Solutions can be found here.
Personal Thoughts: The paper isn’t tedious. Students can do them so long as they know their stuffs. There are several generalising of questions, like question 6 of paper 1. We also saw how conditional probability was actually tested subtly, this tests students’ abilities to reason with guidance (not sure if after this first trial year, will they still guide the students.) Application questions were not tough and well guided. Students can solve it easily if they read it well. Statistics was well crafted and neat.
To be blunt, I’ll give credit to the 9740 H2 Mathematics paper that run concurrently, since it is too tough to set two sets of papers. Its easy to acknowledge that the 9740 (2016) paper was way harder than 9740 (2017). Next year won’t be the same.
Advice: Students should be careful when you revise, make sure you learn, and not do. Understand what you’re doing. The 2017 paper was an inquisitive paper, examiners were watching closely if you pay attention to details, and know your definitions well.
I’ll do an analysis for the paper, you can click on the individual question and read. For students that took the paper, I hope it doesn’t demoralise you.
Paper 1
Easy question, so long as students aren’t careless. I’ll say that this question exists to prime students’ minds for Q9(c).
Firstly, draw on the SAME axes. The modulus function is a throw back to high school.
Next, “hence” tells us that we need to use the curve. Also very easy.
Students CANNOT use a calculator.
The coordinates found must be exact.
Next, the first derivative test is actually very difficult to administer here. Here’s why… First, the question was to determine, we need to substantiate it fully. One will observe that the 
equation contains both x and y. Students need to understand that y coordinate can be obtain using 
. First derivative test suggests we add and minus a small number to x and sub it into , however, students will not be able to figure out if you add or minus a small number to y here. This action assumes the nature of the stationary point. Thus, students are required to do second derivative test here.
Very easy questions.
But for the first part… Just make sure you don’t say “all x values”, because 
. It is all x values defined on domain of C or all points on C.
Remainder and Factor Theorem will save a lot of time here. I know most of my students that were IP tracks, did long division. This wasted a lot of their time, not to mention if they want to check their workings. This question is a clear indicator that Year 3 and Year 4 topics can and will haunt you. R-Formulae!
Next, the gradient being ALWAYS positive. This tests students on their basic knowledge of differentiation. For starters, 
is not an increasing function, but it is STRICTLY increasing function. And this makes a lot of difference.
The rest is GC work.
This is a technical question, testing students on their definitions. Very obvious answers, yet most students might struggle with definitions. And of course 
isn’t really distance since distance is always positive. The more accurate definition is displacement. Then, it was a pure generalisation of our method to solve for the intersection between a line and a plane. Students need to know their stuffs to do this.
This question also paid tribute to the specimen paper, same question number too. ><
Factor Formulae. I’ve always stressed to my students the importance of trigonometry, and how to use the MF26 to their fullest.
Next part, some careless students will actually not read that was given. If you found 
, read again. You probably end up with 2 marks then.
First part, had some students cringing, especially those that did long division in question 5. Cos they did long division again. But I always say in class, quadratic equations can be solved with the quadratic formulae.
Next part, is really specific. They want p and q found first, and then using these values found to factorise the equation. You need to follow their instructions, else you will lose credit.
First part, again paying tribute to specimen paper. Very easy, just make sure you simplified.
Second part, is tutorial question, just make sure you simplified.
Last part, some students tried to be smart and suggest 
is a common ratio, but it is not. Common ratio must be a constant, and this is not. Its just a simple test, use it and show that is less than 1, you’re done. Like I said in previously, question 1 exists to prime students for Q9c. This question actually open up so many possibilities for future sequences and series questions, and I’m very excited.
Simple question, fully guided, just read carefully. Part 2 was neat and tested students’ ability to think out of the box. It is important to note that Q10 and Q6 are the only vectors questions, but all those standard methods to solve for foot of perpendicular, point of reflection, etc were not tested at all. On the other hand, differentiation would have done the trick here. It is interesting since this again open up a lot of possibilities for future questions.
Simple question, fully guided, just read carefully. They could have asked for a graph instead, or even introduce ideas of kinematics from high school. Expect to see more famous scientists appearing in your papers.
Paper 2
Quite simple, simply testing students on their knowledge of parametric equations. Some high school formulae are used here, namely, distance between two points and midpoint theorem.
General question testing on the formulae for APGP. Solving the polynomial of degree 13 will require the use of the GC, with the graph function. In 2016, we saw a polynomial of degree 10, which could be solved with the {apps}{plysmthz} function. Clearly, more demanding of students’ abilities to use technology now.
Last part requires students to read carefully and in depth.
(a) isn’t like a yes or no question, it wants students to state the coordinates, if possible. So read carefully. Give what is necessary.
(b) is very straight forward. Stronger students might observe that the given function is self-inverse.
(a) is a throwback to high school integration. They didn’t ask for graph too, so the entire question can be solved with a GC. Even finding the intersection between the curves. The definite integral can be solved using {MATH}{9}. No modulus should be applied here. Answer is an exact figure so NO rounding off should be done.
(b) is a bit more challenging. (i) should be manageable, the curve can’t be really be observed, although stronger students might realise they can swap x and y axis in the GC. We simply need to integrate. As we can see from the paper, they blend in the techniques with the applications here. (ii) is slightly intuitive, students should realise that 
should behave akin to 
, that is, a constant greater than 1. This is required in order to produce the same shape.
This is a neat question. Some P&C knowledge being tested here, alongside with DRV, ending with a simply binomial distribution. This justifies why P&C is no longer a stand alone topic in this new syllabus, after all, P&C is basic counting principles.
Interesting way to present the questions. (ii) is really specific and students should read really carefully. (iii) is straightforward so long as students are careful that they are dealing with a circle, and its a probability they want.
Very simple. I expected more, given the specimen paper was much harder. Just be precise and careful for the last part, as CLT is an approximation method, not assumption.
(a) is very technical. Students need to understand the product moment correlation coefficient to a fair extent. For (i), a downward sloping line will do, just remember to erase the lines since question wants only points. For (ii), a vertical or horizontal line does not give zero correlation, only a square does. (iii) seems easy, but some students might fall into a trap of producing something that is greater than 0.9, make sure you show some zig-zag pattern.
(b) is quite an alright question. The question just wanted students to state the appropriate model, no justification is required, though students should use the r-value to aid them. (iii) is similar to the question asked in H1 Math, so students should be well prepared for it.
Very good question, especially for (vi) – (viii) provide students a good intuition of what conditional probability is all about. Like I shared in class, a simple example: taking the probability of one being pregnant and the test kit indicates you are pregnant, against the probability of one being pregnant given the test kit indicates you are pregnant. This should be fairly intuitively.
Very simple question, just calculations. And knowing your general formulae well.
,![\frac{\text{P}(X = r + 1)}{\text{P}(X = r)} = \frac{\frac{n!}{[n-(r+1)]!(r+1)!} (p)^{r+1} (1-p)^{n-(r+1)}}{\frac{n!}{(n-r)!r!} (p)^r (1-p)^{n-r}}](http://s0.wp.com/latex.php?latex=%5Cfrac%7B%5Ctext%7BP%7D%28X+%3D+r+%2B+1%29%7D%7B%5Ctext%7BP%7D%28X+%3D+r%29%7D+%3D+%5Cfrac%7B%5Cfrac%7Bn%21%7D%7B%5Bn-%28r%2B1%29%5D%21%28r%2B1%29%21%7D+%28p%29%5E%7Br%2B1%7D+%281-p%29%5E%7Bn-%28r%2B1%29%7D%7D%7B%5Cfrac%7Bn%21%7D%7B%28n-r%29%21r%21%7D+%28p%29%5Er+%281-p%29%5E%7Bn-r%7D%7D&bg=ffffff&fg=000&s=0 "\frac{\text{P}(X = r + 1)}{\text{P}(X = r)} = \frac{\frac{n!}{[n-(r+1)]!(r+1)!} (p)^{r+1} (1-p)^{n-(r+1)}}{\frac{n!}{(n-r)!r!} (p)^r (1-p)^{n-r}}")
![\frac{\text{P}(X = r + 1)}{\text{P}(X = r)} = \frac{\frac{1}{[n-(r+1)]!(r+1)!} (p) (1-p)^{n-r-1}}{\frac{1}{(n-r)!r!} (1-p)^{n-r}}](http://s0.wp.com/latex.php?latex=%5Cfrac%7B%5Ctext%7BP%7D%28X+%3D+r+%2B+1%29%7D%7B%5Ctext%7BP%7D%28X+%3D+r%29%7D+%3D+%5Cfrac%7B%5Cfrac%7B1%7D%7B%5Bn-%28r%2B1%29%5D%21%28r%2B1%29%21%7D+%28p%29+%281-p%29%5E%7Bn-r-1%7D%7D%7B%5Cfrac%7B1%7D%7B%28n-r%29%21r%21%7D+%281-p%29%5E%7Bn-r%7D%7D&bg=ffffff&fg=000&s=0 "\frac{\text{P}(X = r + 1)}{\text{P}(X = r)} = \frac{\frac{1}{[n-(r+1)]!(r+1)!} (p) (1-p)^{n-r-1}}{\frac{1}{(n-r)!r!} (1-p)^{n-r}}")
![\frac{\text{P}(X = r + 1)}{\text{P}(X = r)} = \frac{\frac{1}{[n-(r+1)]!(r+1)!}}{\frac{1}{(n-r)!r!}} \times \frac{p(1-p)^{n-r-1}}{(1-p)^{n-r}}](http://s0.wp.com/latex.php?latex=%5Cfrac%7B%5Ctext%7BP%7D%28X+%3D+r+%2B+1%29%7D%7B%5Ctext%7BP%7D%28X+%3D+r%29%7D+%3D+%5Cfrac%7B%5Cfrac%7B1%7D%7B%5Bn-%28r%2B1%29%5D%21%28r%2B1%29%21%7D%7D%7B%5Cfrac%7B1%7D%7B%28n-r%29%21r%21%7D%7D++%5Ctimes+%5Cfrac%7Bp%281-p%29%5E%7Bn-r-1%7D%7D%7B%281-p%29%5E%7Bn-r%7D%7D&bg=ffffff&fg=000&s=0 "\frac{\text{P}(X = r + 1)}{\text{P}(X = r)} = \frac{\frac{1}{[n-(r+1)]!(r+1)!}}{\frac{1}{(n-r)!r!}} \times \frac{p(1-p)^{n-r-1}}{(1-p)^{n-r}}")
![\frac{\text{P}(X = r + 1)}{\text{P}(X = r)} = \frac{(n-r)!r!}{(n-r-1)!(r+1)!} \times [p(1-p)^{-1}]](http://s0.wp.com/latex.php?latex=%5Cfrac%7B%5Ctext%7BP%7D%28X+%3D+r+%2B+1%29%7D%7B%5Ctext%7BP%7D%28X+%3D+r%29%7D+%3D+%5Cfrac%7B%28n-r%29%21r%21%7D%7B%28n-r-1%29%21%28r%2B1%29%21%7D++%5Ctimes+%5Bp%281-p%29%5E%7B-1%7D%5D&bg=ffffff&fg=000&s=0 "\frac{\text{P}(X = r + 1)}{\text{P}(X = r)} = \frac{(n-r)!r!}{(n-r-1)!(r+1)!} \times [p(1-p)^{-1}]")
.
years by 
.
and 
. State the values of 
and 
.
and 
![x^2 - 9[(y-1)^2 - 1^2] = 18](http://s0.wp.com/latex.php?latex=x%5E2+-+9%5B%28y-1%29%5E2+-+1%5E2%5D+%3D+18&bg=ffffff&fg=000&s=0 "x^2 - 9[(y-1)^2 - 1^2] = 18")
, asymptotes are 
, and vertices 
and 
.
is a graph with asymptotes 
and 
.
![= 2 [2 \times 2 + 3 \times 5 + 4 \times 8 + ... + 21 \times 59]](http://s0.wp.com/latex.php?latex=%3D+2+%5B2+%5Ctimes+2+%2B+3+%5Ctimes+5+%2B+4+%5Ctimes+8+%2B+...+%2B+21+%5Ctimes+59%5D&bg=ffffff&fg=000&s=0 "= 2 [2 \times 2 + 3 \times 5 + 4 \times 8 + ... + 21 \times 59]")
![= 2 [(1+1) \times (3 \cdot 1 - 1) + (2+1) \times (3 \cdot 2 -1) + (3+1) \times (3 \cdot 3 -1) + ... + (20+1) \times (3 \cdot 20 -1) ]](http://s0.wp.com/latex.php?latex=%3D+2+%5B%281%2B1%29+%5Ctimes+%283+%5Ccdot+1+-+1%29+%2B+%282%2B1%29+%5Ctimes+%283+%5Ccdot+2+-1%29+%2B+%283%2B1%29+%5Ctimes+%283+%5Ccdot+3+-1%29++%2B+...+%2B+%2820%2B1%29+%5Ctimes+%283+%5Ccdot+20+-1%29+%5D&bg=ffffff&fg=000&s=0 "= 2 [(1+1) \times (3 \cdot 1 - 1) + (2+1) \times (3 \cdot 2 -1) + (3+1) \times (3 \cdot 3 -1) + ... + (20+1) \times (3 \cdot 20 -1) ]")
![= 2 [\frac{n}{2}(n+1)(2n+1) + n^2 ]](http://s0.wp.com/latex.php?latex=%3D+2+%5B%5Cfrac%7Bn%7D%7B2%7D%28n%2B1%29%282n%2B1%29+%2B+n%5E2+%5D&bg=ffffff&fg=000&s=0 "= 2 [\frac{n}{2}(n+1)(2n+1) + n^2 ]")
![+ \frac{1}{n-1} - \frac{1}{n+1}]](http://s0.wp.com/latex.php?latex=%2B+%5Cfrac%7B1%7D%7Bn-1%7D+-+%5Cfrac%7B1%7D%7Bn%2B1%7D%5D&bg=ffffff&fg=000&s=0 "+ \frac{1}{n-1} - \frac{1}{n+1}]")
![= \frac{1}{2} [1 + \frac{1}{2} - \frac{1}{n} - \frac{1}{n+1}]](http://s0.wp.com/latex.php?latex=%3D+%5Cfrac%7B1%7D%7B2%7D+%5B1+%2B+%5Cfrac%7B1%7D%7B2%7D+-+%5Cfrac%7B1%7D%7Bn%7D+-+%5Cfrac%7B1%7D%7Bn%2B1%7D%5D&bg=ffffff&fg=000&s=0 "= \frac{1}{2} [1 + \frac{1}{2} - \frac{1}{n} - \frac{1}{n+1}]")
, 
and 
, the sum of series tends to 
, a constant. Thus, series is convergent.
by 
. Then we have
![= \frac{3}{4} - \frac{2(n+1)+1}{2(n+1)[(n+1)+1]} - \frac{1}{3}](http://s0.wp.com/latex.php?latex=%3D+%5Cfrac%7B3%7D%7B4%7D+-+%5Cfrac%7B2%28n%2B1%29%2B1%7D%7B2%28n%2B1%29%5B%28n%2B1%29%2B1%5D%7D+-+%5Cfrac%7B1%7D%7B3%7D&bg=ffffff&fg=000&s=0 "= \frac{3}{4} - \frac{2(n+1)+1}{2(n+1)[(n+1)+1]} - \frac{1}{3}")
![\bigg[ \frac{\theta}{2} - \frac{\text{sin}2 \theta}{4} \bigg]_{\frac{\pi}{6}}^{\frac{\pi}{4}}](http://s0.wp.com/latex.php?latex=%5Cbigg%5B+%5Cfrac%7B%5Ctheta%7D%7B2%7D+-+%5Cfrac%7B%5Ctext%7Bsin%7D2+%5Ctheta%7D%7B4%7D+%5Cbigg%5D_%7B%5Cfrac%7B%5Cpi%7D%7B6%7D%7D%5E%7B%5Cfrac%7B%5Cpi%7D%7B4%7D%7D+&bg=ffffff&fg=000&s=0 "\bigg[ \frac{\theta}{2} - \frac{\text{sin}2 \theta}{4} \bigg]_{\frac{\pi}{6}}^{\frac{\pi}{4}}")
is minimum when 
.
onto 
.
![R_g = ( - \infty, 2) \subset ( - \infty, 3 ] = D_f](http://s0.wp.com/latex.php?latex=R_g+%3D+%28+-+%5Cinfty%2C+2%29+%5Csubset+%28+-+%5Cinfty%2C+3+%5D+%3D+D_f&bg=ffffff&fg=000&s=0 "R_g = ( - \infty, 2) \subset ( - \infty, 3 ] = D_f")
, fg exists.
cuts 
at more than one point, 
![\frac{n}{2} [2(15) + (n-1)(0.4)] = 560](http://s0.wp.com/latex.php?latex=%5Cfrac%7Bn%7D%7B2%7D+%5B2%2815%29+%2B+%28n-1%29%280.4%29%5D+%3D+560&bg=ffffff&fg=000&s=0 "\frac{n}{2} [2(15) + (n-1)(0.4)] = 560")
m
![= \big[ 8t^2 - 8 \text{ln}t - \frac{1}{2t^2} \big]_{0.5}^1](http://s0.wp.com/latex.php?latex=%3D+%5Cbig%5B+8t%5E2+-+8+%5Ctext%7Bln%7Dt+-+%5Cfrac%7B1%7D%7B2t%5E2%7D+%5Cbig%5D_%7B0.5%7D%5E1&bg=ffffff&fg=000&s=0 "= \big[ 8t^2 - 8 \text{ln}t - \frac{1}{2t^2} \big]_{0.5}^1")
![= 15 \pi - \pi \big[ \frac{x^3}{3} - 16x \big]_4^5](http://s0.wp.com/latex.php?latex=%3D+15+%5Cpi+-+%5Cpi+%5Cbig%5B+%5Cfrac%7Bx%5E3%7D%7B3%7D+-+16x+%5Cbig%5D_4%5E5&bg=ffffff&fg=000&s=0 "= 15 \pi - \pi \big[ \frac{x^3}{3} - 16x \big]_4^5")
and 
![\frac{dy}{dx} = - \frac{1}{2} \big[ \frac{(16 - 9x^2)^{-0.5}}{-0.5} \big] + C](http://s0.wp.com/latex.php?latex=%5Cfrac%7Bdy%7D%7Bdx%7D+%3D+-+%5Cfrac%7B1%7D%7B2%7D+%5Cbig%5B+%5Cfrac%7B%2816+-+9x%5E2%29%5E%7B-0.5%7D%7D%7B-0.5%7D+%5Cbig%5D+%2B+C&bg=ffffff&fg=000&s=0 "\frac{dy}{dx} = - \frac{1}{2} \big[ \frac{(16 - 9x^2)^{-0.5}}{-0.5} \big] + C")
![y = \frac{1}{3} \text{sin}^{-1} \big[ \frac{3x}{4} \big] + Cx + D](http://s0.wp.com/latex.php?latex=y+%3D+%5Cfrac%7B1%7D%7B3%7D+%5Ctext%7Bsin%7D%5E%7B-1%7D+%5Cbig%5B+%5Cfrac%7B3x%7D%7B4%7D+%5Cbig%5D+%2B+Cx+%2B+D&bg=ffffff&fg=000&s=0 "y = \frac{1}{3} \text{sin}^{-1} \big[ \frac{3x}{4} \big] + Cx + D")
![\big[ \frac{1}{-0.5} \big] \big[\frac{\sqrt{36 - 0.5m}}{0.5} \big] + C= 0.1 t](http://s0.wp.com/latex.php?latex=%5Cbig%5B+%5Cfrac%7B1%7D%7B-0.5%7D+%5Cbig%5D+%5Cbig%5B%5Cfrac%7B%5Csqrt%7B36+-+0.5m%7D%7D%7B0.5%7D+%5Cbig%5D+%2B+C%3D+0.1+t&bg=ffffff&fg=000&s=0 "\big[ \frac{1}{-0.5} \big] \big[\frac{\sqrt{36 - 0.5m}}{0.5} \big] + C= 0.1 t")
.
for some 
![\bigg[ \begin{pmatrix}{3}\\{5}\\{-2}\end{pmatrix} + \lambda \begin{pmatrix}{1}\\{1}\\{-1}\end{pmatrix} \bigg] \cdot \begin{pmatrix}{1}\\{1}\\{-1}\end{pmatrix} = 4](http://s0.wp.com/latex.php?latex=%5Cbigg%5B+%5Cbegin%7Bpmatrix%7D%7B3%7D%5C%5C%7B5%7D%5C%5C%7B-2%7D%5Cend%7Bpmatrix%7D+%2B+%5Clambda+%5Cbegin%7Bpmatrix%7D%7B1%7D%5C%5C%7B1%7D%5C%5C%7B-1%7D%5Cend%7Bpmatrix%7D+%5Cbigg%5D+%5Ccdot+%5Cbegin%7Bpmatrix%7D%7B1%7D%5C%5C%7B1%7D%5C%5C%7B-1%7D%5Cend%7Bpmatrix%7D+%3D+4&bg=ffffff&fg=000&s=0 "\bigg[ \begin{pmatrix}{3}\\{5}\\{-2}\end{pmatrix} + \lambda \begin{pmatrix}{1}\\{1}\\{-1}\end{pmatrix} \bigg] \cdot \begin{pmatrix}{1}\\{1}\\{-1}\end{pmatrix} = 4")
and 
or square both sides. Its easier to do the latter (so I will).
![\bigg[ \frac{2x-3}{x+5} \bigg]^2 \le \bigg[ \frac{2}{3} \bigg]^2](http://s0.wp.com/latex.php?latex=%5Cbigg%5B+%5Cfrac%7B2x-3%7D%7Bx%2B5%7D+%5Cbigg%5D%5E2+%5Cle+%5Cbigg%5B+%5Cfrac%7B2%7D%7B3%7D+%5Cbigg%5D%5E2&bg=ffffff&fg=000&s=0 "\bigg[ \frac{2x-3}{x+5} \bigg]^2 \le \bigg[ \frac{2}{3} \bigg]^2")
for all 
RHS.
.
.
— (1)
— (2)
— (3)
,
, the geometric series converges.
since 
m and 
m (to 3.s.f.).
, where 
is an arbitrary constant.
.
.
![R_{f^{-1}} = D_f = (-5, 2.5) \cup (2.5, 5]](http://s0.wp.com/latex.php?latex=R_%7Bf%5E%7B-1%7D%7D+%3D+D_f+%3D+%28-5%2C+2.5%29+%5Ccup+%282.5%2C+5%5D&bg=ffffff&fg=000&s=0 "R_{f^{-1}} = D_f = (-5, 2.5) \cup (2.5, 5]")
![R_g = (-0.5, 1) \subset D_f = (-5, 2.5) \cup (2.5, 5], ~fg(x)](http://s0.wp.com/latex.php?latex=R_g+%3D+%28-0.5%2C+1%29+%5Csubset+D_f+%3D+%28-5%2C+2.5%29+%5Ccup+%282.5%2C+5%5D%2C+%7Efg%28x%29&bg=ffffff&fg=000&s=0 "R_g = (-0.5, 1) \subset D_f = (-5, 2.5) \cup (2.5, 5], ~fg(x)")
exists.
![= \frac{1}{2} \big[ x^2 e^{x^ +1} \big]_0^n - \frac{1}{2} \int_0^n 2xe^{x^2 + 1} ~dx](http://s0.wp.com/latex.php?latex=%3D+%5Cfrac%7B1%7D%7B2%7D+%5Cbig%5B+x%5E2+e%5E%7Bx%5E+%2B1%7D+%5Cbig%5D_0%5En+-+%5Cfrac%7B1%7D%7B2%7D+%5Cint_0%5En+2xe%5E%7Bx%5E2+%2B+1%7D+%7Edx&bg=ffffff&fg=000&s=0 "= \frac{1}{2} \big[ x^2 e^{x^ +1} \big]_0^n - \frac{1}{2} \int_0^n 2xe^{x^2 + 1} ~dx")
or 
— (1)
— (2)
passes through 
.
.
or 
![\approx \frac{1}{ \sqrt{ 2 + \sqrt{2} \frac{1}{\sqrt{2}} [ 2 \theta - (1 - \frac{(2 \theta )^2}{2!}]} }](http://s0.wp.com/latex.php?latex=%5Capprox+%5Cfrac%7B1%7D%7B+%5Csqrt%7B+2+%2B+%5Csqrt%7B2%7D+%5Cfrac%7B1%7D%7B%5Csqrt%7B2%7D%7D+%5B+2+%5Ctheta+-+%281+-+%5Cfrac%7B%282+%5Ctheta+%29%5E2%7D%7B2%21%7D%5D%7D+%7D&bg=ffffff&fg=000&s=0 "\approx \frac{1}{ \sqrt{ 2 + \sqrt{2} \frac{1}{\sqrt{2}} [ 2 \theta - (1 - \frac{(2 \theta )^2}{2!}]} }")
parallel to the y – axis.
(4dp)
equation contains both x and y. Students need to understand that y coordinate can be obtain using 
. First derivative test suggests we add and minus a small number to x and sub it into 
. It is all x values defined on domain of C or all points on C.
isn’t really distance since distance is always positive. The more accurate definition is displacement. Then, it was a pure generalisation of our method to solve for the intersection between a line and a plane. Students need to know their stuffs to do this.
, read again. You probably end up with 2 marks then.
is a common ratio, but it is not. Common ratio must be a constant, and this is not. Its just a simple test, use it and show that is less than 1, you’re done. Like I said in previously, question 1 exists to prime students for Q9c. This question actually open up so many possibilities for future sequences and series questions, and I’m very excited.
should behave akin to 
, that is, a constant greater than 1. This is required in order to produce the same shape.