Login | Register

Welcome, Guest. Please login or register.

July 10, 2020, 12:57:16 pm

### AuthorTopic: Mathematics Question Thread  (Read 711192 times) Tweet Share

0 Members and 1 Guest are viewing this topic.

#### svnflower

• Adventurer
• Posts: 10
• Respect: 0
##### Re: Mathematics Question Thread
« Reply #4455 on: May 15, 2020, 11:37:36 pm »
0
Ahh thanks heaps, this topic is becoming much more clearer to me now

I've also tried doing this question a few times but keep getting the wrong answer? Would you mind letting me know where I went downhill?

Thanks in advance!!

#### Opengangs

• New South Welsh
• Forum Leader
• Posts: 696
• $\mathbb{O}_\mathbb{G}$
• Respect: +449
##### Re: Mathematics Question Thread
« Reply #4456 on: May 15, 2020, 11:57:59 pm »
+1
Ahh thanks heaps, this topic is becoming much more clearer to me now

I've also tried doing this question a few times but keep getting the wrong answer? Would you mind letting me know where I went downhill?

Thanks in advance!!

Hi there!
Check your bounds :-)) They should be the points of intersection! To find your points of intersection, set $x^2 = 2x + 3$. This is just a quadratic in $x$ that you can solve! This gives you
\begin{align*}
x^2 - 2x - 3 &= 0 \\
(x - 3)(x + 1) &= 0 \\
x = 3, &\quad x = -1.
\end{align*}

So these are your bounds! The next step is to figure out which curve is the top curve! This can be visualised using a diagram and it's pretty easy to see that $y = 2x + 3$ is the top curve as you successfully worked out! So all we need to do is to evaluate the following integral $\displaystyle \int_{-1}^3 (2x + 3 - x^2)\,dx$, which becomes
\begin{align*}
\int_{-1}^3 (2x + 3 - x^2)\,dx &= \left(x^2 + 3x - \frac{1}{3}x^3\right)\bigg|_{-1}^3 \\
&= \left(9 + 9 - 9\right) - \left(1 - 3 + \frac{1}{3}\right) \\
&= 9 + 2 - \frac{1}{3} \\
&= 11 - \frac{1}{3} \\
&= \frac{32}{3} \\
&= 10 \frac{2}{3}.
\end{align*}
I think the mistake here is really figuring out what the bounds are, which are the points of intersection between the two curves. :-))
[2017: HSC]
English (Adv): 86 $\mid$ Mathematics (Adv): 94 $\mid$ Mathematics (Extension 1): 45 $\mid$ Biology: 84 $\mid$ Business Studies: 87 $\mid$ Software Design and Development: 88

[2018 $-$ 2022: UNSW]
Degree: Bachelor of Science (Computer Science) / Bachelor of Science (Mathematics)
Specialisation: Artificial Intelligence

#### svnflower

• Adventurer
• Posts: 10
• Respect: 0
##### Re: Mathematics Question Thread
« Reply #4457 on: May 16, 2020, 01:02:34 am »
+1
Thanks again for all the help today, super helpful!

#### 2020hsc

• Adventurer
• Posts: 19
• Respect: 0
##### Re: Mathematics Question Thread
« Reply #4458 on: June 01, 2020, 03:44:25 pm »
0
hey,

could i get some help on the question below...please and thanks  its in the year 12 unit on sequences and series, looking at solving problems with arithmetic sequences and geometric sequences.

'find a and b if a,b,1 forms a GP and b,a,10 forms and AP'

thanks!

#### fun_jirachi

• MOTM: AUG 18
• Moderator
• Forum Leader
• Posts: 654
• All doom and Gloom.
• Respect: +374
##### Re: Mathematics Question Thread
« Reply #4459 on: June 01, 2020, 05:07:32 pm »
+1
Hey there,

Think about what an AP and a GP define for any three consecutive terms. For $n \in \mathbb{Z}^+, n > 2$, the former states that $T_n - T_{n-1} = T_{n+1} - T_n$, while the latter states that $\frac{T_n}{T_{n-1}} = \frac{T_{n+1}}{T_n}$.

Now, considering this, we have that $\frac{b}{a}= \frac{1}{b} \implies a=b^2$. We also have that $a-b = 10-a \implies 2a = b+10$.

See how you go from here
« Last Edit: June 01, 2020, 05:36:43 pm by fun_jirachi »
Failing everything, but I'm still Flareon up.

HSC 2018: Modern History [88] | 2U Maths [98]
HSC 2019: Physics [92] | Chemistry [93] | English Advanced [87] | Maths Extension 1 [98] | Maths Extension 2 [97]
ATAR: 99.05

UCAT: 3310 - Verbal Reasoning [740] | Decision Making [890] | Quantitative Reasoning [880] | Abstract Reasoning [800]

Quick Link to Guides:
Subject Acceleration (2018)
UCAT Question Compilation/FAQ (2020)

#### 2020hsc

• Adventurer
• Posts: 19
• Respect: 0
##### Re: Mathematics Question Thread
« Reply #4460 on: June 01, 2020, 05:22:05 pm »
0
thanks....but should it not be the other way?

considering the GP is a,b,1 and the AP is b,a,10...then should it not be b/a = 1/b and a-b = 10-a ??

#### fun_jirachi

• MOTM: AUG 18
• Moderator
• Forum Leader
• Posts: 654
• All doom and Gloom.
• Respect: +374
##### Re: Mathematics Question Thread
« Reply #4461 on: June 01, 2020, 05:34:52 pm »
+1
thanks....but should it not be the other way?

considering the GP is a,b,1 and the AP is b,a,10...then should it not be b/a = 1/b and a-b = 10-a ??

Yes, it should! I misread it ~ the point still stands, however - you'll still have a quadratic. I'll edit that now so it's correct. In general, it is handy to know what makes a GP a GP, and what makes an AP an AP, even though these questions are relatively uncommon - the theory is so important to reasoning out further applications
Failing everything, but I'm still Flareon up.

HSC 2018: Modern History [88] | 2U Maths [98]
HSC 2019: Physics [92] | Chemistry [93] | English Advanced [87] | Maths Extension 1 [98] | Maths Extension 2 [97]
ATAR: 99.05

UCAT: 3310 - Verbal Reasoning [740] | Decision Making [890] | Quantitative Reasoning [880] | Abstract Reasoning [800]

Quick Link to Guides:
Subject Acceleration (2018)
UCAT Question Compilation/FAQ (2020)

#### 2020hsc

• Adventurer
• Posts: 19
• Respect: 0
##### Re: Mathematics Question Thread
« Reply #4462 on: June 01, 2020, 05:52:40 pm »
0
Yes, it should! I misread it ~ the point still stands, however - you'll still have a quadratic. I'll edit that now so it's correct. In general, it is handy to know what makes a GP a GP, and what makes an AP an AP, even though these questions are relatively uncommon - the theory is so important to reasoning out further applications

ok, sweet. thanks a lot  from there then do you solve it as a simultaneous equation or how would be best?

#### Einstein_Reborn_97

• MOTM: April 20
• Forum Regular
• Posts: 89
• There is no substitute for hard work.
• Respect: +38
##### Re: Mathematics Question Thread
« Reply #4463 on: June 01, 2020, 07:15:16 pm »
0
ok, sweet. thanks a lot  from there then do you solve it as a simultaneous equation or how would be best?
Yes, solve it using a simultaneous equation. Let me know what answers you get
HSC 2020: Advanced English | Advanced Mathematics | Physics | Chemistry | Biology | Studies of Religion 2

My HSC Journey Journal

#### 2020hsc

• Adventurer
• Posts: 19
• Respect: 0
##### Re: Mathematics Question Thread
« Reply #4464 on: June 01, 2020, 07:27:11 pm »
0
Yes, solve it using a simultaneous equation. Let me know what answers you get

thanks!!

coming out with b= -2, a = 4 or b= 5/2, a = 25/4

sound about right?

#### Einstein_Reborn_97

• MOTM: April 20
• Forum Regular
• Posts: 89
• There is no substitute for hard work.
• Respect: +38
##### Re: Mathematics Question Thread
« Reply #4465 on: June 01, 2020, 08:00:28 pm »
0
thanks!!

coming out with b= -2, a = 4 or b= 5/2, a = 25/4

sound about right?
Yep, correct. Both solutions work for both the GP and the AP.
HSC 2020: Advanced English | Advanced Mathematics | Physics | Chemistry | Biology | Studies of Religion 2

My HSC Journey Journal

#### RuskiBrah

• Adventurer
• Posts: 6
• Respect: 0
##### Re: Mathematics Question Thread
« Reply #4466 on: June 02, 2020, 09:04:26 pm »
0
hey guys, was wondering what formula to use for this question and why. cheers

#### Einstein_Reborn_97

• MOTM: April 20
• Forum Regular
• Posts: 89
• There is no substitute for hard work.
• Respect: +38
##### Re: Mathematics Question Thread
« Reply #4467 on: June 02, 2020, 09:30:37 pm »
0
hey guys, was wondering what formula to use for this question and why. cheers
Hey RuskiBrah,

For (a): Use the formula for the sum of a geometric series:
a=45, n=6, r=0.4 (so make sure to use to the correct formula; -1<r<1).
Why? Initial height (first term) is 45 cm. Common ratio, r is 0.4: "grows by 2/5 of its previous growth each month". n in this case is the number of months so is equal to 6.
Your answer should be 74.7 cm.

For (b): Use the formula for limiting sum of an infinite geometric series:
a=45 and r=0.4
Why? The question is asking for the lamb's "final height" and for geometric series where -1<r<1, you can find its limiting sum because the series converge.
Your answer should be 75 cm.

Hope that helps! Let me know if you need any further explanations
HSC 2020: Advanced English | Advanced Mathematics | Physics | Chemistry | Biology | Studies of Religion 2

My HSC Journey Journal

#### RuskiBrah

• Adventurer
• Posts: 6
• Respect: 0
##### Re: Mathematics Question Thread
« Reply #4468 on: June 02, 2020, 10:11:00 pm »
0
ohh i forgot about that formula. that clears it up, thank you

#### RuskiBrah

• Adventurer
• Posts: 6
• Respect: 0
##### Re: Mathematics Question Thread
« Reply #4469 on: June 02, 2020, 10:18:04 pm »
0
Hey RuskiBrah,

For (a): Use the formula for the sum of a geometric series:
a=45, n=6, r=0.4 (so make sure to use to the correct formula; -1<r<1).
Why? Initial height (first term) is 45 cm. Common ratio, r is 0.4: "grows by 2/5 of its previous growth each month". n in this case is the number of months so is equal to 6.
Your answer should be 74.7 cm.

For (b): Use the formula for limiting sum of an infinite geometric series:
a=45 and r=0.4
Why? The question is asking for the lamb's "final height" and for geometric series where -1<r<1, you can find its limiting sum because the series converge.
Your answer should be 75 cm.

Hope that helps! Let me know if you need any further explanations

so in other words, its a limiting sum because its height could go on infinitely?