1st August Discussion Questions

Q4iii: Functions

$R_{f^{-1}} = D_f$ is $(-\infty, \infty)$, isn’t it (we combine the domains of the two “pieces” to get t he overall domain)

Q8iii: Differentiation

The main theoretical idea (which came out in the A levels last year), is the link between the gradient and angle of a line.

Note that the gradient is “rise over run”. For example, the line $y=4x$ has gradient 4 because if $x$ increases by 1, $y$ increases by 4.

Meanwhile, if we let $\theta$ be the angle between the $x$-axis and the line, then $\tan \theta$ is “opposite over adjacent” which is the same as “rise over run”. This gives us an important observation:

$$\tan \theta = \textrm{gradient of line}$$

The details to show the angle is twice is a bit tricky, but can you first find the gradients of $PM$ and $PN$?

Q9: Integration

(a)(ii) Area

Area of $R$ will be given by $\int y \, \mathrm{d}x$, which for our question becomes $\displaystyle \int_0^2 \frac{\sqrt{16-x^2}}{4} \, \mathrm{d}x$. The trick to integrating $\sqrt{16-x^2}$ exactly comes form part (a)(i): can you see the link?

(a)(iii) Volume

What is the problem you encountered here? Thought this will be a relatively straightforward application of the volume formula. Send me your working?

(b): Volume

The trick to handling this question is to observe that the region can be split up into two parts based on the intersection. The “green” part only touches the ellipse, so we have $\displaystyle \pi \int_a^1 x^2_{\textrm{ellipse}} \, \mathrm{d}y$ for the volume, where $a$ is the intersection point.

Meanwhile, the “pink” part only touches $y=x^2$, so we have $\displaystyle \pi \int_0^a x^2_{\textrm{quadratic}} \, \mathrm{d}y$ for the volume.

Note that since there was no mention of exact/no GC, we can use our GC to get $a$ and perform the integration.

Q10: AP/GP

(iii)

What’s the problem with this? I believe this should be application of the AP $S_n$ formula. To get the answer of the form $0.005nk$, we likely have to drag out a common factor of $0.01k$ from our brackets. Write out what you can and send it to me and I will guide you from there.

(iv), (v)

I believe these two parts make use of the part (i)(a) and (iii) answers: we’d equate the two with $n=240$ (20 years) to find $k$, and then use the value of $k$ to find $n$ when the value is $150000.

Q11: Vectors

(ii)

The key observation here is that we are given the equations of planes $EFIH$ (part (i)) and $DGIH$. If you look closely at the two planes (their names), they contain $IH$. So the intersection between the two planes is our line $HI$.

(iv)

With the equation of $HI$ from (ii), are you able to see that this question is about the perpendicular distance from point to line? We can use the $|\mathbf{a}\times \hat{\mathbf{b}}|$ formula modified to this question:
$\left|\frac{\overrightarrow{JI}\times\overrightarrow{HI}}{|\overrightarrow{HI}|}\right|$.

Q2: Maclaurin/Binomial

Trigo gives us $\tan \left(\frac{\pi}{4} \right) = \frac{h}{QT}$. From here we will have to make use of the $\tan(A \pm B)$ formula in MF26 and the less commonly used fact of small angle approximation that $\tan x \approx x$. These two hints should give us the first show for $QT$.

For $PQ^2$ we’d have to find $PT$ through $\tan \frac{\pi}{6}$ and then use Pythagoras Theorem. Along the way we will have to use the binomial expansion to handle expanding negative powers. Give it a try and send me what you have and I’d guide you from there.

Q3(iv)

Parts (ii)-(iv) of this question involve pattern finding which is quite rare. It’s fine to skip this question actually. If you’re interested in the answer, then we should observe that the coefficients of odd powers form a GP with first term $-a$ and common ratio $a^2$. So we can apply the $S_\infty$ formula to them.

Q5: AP/GP

(iii), (iv)

I’d go out of order to talk about parts (iii) and (iv) first.

Very important exam technique: make use of results from “show” parts even if we couldn’t do them

So to tackle part (iii) we use the result in part (ii): we want $160-120(0.8)^n > 180$ for overflowing.

Then part (iv) is to consider if $n \to \infty$.

(ii)

The trick to this question is to identify that it is actually a “bank compound interest” type question in disguise! Two things are happening: we fill water every day (similar to deposit money every month) and then a percentage is lost at the end of each day (similar to earning interest).

In solving part (i) you were on the way to find the pattern.
For $n=1$, the volume of water is $80(0.8)$.
For $n=2$, the volume of water is $40(0.8)+80(0.8)^2$.
We will need to investigate the pattern for $n=3, 4, 5$ etc (this pattern is slightly trickier than usual so write out more: cannot be lazy here) to see if we spot anything. Give it a try!

Q6: AP/GP

Consider the AP $$a, a+d, a+2d, a+3d, a+4d, a+5d, \ldots$$

The even-numbered terms (2nd, 4th, 6th, etc) are $a+d, a+3d, a+5d$ so they form a new AP with first term $a+d$ and common difference $2d$. Let’s use that to form the first equation involving 408480.

The next part of the question (1st, 9th, 21st) form a GP is pretty standard: are you able to form the appropriate equations and find $r$ (which will help us find $d$ in terms of $a$) from there?

Q8

What’s the problem with this question? The trick is understand that the line $a_{n+1} = a_n + 0.15$ means it is an AP (can understand why?).

And for the second half, $\frac{b_m}{b_{m-1}} = 0.98$ means it is a GP. Thereafter it’s about using the AP and GP formulas.

Q9

What’s the problem with this?