# 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?

# 2016 Discussion Questions

## Q3

Hopefully you were able to spot that this question is about **transformations**. So let's see how we can use the ideas we learn there to tackle this weird question

The turning point on $y=x^4$ corresponds to the point with coordinates $(a,b)$ on $y=f(x)$

The turning point on $y=x^4$ is the origin $(0,0)$. So what two transformations will we need to do to get to $(a,b)$? Starting from $y=x^4$, what will the resultant equation be?

## Answer to this part

After translations, we have $y = (x-a)^2 + b$. That means that $l = a$ and $m=b$ by comparison with $f(x) = k(x-l)^4 + m$.

After that, we still have the $k$ in our equation and the point $(0, c)$. We will substitute that into $f(x) = k(x-l)^4 + m$ to find $k$.

The second half of the question is the $\displaystyle \frac{1}{f(x)}$ transformation. We first draw the $y=f(x)$ graph: it's a "quadratic"-like graph (because the original graph $y=x^4$ looks like $y=x^2$ that has a minimum point at $(a,b)$ and $y$-intercept $(0, c)$. Now we apply the $\displaystyle \frac{1}{f(x)}$ rules.

*Analysis*: the second half of the question is something we want to be able to attempt even if the first part is tricky.

## Q10

Part (ii) the show part is super tedious but we want to be able to carry out with practice: $$\begin{gather} f(1+\sqrt{x}) = x \\ 1 + \sqrt{1 + \sqrt{x}} = x \tag{1} \\ \sqrt{1 + \sqrt{x}} = x - 1 \\ 1 + \sqrt{x}=x^2 - 2x + 1 \\ \sqrt{x} = x^2 - 2x \\ x = x^4 - 4x^3 + 4x^2 \\ x^3 - 4x^2 + 4x - 1 = 0 \end{gather}$$

After pressing our GC to solve the equation we will have 3 answers but the question seem to suggest only 1. We reject $x=0.381966$ because it's not in the domain of $f^{-1}$. Meanwhile $x=1$ is not valid in equation (1) above (sub it in and see LHS vs RHS). So $x=2.62$ is the only acceptable answer.

To explain the last part, $$\begin{gather} ff(x) = x \\ f^{-1}ff(x) = f^{-1}(x) \\ f(x) = f^{-1}(x) \end{gather}$$

Reason for first to second line is we apply $f^{-1}$ on both sides. Reason for second to third line is because $f^{-1}f (x) = x$.

### Part b

For part (i) we want to be able to try using unfamiliar math notation they introduce to us. For $g(4)$, notice that 4 is even so we will use the second line in the $g(n)$ piecewise definition. So $g(4) = 2 + g(2)$. 2 is even again in $g(2)$ so we apply the second line $g(2) = 2 + g(1)$. Now 1 is odd so we use the third line $g(1) = 1 + g(0)$. and Finally 0 is the first line, $g(0) = 1$.

Putting it all together, $g(0) = 1$, $g(1) = 2$, $g(2) = 4$ and $g(4) = 6$.

Think you'd be able to try out for $g(7)$ and $g(12)$?

I think I'd discuss (ii) with you since it's quite weird.

## Question 11

Do you recall an idea about $\mathbf{r}\cdot \hat{\mathbf{n}}$ for equation of planes?

Let's take a look at the plane (a separate example) $$\mathbf{r} \cdot \begin{pmatrix} -2 \\ 1 \\ 2 \end{pmatrix}= 5$$

The number 5 on the RHS has not much meaning except that points on the plane have to satisfy the equation to get 5. But let's transform this equation to the $\mathbf{r}\cdot \hat{\mathbf{n}}$ form. Since $| \mathbf{n} | = 3$, we divide our equation of the plane by 3 on both sides to get $$\mathbf{r} \cdot \frac{\begin{pmatrix} -2 \\ 1 \\ 2 \end{pmatrix}}{3}= \frac{5}{3}$$

Now $\frac{5}{3}$ has a meaning: it is the distance between the plane and the origin is $\frac{5}{3}$.

If a plane has equation $\mathbf{r}\cdot \hat{\mathbf{n}} = k$, then $k$ is the perpendicular distance between the plane and the origin.

In your earlier working you should have found an equation of $p$ similar to $\mathbf{r} \cdot \begin{pmatrix} -2 \\ 1 \\ 2 \end{pmatrix}= -1$. Converting to the form we discussed we get $\mathbf{r} \cdot \hat{\mathbf{n}}= -\frac{1}{3}$, which means that $p$ is $-\frac{1}{3}$ units away from the origin.

To get to the planes the question one, we have to travel 12 units away from $p$. If we travel 12 units negative, then the new plane is $-12\frac{1}{3}$ units away from the origin. If we travel 12 units positive, then the new plane is $11\frac{2}{3}$ units away from the origin.

Thus we have $$\mathbf{r} \cdot \frac{\begin{pmatrix} -2 \\ 1 \\ 2 \end{pmatrix}}{3}= -\frac{37}{3}$$ as the equation of one of the planes. Converting to cartesian gives us $-2x + y + 2z = -37$. We can then do the same for the other plane.

## Question 7

Things to take note: the difference between population variance $\sigma^2$, the unbiased estimate of population variance $s^2$ and the sample variance where we have to use $s^2 = \frac{n}{n-1} (\textrm{sample variance})$.

Did you spot that 6.82 is the sample variance?

Then were you able to distinguish between 108 and 110? Which one is $\mu$ vs which one is $\overline{x}$?

## Question 8 last part

This is the case of "template C" in my summary notes with $\sigma$ unknown.

This question is particular tricky also because they left it open-ended: instead of telling you the conclusion they want, they just asked how it will be affected. So we just choose whatever conclusion we want (e.g. reject $H_0$ and find out the inequality/range for $\sigma$.

Able to give that a try?

## Question 9 first part

### Manual way

We can find $\sum x$ by taking $78\times 4 + 79 \times 5 + \cdots + 82 \times 7$.

We can find $\sum x^2$ by taking $78^2 \times 4 + 79^2 \times 5 + \cdots + 82^2 \times 7$.

Thereafter we can apply our usual formula

### GC way

Key $78, \ldots 82$ into $L_1$ in the list `stat`

->`edit...`

in your GC
and $4, \ldots 7$ into $L_2$.

Then go to `stat`

->`calc`

->`1-var stat`

and key in $L_1$ as `xList`

and $L_2$ as `freqList`

. Thereafter we can get $\overline{x}$ and $s$ (your calculator shows this as $s_x$. (note: we do not take $\sigma_x$ on this page for our syllabus$.

For the second part we can just do our usual keying in of $Z$-test, but we could also use the `data`

option.

## Question 10 second part

### Combining the two samples

To combine the two samples, we can add up $\sum x$ and $\sum y$ to get the new "$\sum x$". Similarly for $\sum x^2$. Then we can proceed to find our usual $\overline{x}$ and $s^2$ with the new combined sample with $n=120$.

# Plan for 25th July

We'd go through those questions you sent to me (I've typed out comments below).

Thereafter I think we work on past revision first (you mention planning to start on the 2016 paper right?).

For the next and last topic linear regression and correlation I plan to only just go through the rough idea/big picture and leave the details to next week (usually can cover everything in 1-1.5 lessons).

Similar to hypothesis testing this last topic is quite template-y with a lot of keying GC, just that while hypothesis has a few moderately lengthy templates, we now have a lot of short templates.