximum area of the trapezium.
I note that `x` can not be greater than 4 because the parabola returns to the `x`-axis at `x = 8` (so `2x <= 8`).
Keep the working nice and methodical. Taking shortcuts is the best way to make mistakes.
`A = 1/2 (a + b) h`
`A = 1/2 (f(x) + f(2x)) (2x - x)`
`A = 1/2 (8x - x^2 + 8(2x) - (2x)^2) x`
`A = 1/2 (8x - x^2 + 16x - 4x^2) x`
`A = 1/2 (24x - 5x^2) x`
`A = 1/2 (24x^2 - 5x^3)`
`A = 12x^2 - 5/2x^3`
At this point we want to differentiate.
`(dA)/(dx) = 24x - 15/2x^2 = 0` at maximum or minimum area
`x(24 - 15/2x) = 0`
`x = 0` gives minimum area
`24 - 15/2x = 0` gives maximum area
`15/2x = 24`
`x = 48/15 = 16/5 = 3.2`
We have a result. Don't forget to reread the question; we are asked to find the maximum area.
maximum area `= 12(16/5)^2 - 5/2(16/5)^3`
`= 1024/25 = 40.96` (exact value)
QUESTION THREE (d)
If `y = k - x`, where `k` is a number, use calculus to show that the minimum value of `x^2 + 2y^2` is equal to `(2k^2)/3`.
The biggest problem here might be what we're going to call the expression `x^2 + 2y^2`. I definitely want to be able to refer to it. How about `E` for expression? We want to find the minimum value of `E`.
`E = x^2 + 2y^2`
This is a calculus exam, so to find a minimum we'll no doubt have to differentiate. But to do that I don't really want both `x` and `y`. I'll replace `y` using the first equation given.
`E = x^2 + 2(k - x)^2`
`E = x^2 + 2(k^2 - 2kx + x^2)`
`E = x^2 + 2k^2 - 4kx + 2x^2`
`E = 3x^2 - 4kx + 2k^2`
Time to differentiate.
`(dE)/(dx) = 6x - 4k = 0` at minimum*
`6x = 4k`
`x = (4k)/6`
`x = (2k)/3`
Don't mistake this for the `(2k^2)/3` which is the minimum value of `E`. They're not the same thing. This is simply the value of `x` where the minimum value of `E` occurs.
* We know this is going to be a minimum value of `E` because the coefficient of the `x^2` term is positive, so the parabola vertex is a minimum. It would be good to state this, but we could also confirm it by differentiating again to get the second derivative, the rate of change of the gradient.
`(d^2E)/(dx^2) = 6 > 0 =>` minimum
Right, that works. Let's substitute that value of x back into the expression. I'll use my simplified expression from earlier.
`E = 3x^2 - 4kx + 2k^2`
`E_"min" = 3((2k)/3)^2 - 4k(2k)/3 + 2k^2`
`E_"min" = 3(4k^2)/9 - (8k^2)/3 + 2k^2`
`E_"min" = (4k^2)/3 - (8k^2)/3 + (6k^2)/3`
`E_"min" = (4k^2)/3 - (8k^2)/3 + (6k^2)/3`
`E_"min" = (2k^2)/3`
And that's it. QED.
Comment: This one is also pretty straight forward.
Conclusion
The algebra and calculus questions that I looked at were not particularly difficult or unclear. On that basis they seem like OK exams (apart from using the unit symbol "ml" instead of "mL".
These exams were held on 5 November. The NCEA Level 2 calculus exam in 2016 was on 24 November. Why such a big loss of learning time?
I've always thought it extremely strange that the NCEA exam rules/guidelines on the front page of each exam do not mention the duration of the exam, or mention how long all three should take.
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