My solutions to the 2019 Mathematics Extension 2 Paper (1 Viewer)

[sharky564]

Of course, I can make mistakes, so any errors found would be great.

EDIT: Newest version now a file on google drive as my last edit caused it to become too large: https://drive.google.com/open?id=1U_buMWY-R6i_ti3HunPg9TbSj_o74zFR

 

[sharky564]

Hope you don't mind but I repackaged your solutions into a single pdf.

It's a bit easier to download that way.

Thanks! I've replaced the images with the pdf.

 

[psmao]

Of course, I can make mistakes, so any errors found would be great.

Isn’t question 10 B since it’s (4c2+4c3)(10C1*9C1)

 

[Jimmy9945]

Isn’t question 10 B since it’s (4c2+4c3)(10C1*9C1)

(10C1*9C1/2)(4C2+2*4C3)

 

[sharky564]

Isn’t question 10 B since it’s (4c2+4c3)(10C1*9C1)

That counts codes like 2211 twice?

 

[psmao]

(10C1*9C1/2)(4C2+2*4C3)

why would you divide by 2? Isn’t 4455 differ to 5544

 

[quickoats]

Isn’t question 10 B since it’s (4c2+4c3)(10C1*9C1)

I got A for question 10 (10C2)(4!/3! + 4!/2!2! + 4!/3!)

 

[psmao]

That counts codes like 2211 twice?

Isn’t it counting 2211 and then 1122?
As u r choosing 2 spots, and then fit in one spot with a digit, fit in another with another digit

 

[Nash__]

after looking at these solutions, what do you think 93 will align to?

 

[Jimmy9945]

Well how I did it was 10C2(2^4-2).
The 10C2 to get the two numbers, and the 2^4 to choose either of the 2 numbers in each of the 4 spots. The -2 us to exclude the combinations where they are all the same number, eg 2222

 

[sharky564]

Isn’t it counting 2211 and then 1122?
As u r choosing 2 spots, and then fit in one spot with a digit, fit in another with another digit

No, what you're saying is you're either choosing any two of the 4 possible spots for one digit (e.g. you're choosing the last two digits to be 1 if your initial 2 digits are 2, 1) and iterating over all possible ordered pairs of numbers, meaning you also count the time you choose the first two digits to be 2, which is identical to the previous case. As a result, you double count these numbers, resulting in too many numbers. I have verified with a Python program and it has also given 630 as the answer:

count = 0
numbers = [str(a) + str(b) + str(c) + str(d) for a in range(0, 10) for b in range(0, 10) for c in range(0, 10) for d in range(0, 10)]
for x in numbers:
    if len(set(list(x))) == 2:
        count += 1

print(count)

 

[Jimmy9945]

after looking at these solutions, what do you think 93 will align to?

Probs 97

 

[psmao]

Well how I did it was 10C2(2^4-2).
The 10C2 to get the two numbers, and the 2^4 to choose either of the 2 numbers in each of the 4 spots. The -2 us to exclude the combinations where they are all the same number, eg 2222

Wouldn’t u need 10P2 coz these two numbers can swap their positions

 

[sharky564]

Wouldn’t u need 10P2 coz these two numbers can swap their positions

No because that swapping is taken into account in 2^4 - 2

 

[psmao]

What would 96 align to
I think I fked up my state rank holly sht

 

[blyatman]

Guys, its over. Go out and enjoy yourselves lol, or study for your next exam.

 

[RayLn]

answers all look right to me, but there is a stationary point at x=0 for 12d)ii

 

[Jimmy.Chen]

answers all look right to me, but there is a stationary point at x=0 for 12d)ii

The question said without calculus so I don't think stationary point is important.

 

[sharky564]

answers all look right to me, but there is a stationary point at x=0 for 12d)ii

Oh right, that makes sense. I'll edit that.

 

[Minecraft]

The question said without calculus so I don't think stationary point is important.

You don't need calculus to work that out. But I didn't notice that as well