Past Paper Questions for 3u (1 Viewer)

frankfurt · Oct 10, 2020

Hi,

Could you please help me out with these past paper questions?

The answer is C but the answer i get is 362304

There was actually no answer for this one in the answer key for this one. i think they might've skipped it

thanks in advance

)

devtrivedi · Oct 10, 2020

Yeh the answer is C
To find the arrangement, do the total number of ways (9!) - the number of arrangements where no boys are together (4! x 5!)

devtrivedi · Oct 10, 2020

and for Part (e) f(g(x))=x for every function where g(x) is the inverse function of f(x)

YonOra · Oct 11, 2020

Why isn't it 2! (the two different ways you can seat the students) x 8! (the remaining seats and how you can seat them). Admittedly, i've half forgotten this stuff and have only done Yr11 content with it, and its a bit sus that im' not using the (n-1)! formula. Can anyone explain?

Pikapizza · Oct 11, 2020

YonOra said:
Why isn't it 2! (the two different ways you can seat the students) x 8! (the remaining seats and how you can seat them). Admittedly, i've half forgotten this stuff and have only done Yr11 content with it, and its a bit sus that im' not using the (n-1)! formula. Can anyone explain?

The question states how many ways you can do if at least two students sit together so either you take cases (if 2 students sit together, if 3 sit together, if 4..., if 5... and so on)
OR you do as @devtrivedi did which is much quicker, you do: total - arrangements of the boys all separated (do insertion method like above)

Edit: there would be many more ways to do these questions but when you see an “at least” it is most common for it to be either cases or total-unfavourable method

CM_Tutor · Oct 11, 2020

YonOra said:
Why isn't it 2! (the two different ways you can seat the students) x 8! (the remaining seats and how you can seat them). Admittedly, i've half forgotten this stuff and have only done Yr11 content with it, and its a bit sus that im' not using the (n-1)! formula. Can anyone explain?

Had the question been to seat the 10 people around a circular table with two specific people seated together, the solution would match your suggestion:

Seat the first of the two specific people in 1 way (as all positions at an empty circular table are identical under rotation)
Seat the second of the two specific people in 2 ways (in one of the two seats adjacent to the first specific person)
There are 8 people remaining and 8 seats, so there are 8! arrangements.
Thus, total number of arrangements is 2 x 8! = 80,640

However, the question has two groups of 5 people (students and parents) and we need at least two students together. We could do this by counting up the cases:

with exactly two students together (and the other three students not together with the first two, but possibly together with each other)
then, with exactly three students together (and the other two together or separate but not together with the first three)
then, with four students together
then, with all five students together

but that is long and involved.

We are better off looking to the complementary event, which is the single case with no students together. This can only be done if the seating alternates parent - student - parent - student etc.

Total number of arrangements of 10 people at a circular table is 9! = 362,880
Seating alternately, we seat a student first in 1 way, then the other four students have only four possible seats, so seat them in 4! ways. The remaining five seats go to the five parents in 5! ways
So, seating alternatively can be done in 1 x 4! x 5! = 2,880 ways
So, at least two students together has 9! - 4! x 5! = 360,000

Intuitively, the likelihood of there being at least one pair of students together if seated randomly is quite high as the only way that it doesn't happen is if the result is an alternate seating pattern - and alternate seating is unlikely. Two students together is also less likely if it is a specific pair who must be together. Looking at the above answers, we have:

$\begin{align*} P(\text{alternating -P-S-P-S- seating}) &= \frac{2880}{362880} = \frac{1}{126} \\ \\ P(\text{two specific students together}) &= \frac{80640}{362880} = \frac{2}{9} \\ \\ P(\text{at least two students together}) &= \frac{360000}{362880} = \frac{125}{126} \end{align*}$

This fits the pattern that is expected, and shows that the specific two together interpretation that you are thinking about has a probability of around 22% compared to over 99% for at least two of the students, not specifying which, are together.

CM_Tutor · Oct 11, 2020

Pikapizza said:
The question states how many ways you can do if at least two students sit together so either you take cases (if 2 students sit together, if 3 sit together, if 4..., if 5... and so on)
OR you do as @devtrivedi did which is much quicker, you do: total - arrangements of the boys all separated (do insertion method like above)

Edit: there would be many more ways to do these questions but when you see an “at least” it is most common for it to be either cases or total-unfavourable method

Pikapizza is correct that there is a difference between exactly and at least two students. However, the approach I addressed above leading to 2 x 8! requires the two students in question to be specified. The number of arrangements with the two together being any pair of students would be different again.

YonOra · Oct 11, 2020

CM_Tutor said:
Had the question been to seat the 10 people around a circular table with two specific people seated together, the solution would match your suggestion:

Seat the first of the two specific people in 1 way (as all positions at an empty circular table are identical under rotation)

Seat the second of the two specific people in 2 ways (in one of the two seats adjacent to the first specific person)

There are 8 people remaining and 8 seats, so there are 8! arrangements.

Thus, total number of arrangements is 2 x 8! = 80,640

However, the question has two groups of 5 people (students and parents) and we need at least two students together. We could do this by counting up the cases:

with exactly two students together (and the other three students not together with the first two, but possibly together with each other)

then, with exactly three students together (and the other two together or separate but not together with the first three)

then, with four students together

then, with all five students together

but that is long and involved.

We are better off looking to the complementary event, which is the single case with no students together. This can only be done if the seating alternates parent - student - parent - student etc.

Total number of arrangements of 10 people at a circular table is 9! = 362,880

Seating alternately, we seat a student first in 1 way, then the other four students have only four possible seats, so seat them in 4! ways. The remaining five seats go to the five parents in 5! ways

So, seating alternatively can be done in 1 x 4! x 5! = 2,880 ways

So, at least two students together has 9! - 4! x 5! = 360,000

Intuitively, the likelihood of there being at least one pair of students together if seated randomly is quite high as the only way that it doesn't happen is if the result is an alternate seating pattern - and alternate seating is unlikely. Two students together is also less likely if it is a specific pair who must be together. Looking at the above answers, we have:

$\begin{align*} P(\text{alternating -P-S-P-S- seating}) &= \frac{2880}{362880} = \frac{1}{126} \\ \\ P(\text{two specific students together}) &= \frac{80640}{362880} = \frac{2}{9} \\ \\ P(\text{at least two students together}) &= \frac{360000}{362880} = \frac{125}{126} \end{align*}$

This fits the pattern that is expected, and shows that the specific two together interpretation that you are thinking about has a probability of around 22% compared to over 99% for at least two of the students, not specifying which, are together.

Thank you, this was really helpful!

frankfurt · Oct 11, 2020

devtrivedi said:
and for Part (e) f(g(x))=x for every function where g(x) is the inverse function of f(x)

so then is the answer just 3?

shashysha · Oct 11, 2020

frankfurt said:
so then is the answer just 3?

not too sure but I think you would differentiate both sides of f(g(x)) = x so you get f'(g(x)).g'(x)=1 then rearrange to find g'(x)

CM_Tutor · Oct 11, 2020

frankfurt said:
so then is the answer just 3?

No.

Since $g(x)$ is the inverse of $f(x)$ , $f(g(x)) = g(f(x)) = x$ so long as $x$ is within the appropriate domain. It follows, so long as $x=3$ is in the domain, that $f(g(3)) = g(f(3)) = 3$ , but we seek to find $g'(3)$ .

First, checking the domain, in this case we have

$\begin{align*} f(x) &= x + e^{5x} + 2 \\ f'(x) &= 1 + 5e^{5x} > 0 \text{ for all } x \end{align*}$

and so $f(x)$ increases throughout its domain, which is $x\in\mathbb{R}$ - and so can be inverted for all $x$ .

The question directs that you consider the derivative of $f(g(x))$ . Since $x=3$ is within the domain of inversion, we know that $f(g(x))=x$ . Our goal, remember, is the value of $g'(3)$ :

$\begin{align*} f(g(x)) &= x \\ \text{On differentiating, } \frac{d}{dx} f(g(x)) &= \frac{d}{dx} x \\ \text{Applying the Chain Rule, } f'(g(x)) \times g'(x) &= 1 \\ g'(x) &= \frac{1}{f'(g(x))} \\ \text{Thus, at $x=3$: } g'(3) &= \frac{1}{f'(g(3))} \end{align*}$

So, we need the value of $f'(g(3))$ , and given $f'(x) = 1 + 5e^{5x}$ , we need the value of $g(3)$ .

$\begin{align*} \text{Let } x &= g(3) = f^{-1}(3) \\ f(x) &= 3 \\ x + e^{5x} + 2 &= 3 \\ x + e^{5x} &= 1 \\ \text{By observation, the solution is } x &= 0 \end{align*}$

$\text{Thus, } f'(g(3)) = f'(0) = 1 + 5e^0 = 6$

$\text{Thus, } g'(3) = \frac{1}{f'(g(3))} = \frac{1}{{6}$

Past Paper Questions for 3u (1 Viewer)

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