NCEA Level 2 Mathematics, Algebra Exam 2020

A review of the excellence questions in this exam.

Pictures of diagrams coming soon.

QUESTION ONE (c) (i)

Merit question which leads on to two excellence questions.

In the 16th century, mathematicians we developing a formula to solve any cubic equation. They used expressions in the form of `y = x^3 - 12Px + R`, where `P` and `R` are positive constants.

(i) The graph of `y = x^3 - 12Px + R`, for some values of `P` and `R`, passes through the point A (4,40) and is sketched below.

[pic]

Find an expression for `P` in terms of `R`.

We can substitute in the `x` and `y` values of point A then rearrange the equation.

`y = x^3 - 12Px + R`

`40 = (4)^3 - 12P(4) + R`

`40 = 64 - 48P + R`

`48P = 64 - 40 + R`

`48P = 24 + R`

`P = (R + 24)/48`

You don't need to break it down into two separate terms.

Comment:

Pretty straight forward.

The form of cubic used in called a "suppressed" cubic, because it does not have an `x^2` term. (This is done by remapping the axes using a substitution.) It makes the cubic much easier to solve.

QUESTION ONE (c) (ii)

(ii) At point B it is true that `3x^2 - 12P = 0`.

Using algebra, show that `x = -2P^0.5` at B.

The equation comes from differentiating the cubic, and B is a stationary point. But this is an algebra exam.

`3x^2 - 12P = 0`

`3x^2 = 12P`

`x^2 = (12P)/3 = 4P`

`x = +-sqrt(4P)`

`x = +-2sqrtP = +-2P^0.5`

But B is the first stationary point.

`=> x = -2P^0.5`

QED.

Comment:

Easy – just make sure you include the reason why you only take the negative root.

QUESTION ONE (c) (iii)

(iii) Consider again the curve with the equation `y = x^3 - 12Px + R`. As the values of P and R vary, the shape of the curve changes, and the lengths of the orange lines and of the central blue line (below) vary. However, by symmetry, the two orange lines remain the same length as each other.

...

Find the condition that `P` and `R` must satisfy for the graph to have three distinct `x`-intercepts.

Write your answer in a form where the exponents of `P` and `R` are both positive integers.

The equation is centred on the `y`-axis, as we showed with part (c) (ii), and `R` is the `y`-intercept, which is the centre of the cubic's rotational symmetry (it's a point of inflexion, where the graph changes from curving downward to curving upward.

Remember that we were told at the start of the question that `R` is a positive constant (along with `P`).

If the graph is going to have three points where it intersects the `x`-axis then the second stationary point, at C, needs to dip below the `x`-axis, which basically means C needs to have a negative `y` value.

We know the `x` value of point C – we worked it out in part (c) (ii) – and we can get the `y` value by substituting the `x` value into the expression for `y`.

`x^3 - 12Px + R < 0`

`(2P^0.5)^3 - 12P(2P^0.5) + R < 0`

`8P^1.5 - 24P^1.5 + R < 0`

`-16P^1.5 + R < 0`

`R < 16P^1.5`

But we are required to use positive integer exponents. This can be done just by squaring both sides.

`R^2 < (16P^1.5)^2`

`R^2 < 256P^3`

QED.

Comment:

This was trickier, and some thought should be given to finding the easiest way to tackle the problem. This is one of those times that you should try to understand the question before starting to answer it.

If `R` was not a positive constant then the cubic would be mostly below the `x`-axis, and we would have to consider the condition that B is above the `x`-axis, that it has a positive `y` value. By squaring both sides of the inequality we actually end up covering this option anyway. (Substitute in `x = -2P^0.5` and let the inequality be `> 0`. It should end up the same except for a negative sign, which vanishes when squaring the inequality.)

QUESTION TWO (b) (ii)

Solve each of the following equations:

...

(ii) `log_5(x) + log_5(2x) = 4`.

Hm, a Level 2 log excellence question. Curious.

`log_5(x) + log_5(2x) = 4`

`log_5(x times 2x) = 4`

`log_5(2x^2) = 4`

`2x^2 = 5^4`

`x^2 = 1/2 times 5^4`

`x = +-sqrt(1/2 times 5^4)`

We can take only the positive square root because it's not possible to take the log of a negative number.

`x > 0` to be able to take `log_5(x)`

`=> x = sqrt(1/2 times 5^4)` only)

`x = 17.6776...`

`x ≈ 17.68` (4sf)

Comment:

The log rule needed for this question was not included on the Level 2 Mathematics formulae sheet until 2020 (the year of this exam) – perhaps that's why it's an excellence question. Or perhaps the questions are graded after the examiners see how many people answer the questions correctly. (We see an example of the answers being updated to fit the exam in the 2020 Statistics exam, where the assessment schedule has a mean calculated for n=69 and n=70.)

It's interesting that the extra log rules were introduced to the formulae sheet after the first year of pandemic disruption. That's a bit rough, although the new rules are not particularly hard to learn.

Logarithms are my easier to get to grips with if you understand that a logarithm is the inverse function of the log's base to the power of. That means they cancel each other out whichever way around they are applied to each other. For example, `log_5(5^x) = x` and `5^(log_5(x)) = x`.

QUESTION TWO (d)

Consider two parabolas:

• Parabola One given by `y = ax^2 + bx + c` and

• Parabola Two given by `y = dx^2 + ex + c`, where `a, b, c, d,` and `e` are constants.

Use algebra to determine the restrictions on the values of `a, b, c, d,` and `e` that would ensure that the parabolas meet at two distinct points.

Note that both parabolas have the value `c`, which is the `y` intercept. This means the parabolas will always intersect at `(0, c)`.

When two functions intersect, they have the same `x` and `y` values at the points of intersection. Because the parabolas are both expressions for their `y` values, we can start by letting them equal each other, then start boiling it down.

`ax^2 + bx + c = dx^2 + ex + c`

`ax^2 + bx + c - (dx^2 + ex + c) = 0`

`ax^2 + bx + c - dx^2 - ex - c = 0`

`ax^2 + bx - dx^2 - ex = 0`

`ax^2 -dx^2 + bx - ex = 0`

`(a - d)x^2 + (b - e)x = 0`

`x((a - d)x + (b - e)) = 0`

`=> x = 0` or `(a - d)x + (b - e) = 0`

The foretold perpetual point of intersection on the `y` axis! Make it clear that you recognise it for what it is.

`x = 0` gives a permanent point of intersection on the `y` axis, at `(0, c)`

The second solution takes a little more rearranging.

`(a - d)x = - (b - e)`

`(a - d)x = e - b`

`x = (e - b) / (a - d)`

This solution for `x` has a couple of limitations on it, which should be stated.

For a second point of intersection

`a ≠ d` because we cannot divide by zero;

and `b ≠ e` so that the point is distinct from the intersection on the `y` axis.

Comment:

The explanation of the limitations in the assessment schedule really isn't clear because of the lack of punctuation. Could do better, NCEA. Make it easy for students to learn from past exams!

QUESTION THREE (c) (i)

Zahra sells zips. Zahra notices that the higher the price of a zip, the fewer zips are sold. As an experiment, Zahra increases the price of a zip by $2 each day (starting at $7) and keeps a record of how many zips are sold each day. She does this for 6 days and find that the number of zips sold each day started at 98 and is dropping by 3 each day.

The total amount of money Zahra received each day for zips, the turnover, is also recorded in the table below.

Day `d` 1 2 3 4 5 6

Price of a zip ($) `= 2d + 5` 7 9 11 13 15 17

Number of zips sold `= 101 - 3d` 98 95 92 89 86 83

Turnover ($) 686 855 1012 1157 1290 1411

(i) If all the patterns continue to be valid, is there any day on which the turnover is exactly $445? Use algebra to justify your answer and explain your conclusions.

Sounds like some factorising is required. The price for a zip `times` the number of zips sold `=` the turnover. We can make an equation representing that, expand and shift every term to the left, then refactorise or use the quadratic equation.

`(2d + 5)(101 - 3d) = 445`

`202d - 6d^2 + 505 - 15d - 445 = 0`

`-6d^2 + 202d - 15d + 505 - 445 = 0`

`-6d^2 + 187d + 60 = 0`

`6d^2 - 187d - 60 = 0`

I think the quadratic equation sounds like a good idea here.

`d = (187 +- sqrt((-187)^2 - 4(6)(-60)))/(2(6))`

`d = (187 +- sqrt(34969 + 1440))/12`

`d = (187 +- sqrt(36409))/12`

`d = -0.3176` or `d = 31.48`

Neither of these is a positive integer, so no, there is no day on which the turnover is exactly $445.

Comment:

This one is not particularly hard. It would be nice to have it tied to the next part of the question more clearly.

QUESTION THREE (c) (ii)

(ii) Zahra realises that not every whole number is a possible turnover value for a given day.

Using algebra, find at least three conditions a whole number `k` must satisfy for it to be a possible turnover for a given day.

Since the question is so vague it's not possible to do more than guess at what's required. However, using the previous question as a hint, let's start with the quadratic equation and set it equal to k. This is a sensible move to at least get started.

`(2d + 5)(101 - 3d) = k`

`202d - 6d^2 + 505 - 15d - k = 0`

`- 6d^2 + 202d - 15d + 505 - k = 0`

`- 6d^2 + 187d + 505 - k = 0`

`6d^2 - 187d + k - 505 = 0`

`d = (187 +- sqrt(187^2 - 4(6)(k - 505))) / (2(6))`

`d = (187 +- sqrt(34969 - 24(k - 505))) / 12`

`d = (187 +- sqrt(34969 - 24k + 24 times 505)) / 12`

`d = (187 +- sqrt(47089 - 24k)) / 12`

Now, some conditions...

For there to be at least one solution, the discriminant `>= 0`

`=> 47089 - 24k >= 0`

`47089 >= 24k`

`24k <= 47089`

`k <= 47089 / 24`

`k <= 1962.04` (Condition 1)

The next "conditions" need more thought. To start with, the square root needs to give a whole number. When anything other than a square number is square rooted, it gives an irrational number.

`sqrt(47089 - 24k)` must be a whole number
`=> 47089 - 24k` must be a square number (Condition 2)

`d` is an integer
`=> 187 +- sqrt(47089 - 24k)` must be a multiple of 12 (Condition 3)

`d` is positive
`=> 187 +- sqrt(47089 - 24k) > 0` (Condition 4)

And some bonus conditions:

`187 / 12 = 15 7/12`
`=> sqrt(47089 - 24k)` must be `5` more than a multiple of `12` (Condition 5)
and: `-sqrt(47089 - 24k)` must be `7` less than a multiple of `12` (Condition 6)

Getting silly now but that pair are still valid conditions for `k`. Going further, this is the full list:

`k = 142` (Condition 7)
`=> d = 33`

`k` `d`

142 33

345 32

536 31

686 1

715 30

855 2

882 29

1012 3

1037 28

1157 4

1180 27

1290 5

1311 26

1411 6

1430 25

1520 7

1537 24

1617 8

1632 23

1720 9

1715 22

1775 10

1786 21

1836 11

1845 20

1885 12

1892 19

1922 13

1927 18

1947 14

1950 17

1960 15

1961 16

The turnover hits zero (goes negative) after 33 days, so there are no higher values of `k` than 33.

Comment:

This was a very ill-defined question. It's really not clear what the answer is supposed to be, or even what sort of answer is expected. Figuring out what might be desired was the hardest part of this question. Mathematics exam questions are not normally this vague, even in NCEA.

The assessment schedule seems to show how ill-defined the question was with yet another example of apparently tweaking the marking after the exam, with just one (not three!) conditions needing to be found for an E1 grade. Just two conditions is enough for an E2 grade. Listing all four of the conditions (called "conclusions" is the schedule) does not get any extra credit.

The first copy of the assessment schedule I saw had no negative signs in the quadratic formula for this answer – just spaces where a negative sign should have been. I'm unclear if it was another mistake or a display error on that particular computer.

I think this adds up to an NCEA exam failure.

Conclusion

There are a lot of "find the condition" and "determine the restrictions" questions in this exam.