Mathematics Question Thread

[jamonwindeyer]

Hi everyone!

I'm pretty new here, this is my first post! Hope I'm doing it properly haha.

I've been having trouble with these questions I have attached below from the 2017 CSSA Trial Paper. It provides solutions but I don't know how they got there!

In 13.a.i, I don't understand how they got the answer (I thought your were meant to use the product rule?)

In 14.a.ii, I don't understand where the π comes from for (5-π)/2

I'm sure there are other people out there that have been struggling with these questions because my maths class have been talking about them non-stop!

Hi Isabella! You are indeed doing it right, welcome!

- That first one is a trick - That first exponential isn't a function. It is just a number! Like, put \(e^\pi\) in your calculator, it has a value! So this is just a case like \(y=2x\) or \(y=\frac{5}{4}x\). It just so happens the number out front is an exponential. Just like those other cases, the derivative of a number out the front of \(x\) is just the number itself.

- In that second one, the angle they are talking about is the angle in that little right angled triangle. To get that, we need the answer we had from Part (i), the whole angle, the 5/2. Then we subtract 90 degrees (or \(\frac{\pi}{2}\)) to just leave the triangle.



I hope this helps

[SanaBanana]

hiya, so ihave a question.
say you do a math question and it has parts to it, like (i),(ii), and so on, and you figure out how to solve the latter parts before the former, and then you USE the answers from the latter parts to solve the former parts. can you be marked down or not get the marks for the former parts?
(do i make any sense at all??)

[RuiAce]

hiya, so ihave a question.
say you do a math question and it has parts to it, like (i),(ii), and so on, and you figure out how to solve the latter parts before the former, and then you USE the answers from the latter parts to solve the former parts. can you be marked down or not get the marks for the former parts?
(do i make any sense at all??)

Yes. Whilst you're always allowed to use the former parts to solve the latter, in general you cannot use the latter to solve the former.

The only exception is the oddball case where you actually solved both the former and the latter at the same time, possibly without realising it.

(If I were an examiner, I probably would've just said whatever. But questions are designed so that each part is supposed to somehow follow from the previous ones unless explicitly stated otherwise. So other examiners do have every right to penalise you for it.)

[billy.ohlmeyer]

my textbook has this question:
Given that the wingspan of an aeroplane is 30m, find the plane's altitude to the nearest metre if the wingspan subtends an angle of 14' when it is directly overhead.
how do i do this?

thanks

[isabella104]

Hi again everyone!

Thank you so much Jamon for your help with the previous questions I posted on this thread!

These multiple choice questions are the death of me...without the worked solutions I am lost!!! Does anyone know how to solve any of these questions I have attached below?

The answers are:
5. C
7. D
8. C
9. A

Thanks!

Also, just a question in terms of how to approach mathematics exams...I've overheard some people with tutors recommend finishing the 3 hr exam in 2 hrs so that there is 1 hr remaining to go through and double check your work. I tend to be very slow in maths exams, with 5-10 mins (if any) to go back through my work. Should I be concerned?? Or is it unnecessary to finish the exam 1 hour early?? I try to go slowly so that I don't make mistakes the first time round, and in the time remaining I tend to quickly go through the questions I couldn't complete previously because they were too hard. Any thoughts?

[isabella104]

hiya, so ihave a question.
say you do a math question and it has parts to it, like (i),(ii), and so on, and you figure out how to solve the latter parts before the former, and then you USE the answers from the latter parts to solve the former parts. can you be marked down or not get the marks for the former parts?
(do i make any sense at all??)

In my experience, if you get the wrong answer in (i) but use the correct processes in (ii), (iii) and so on, you can still get the marks for (ii), (iii), etc. because you understood the process. But in my experience this only tends to happen if you accidentally copy down the question wrong or something in (i). Hope this helps

[Never.Give.Up]

Hi again everyone!

Thank you so much Jamon for your help with the previous questions I posted on this thread!

These multiple choice questions are the death of me...without the worked solutions I am lost!!! Does anyone know how to solve any of these questions I have attached below?

The answers are:
5. C
7. D
8. C
9. A

Thanks!

Also, just a question in terms of how to approach mathematics exams...I've overheard some people with tutors recommend finishing the 3 hr exam in 2 hrs so that there is 1 hr remaining to go through and double check your work. I tend to be very slow in maths exams, with 5-10 mins (if any) to go back through my work. Should I be concerned?? Or is it unnecessary to finish the exam 1 hour early?? I try to go slowly so that I don't make mistakes the first time round, and in the time remaining I tend to quickly go through the questions I couldn't complete previously because they were too hard. Any thoughts?

Hey
For q 5
You can use null factor law between o and pi
so,
tanx-2=0 and 2sinx-1=0
therefore tanx= 2 between 0 and pi (gives one solution as tan is only positive in this domain in first quadrant)
and 2sinx=1, sinx=1/2 (gives 2 solutions as sin is positive in both 1st and 2nd quadrants which is in domain of 0 and pi)
therefore, there are 3 solutions.

for question 7
this is a geometric series where n= p-q+1, which is 15-3+1, n=13
t1= 2^3 =8
t2 = 2^4 = 16
t3 = 2^5 = 32 (and so on...)
from this a=8, r=2
therefore terms of geometric series --> a(r^n -1)
so, 8 (2^13 -1)
= 2^3 (2^13 -1)
= 2^16 -8 (with index law m^a (m^b) = m^a+b
therefore it is equal to d.

for question 8
you can find the area of the trapezium a= 1/2h(a+b) --> you must find area of trapezium above x-axis first so, 1/2(10)(2+5) = 35 (you need this later)
therefore because it is between -6 and k, we know it is the trapezium below the x-axis
so, a=1/2 (k-4) (4+6) h=k-a a=6 and b=4 (these are positive because they are a distance)
therefore a= 1/2 (k-4) (10)
=5(k-4)
=5k-20 =35 (because it is equal to zero you need to cancel them out by making them equal to each other)
5k= 55
k=11
therefore, C.

for question 9
I was also confused by this but i found if you sub in points for x, you can solve the equation (it must equal 0)

hope this makes sense

[isabella104]

Hey
For q 5
You can use null factor law between o and pi
so,
tanx-2=0 and 2sinx-1=0
therefore tanx= 2 between 0 and pi (gives one solution as tan is only positive in this domain in first quadrant)
and 2sinx=1, sinx=1/2 (gives 2 solutions as sin is positive in both 1st and 2nd quadrants which is in domain of 0 and pi)
therefore, there are 3 solutions.

for question 7
this is a geometric series where n= p-q+1, which is 15-3+1, n=13
t1= 2^3 =8
t2 = 2^4 = 16
t3 = 2^5 = 32 (and so on...)
from this a=8, r=2
therefore terms of geometric series --> a(r^n -1)
so, 8 (2^13 -1)
= 2^3 (2^13 -1)
= 2^16 -8 (with index law m^a (m^b) = m^a+b
therefore it is equal to d.

for question 8
you can find the area of the trapezium a= 1/2h(a+b) --> you must find area of trapezium above x-axis first so, 1/2(10)(2+5) = 35 (you need this later)
therefore because it is between -6 and k, we know it is the trapezium below the x-axis
so, a=1/2 (k-4) (4+6) h=k-a a=6 and b=4 (these are positive because they are a distance)
therefore a= 1/2 (k-4) (10)
=5(k-4)
=5k-20 =35 (because it is equal to zero you need to cancel them out by making them equal to each other)
5k= 55
k=11
therefore, C.

for question 9
I was also confused by this but i found if you sub in points for x, you can solve the equation (it must equal 0)

hope this makes sense

Thank you so much! You are a legend!

I'm still a bit confused on Q7,

where you wrote:
therefore terms of geometric series --> a(r^n -1)
so, 8 (2^13 -1)
= 2^3 (2^13 -1)
= 2^16 -8 (with index law m^a (m^b) = m^a+b
therefore it is equal to d.

I don't quite understand the those three lines. I think I'm misreading it, I keep interpreting it as:
8 x (2^[13-1])
=2^3 x 2^12
=2^15  ?

I don't understand where the 2^16 and the -8 comes from?? I thought they were all positive???

If you could clarify that would be great, if not, you were a MASSIVE help anyway. Thank you!!!

[fun_jirachi]

Hi again everyone!

Thank you so much Jamon for your help with the previous questions I posted on this thread!

These multiple choice questions are the death of me...without the worked solutions I am lost!!! Does anyone know how to solve any of these questions I have attached below?

The answers are:
5. C
7. D
8. C
9. A

Thanks!

Also, just a question in terms of how to approach mathematics exams...I've overheard some people with tutors recommend finishing the 3 hr exam in 2 hrs so that there is 1 hr remaining to go through and double check your work. I tend to be very slow in maths exams, with 5-10 mins (if any) to go back through my work. Should I be concerned?? Or is it unnecessary to finish the exam 1 hour early?? I try to go slowly so that I don't make mistakes the first time round, and in the time remaining I tend to quickly go through the questions I couldn't complete previously because they were too hard. Any thoughts?

so for your q9,
i guess maybe its expected for 2u that you know your basic trigonometric functions as graphed on a cartesian plane, but anyway
the regular y = tan x tends to positive infinity at pi/2 and negative infinity at -pi/2, and this continues at intervals of pi.
From knowing our tan graph, we know the answer isnt b
seeing that it cuts the x-axis at pi/4 also rules out C
now we notice that the asymptotes are in increments of pi/2, not pi, so we go to A, which is the only option left with a transformation that allows for this change -- you could look inside the brackets and simplify it down from y = tan (2x-pi/2) to y = tan (2(x-pi/4))
 Hope this helps

just gonna edit in an explanation for q7 so you fully understand
the expression is essentially 2^3+2^4+2^5+...+2^14+2^15
from the sum of a geometric series we get the sum equal to (8(2^13-1))/(2-1) from a(r^n-1)/(r-1)
because there is a 1 on the denominator, the sum is equal to 8(2^13-1) = 2^3(2^13-1)=2^16-1 as previously stated
i think you're getting confused because you're inserting brackets from nowhere; you're combining the indice 13 and the number 1 into a single expression, like so
8(2^13-1) --> 8(2^(13-1))
which is wrong because BIMA
the 2^16 and the -8 come from just expanding the brackets and using indice laws
im gonna insert a paint picture just in case it helps, the not thing is what you're doing and that's how you're getting the 2^15, while the correct answer is derived from the thing on the left
Hope this helps

[isabella104]

so for your q9,
i guess maybe its expected for 2u that you know your basic trigonometric functions as graphed on a cartesian plane, but anyway
the regular y = tan x tends to positive infinity at pi/2 and negative infinity at -pi/2, and this continues at intervals of pi.
From knowing our tan graph, we know the answer isnt b
seeing that it cuts the x-axis at pi/4 also rules out C
now we notice that the asymptotes are in increments of pi/2, not pi, so we go to A, which is the only option left with a transformation that allows for this change -- you could look inside the brackets and simplify it down from y = tan (2x-pi/2) to y = tan (2(x-pi/4))
 Hope this helps

just gonna edit in an explanation for q7 so you fully understand
the expression is essentially 2^3+2^4+2^5+...+2^14+2^15
from the sum of a geometric series we get the sum equal to (8(2^13-1))/(2-1) from a(r^n-1)/(r-1)
because there is a 1 on the denominator, the sum is equal to 8(2^13-1) = 2^3(2^13-1)=2^16-1 as previously stated
i think you're getting confused because you're inserting brackets from nowhere; you're combining the indice 13 and the number 1 into a single expression, like so
8(2^13-1) --> 8(2^(13-1))
which is wrong because BIMA
the 2^16 and the -8 come from just expanding the brackets and using indice laws
im gonna insert a paint picture just in case it helps, the not thing is what you're doing and that's how you're getting the 2^15, while the correct answer is derived from the thing on the left
Hope this helps

Aaaaaaaaah, got it! Thanks so much!!!!!

[amelia20181]

how do you do this question

The combination lock on a safe has three concentric circular discs, each showing the digits 0 to 9. Only one combination of digits will open the safe. what is the probability of opening the safe at my first attempt if i do not know the combination?

[fun_jirachi]

how do you do this question

The combination lock on a safe has three concentric circular discs, each showing the digits 0 to 9. Only one combination of digits will open the safe. what is the probability of opening the safe at my first attempt if i do not know the combination?

I'm not too sure on this one, but I think this is how it works;
Because there are three concentric circular discs each with digits 0 to 9, you're essentially looking at the chance of guessing correctly a number for 000 to 999, in which case the answer is 1 in 1000.

[jamonwindeyer]

I'm not too sure on this one, but I think this is how it works;
Because there are three concentric circular discs each with digits 0 to 9, you're essentially looking at the chance of guessing correctly a number for 000 to 999, in which case the answer is 1 in 1000.

Perfect example of breaking down a wordy problem!! Love your work

[darlo69]

Is it too late to drop to General? Trials is in less than a week and I can't even do the questions from the year 11 course ahhhh

[jamonwindeyer]

Is it too late to drop to General? Trials is in less than a week and I can't even do the questions from the year 11 course ahhhh

You'd be relearning an entire course in three months! Definitely not a good call - Stick with it, pick a topic at a time and work really hard to understand it. I promise it will get easier! You just need to be willing to really invest a heap of time to make it happen