Mathematics Extension 1 Exam Predictions/Thoughts (1 Viewer)

[fwhite]

I think both k=5 and k=14 work depending on how you wrote it and if you made it 5^k * 2^(n-k) or 5^(n-k) * 2^k.

 

[fwhite]

Wait so k=5 is wrong?
if so would I lose 1 mark on that or more?

k=5 is right if you wrote it the right way

 

[Trebla]

I gave a worded solution for 14biii (the x1<a<x2 question). I am very sure it is valid, but i'm not sure the markers accept that sort of thing. Will I still get full marks if I proved it graphically? Will the marker even bother trying to interpret what I was saying?

Depends on what it is. If it’s just “hey just look at the graph therefore it is true” then no.

 

[luckystrike826]

hey lads what do we reckon the raw mark needs to be for a state rank?

 

[Boaard]

69ish

 

[Trebla]

How did you guys do the 14bi and iii?

For i) since the two graphs intersect only once then there is only one real solution where the y-values are equal -then rearrange to get the result.

For iii), I came up with this:
a^3 - ka^2 - 1 = 0
a^2(a - k) = 1
a = k + 1/a^2
Since a^2 is positive then a > k and it follows that k + 1/a^2 < k + 1/k^2

 

[IforgotMyPassword]

Predictions for e4 cut off this year?

 

[Trebla]

I think both k=5 and k=14 work depending on how you wrote it and if you made it 5^k * 2^(n-k) or 5^(n-k) * 2^k.

Not really, because the x is attached to the 5. The former is the coefficient of x^k and the latter is the coefficient of x^(n-k).

 

[fwhite]

Not really, because the x is attached to the 5. The former is the coefficient of x^k and the latter is the coefficient of x^(n-k).

So was it 5 or 14

 

[jeception177]

Not really, because the x is attached to the 5. The former is the coefficient of x^k and the latter is the coefficient of x^(n-k).

I had a similar q for trials and I did
Tk+1/Tk = 1
Then solved from there surely k=5 is fine from that

 

[akkjen]

For i) since the two graphs intersect only once then there is only one real solution where the y-values are equal -then rearrange to get the result.

For iii), I came up with this:
a^3 - ka^2 - 1 = 0
a^2(a - k) = 1
a = k + 1/a^2
Since a^2 is positive then a > k and it follows that k + 1/a^2 < k + 1/k^2

For iii)
Can you prove that it lies between a negative and positive sign
third part I showed f(x1) was negative and f(x2) was positive. a is in between.

 

[akkjen]

Do you lose marks for having a decimal value instead of exact value even when the question doesn't ask for exact value?

 

[luckystrike826]

For iii)
Can you prove that it lies between a negative and positive sign
third part I showed f(x1) was negative and f(x2) was positive. a is in between.

yep but i think it was important that you had to mention:
1. alpha is the only root, hence it its alpha thats between the region
2. the function is continuous

 

[Trebla]

I had a similar q for trials and I did
Tk+1/Tk = 1
Then solved from there surely k=5 is fine from that

The question specifies a particular term. Therefore the order matters and because the equation is linear, there can only be one value for k, not two.

If you did (5x)^k*2^(n-k) the solution you find for k identifies the coefficient of x^k.

If you did (5x)^(n-k)*2^k then the solution you have found for k identifies the cofficient of x^(n-k) (because you have defined k differently to the question) which is NOT the same as x^k.

For iii)
Can you prove that it lies between a negative and positive sign
third part I showed f(x1) was negative and f(x2) was positive. a is in between.

Technically this requires a bit more rigour because you’ve made some assumptions about f(x). If you have proved
f(x1) < f(a) < f(x2)
You can only conclude
x1 < a < x2
if f(x) is increasing in that domain. If f(x) was decreasing in that domain, the signs would reverse. If there are stationary points then you run into more complications (which prob doesn’t happen here). Without knowing exactly what you wrote maybe you could get away with it.

 

[akkjen]

The question identifies a particular term, not coefficient. Therefore the order matters and because the equation is linear, there can only be one value for k, not two.

If you did (5x)^k*2^(n-k) the solution you find for k identifies the coefficient of x^k.

If you did (5x)^(n-k)*2^k then the solution you have found for k identifies the cofficient of x^(n-k) (because you have defined k differently to the question) which is NOT the same as x^k.

Technically this requires a bit more rigour because you’ve made some assumptions about the inverse. If you have proved
f(x1) < f(a) < f(x2)
You can only conclude
x1 < a < x2
if f(x) is increasing in that domain. If f(x) was decreasing in that domain, the signs would reverse. If there are stationary points then you run into more complications (which prob doesn’t happen here).

do you think the markers are going to be that picky?

 

[luckystrike826]

The question specifies a particular term. Therefore the order matters and because the equation is linear, there can only be one value for k, not two.

If you did (5x)^k*2^(n-k) the solution you find for k identifies the coefficient of x^k.

If you did (5x)^(n-k)*2^k then the solution you have found for k identifies the cofficient of x^(n-k) (because you have defined k differently to the question) which is NOT the same as x^k.

Technically this requires a bit more rigour because you’ve made some assumptions about f(x). If you have proved
f(x1) < f(a) < f(x2)
You can only conclude
x1 < a < x2
if f(x) is increasing in that domain. If f(x) was decreasing in that domain, the signs would reverse. If there are stationary points then you run into more complications (which prob doesn’t happen here). Without knowing exactly what you wrote maybe you could get away with it.

hold a sec -- couldn't it be as simple as there must be a root between x1 and x2 given the change of signs and the fact that fx is continuous?

 

[jeception177]

The question specifies a particular term. Therefore the order matters and because the equation is linear, there can only be one value for k, not two.

If you did (5x)^k*2^(n-k) the solution you find for k identifies the coefficient of x^k.

If you did (5x)^(n-k)*2^k then the solution you have found for k identifies the cofficient of x^(n-k) (because you have defined k differently to the question) which is NOT the same as x^k.

Technically this requires a bit more rigour because you’ve made some assumptions about f(x). If you have proved
f(x1) < f(a) < f(x2)
You can only conclude
x1 < a < x2
if f(x) is increasing in that domain. If f(x) was decreasing in that domain, the signs would reverse. If there are stationary points then you run into more complications (which prob doesn’t happen here). Without knowing exactly what you wrote maybe you could get away with it.

Ahh I see would I be looking at 2/3 or because I defined it differently would I be getting no marks for that

 

[Trebla]

hold a sec -- couldn't it be as simple as there must be a root between x1 and x2 given the change of signs and the fact that fx is continuous?

The change of sign alone doesn’t tell you which way around the inequality is. You need to know the behaviour of the function in that domain to be able make that conclusion. You are slightly saved by the fact that x2 can be explictly shown to be greater than x1 in the particular example.

 

[greetings]

anyone else accidentally get an inverse tan integral for the u=(cosx)^2 by subbing u^2 lol

 

[gunnerfan99]

Ahh I see would I be looking at 2/3 or because I defined it differently would I be getting no marks for that

2/3 I did the same thing