3U Maths Question Thread

[RuiAce]

Help on e&f please! 

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[Dragomistress]

When a question asks you to draw a circle, do you have to draw it when it is already provided in the question booklet.

[RuiAce]

When a question asks you to draw a circle, do you have to draw it when it is already provided in the question booklet.

Please clarify. Post the question as a screenshot/photo in its entirety.

[yuriques]

Quick question: For circle geometry, do they deduct marks for not writing the converse theorem and instead just writing the normal one on the syllabus?

[RuiAce]

Quick question: For circle geometry, do they deduct marks for not writing the converse theorem and instead just writing the normal one on the syllabus?

If you plan to use the converse of a theorem, you should always be writing it the way the converse SHOULD be stated. Otherwise, your working out is incorrect, because your reasoning is reversed.

E.g. the usual one to the angle in a semicircle is just that "the angle in a semicircle is a right angle". If you want to apply its converse, you will need to state "since this is a right angle, it is the angle in a semicircle".

[yuriques]

If you plan to use the converse of a theorem, you should always be writing it the way the converse SHOULD be stated. Otherwise, your working out is incorrect, because your reasoning is reversed.

E.g. the usual one to the angle in a semicircle is just that "the angle in a semicircle is a right angle". If you want to apply its converse, you will need to state "since this is a right angle, it is the angle in a semicircle".

Ah I see, I guess I can't be lazy then haha Thanks!

[kemi]

For HSC 2009, Q6 b) Part (i), why are the number of terms in the series

n - r + 1

Thanks!

[RuiAce]

For HSC 2009, Q6 b) Part (i), why are the number of terms in the series

n - r + 1

Thanks!

If you are going from 0 to n, there are n+1 terms.

That is something you should be well aware of by now. For example, if you are going from 0 to 3, you will have 1, a, a^2, and a^3. There are 4 terms, not 3.
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And then we truncate.

In the above example, if we go from 1 to 3, we will have 3 terms. Note that (3 - 1 + 1) = 3 as well.
In the above example, if we go from 2 to 3, we will have 2 terms. Note that (3 - 2 + 1) = 2 as well.

In a similar way, if we go from r to n, we will have (n - r + 1) terms.

[kemi]

If you are going from 0 to n, there are n+1 terms.

That is something you should be well aware of by now. For example, if you are going from 0 to 3, you will have 1, a, a^2, and a^3. There are 4 terms, not 3.
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And then we truncate.

In the above example, if we go from 1 to 3, we will have 3 terms. Note that (3 - 1 + 1) = 3 as well.
In the above example, if we go from 2 to 3, we will have 2 terms. Note that (3 - 2 + 1) = 2 as well.

In a similar way, if we go from r to n, we will have (n - r + 1) terms.

Thanks! I understood the '+1' but for some reason the 'n - r' confused me? :S but thanks for clarifying

[RuiAce]

Thanks! I understood the '+1' but for some reason the 'n - r' confused me? :S but thanks for clarifying

You don't start from what's theoretically the very first term. You start halfway through.
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It's sort of like a geometric series, except you don't start at \( T_1\), which is \(a\). You start at \( T_k\) instead, which is \( ar^{k-1} \)

[seventeenboi]

hiiiii i need help with these two questions plsss thank you!!

[legorgo18]

hiiiii i need help with these two questions plsss thank you!!

Hey for 31 for part i) i think just x=-1 then part ii integrate then you have the + c then you can solve for c then x=-1 again and get it sorry ceebs math atm

[RuiAce]

hiiiii i need help with these two questions plsss thank you!!

Note, however, performing an indefinite integral and then explicitly finding the constant of integration is perfectly valid.
\begin{align*}\int_{-1}^0\left[1 + \binom{n}{1}x+\binom{n}{2}x^2+\dots + \binom{n}{n}x^n\right]dx& =\int_{-1}^0 (1+x)^n\,dx\\ \left[x+\binom{n}{1}\frac{x^2}2+\binom{n}2\frac{x^3}3+\dots + \binom{n}{n}\frac{x^{n+1}}{n+1}\right]_{-1}^0&= \left[\frac{(1+x)^{n+1}}{n+1}\right]_{-1}^0\\ -\left[-1+\binom{n}{1}\frac12 - \binom{n}{2}\frac13 + \dots +(-1)^{n+1}\binom{n}{n} \frac{1}{n+1} \right] &= \frac{1}{n+1}\\ 1-\frac12 \binom{n}{1}+ \frac13\binom{n}{2} - \dots +(-1)^{n}\frac{1}{n+1}\binom{n}n&= \frac1{n+1} \end{align*}
Note: \( (-1)^2 = 1\text{ so }(-1)^{n+2} = (-1)^n \)

[winstondarmawan]

Hello! Would appreciate help with this question from the 2002 HSC.
https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/22685151_1355403897918463_1221729249_n.png?oh=ac5057d687a2f980ee8080a4afa26589&oe=59EB4912
TIA!
Note: Previous parts were show the max height and show the range.

[RuiAce]

hiiiii i need help with these two questions plsss thank you!!

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\begin{align*}\text{We extract }\binom{n}{0} \text{from the first bracket }&\text{and pair with }\binom{n}{n}x^n\text{ from the second}\\ \text{We extract }\binom{n}{1}x \text{from the first bracket }&\text{and pair with }\binom{n}{n-1}x^{n-1}\text{ from the second}\\ \text{We extract }\binom{n}{2}x^2 \text{from the first bracket }&\text{and pair with }\binom{n}{n-2}x^{n-2}\text{ from the second}\\ &\vdots\\ \text{We extract }\binom{n}{n}x^n \text{from the first bracket }&\text{and pair with }\binom{n}0\text{ from the second}\end{align*}