Author Topic: Mathematics Question Thread (Read 1626448 times) Tweet Share

jamesrandom · « **Reply #3825 on:** October 19, 2018, 03:54:10 pm »

Hey, Im just starting Year 12 now so dont think im doing my HSC next week and dont know how to do this haha.

Anyway, im reviewing product rule and forgot how to do this.

Differentiate:

(x + 1)(2x + 5)^4

I know its probably easy im just confused as to how my textbook got the answer

(10x + 13)(2x + 5)^3

RuiAce · « **Reply #3826 on:** October 19, 2018, 04:14:48 pm »

Quote from: jamesrandom on October 19, 2018, 03:54:10 pm

Hey, Im just starting Year 12 now so dont think im doing my HSC next week and dont know how to do this haha.

Anyway, im reviewing product rule and forgot how to do this.

Differentiate:

(x + 1)(2x + 5)^4

I know its probably easy im just confused as to how my textbook got the answer

(10x + 13)(2x + 5)^3

$\text{You need the chain rule as well as factorisation techniques}\\ \text{on top of the product rule.}$
\begin{align*} \frac{d}{dx} (x+1)(2x+5)^4 &= 1 \times (2x+5)^4 + (x+1) \times 8(2x+5)^3\\ &= (2x+5)^4 + (8x+8)(2x+5)^3\\ &= (2x+5)^3 \left[ (2x+5) + (8x+8) \right] \\ &= (10x+13)(2x+5)^3\end{align*}

Mate2425 · « **Reply #3827 on:** October 19, 2018, 05:55:01 pm »

Hey Rui, could you please explain why the answer is pi/3 and 5pi/3 in HSC 2016 Advanced Math, rather than pi/3 and 2pi/3.
I thought that you consider where sin is pos in ASTC which would be in the first and second quadrant but my answers do not relate.

Thank you.

RuiAce · « **Reply #3828 on:** October 19, 2018, 06:00:03 pm »

Quote from: Mate2425 on October 19, 2018, 05:55:01 pm

Hey Rui, could you please explain why the answer is pi/3 and 5pi/3 in HSC 2016 Advanced Math, rather than pi/3 and 2pi/3.
I thought that you consider where sin is pos in ASTC which would be in the first and second quadrant but my answers do not relate.

Thank you.

Which question of the 2016 paper was this?

fun_jirachi · « **Reply #3829 on:** October 19, 2018, 06:17:25 pm »

Quote from: Mate2425 on October 19, 2018, 05:55:01 pm

Hey Rui, could you please explain why the answer is pi/3 and 5pi/3 in HSC 2016 Advanced Math, rather than pi/3 and 2pi/3.
I thought that you consider where sin is pos in ASTC which would be in the first and second quadrant but my answers do not relate.

Thank you.

You should probably provide the question with your question, so we know what we're looking at! I found it anyway, but maybe do that next time, it really helps us help you!

I'm assuming this is 11(g) in 2016?
Assuming I'm correct, the answer is right. You solve $\sin\frac{x}{2}=\frac{1}{2}$ and of course you get your $\frac{\pi}{6}, \frac{5\pi}{6}$ (from your angles of any magnitude, or ASTC or whatever you want to call it) since $\frac{13\pi}{6}$ is far too big. Since $\frac{\pi}{6}, \frac{5\pi}{6} = \frac{x}{2}$ x is equal to twice those, so you halve the denominator to get the suggested answers. Im not sure where you pulled the 2pi/3 from, can you maybe explain how you got there?

Hope this helps

raghav_singh · « **Reply #3830 on:** October 19, 2018, 08:07:31 pm »

Hi so I was wondering whether the 2U Maths work done in Year 11 will be tested in the HSC exams. OR, will it it just come back in the form of harder HSC topics, and be somewhat of a foundation to complete questions in these topics.

RuiAce · « **Reply #3831 on:** October 19, 2018, 08:11:14 pm »

Quote from: raghav_singh on October 19, 2018, 08:07:31 pm

Hi so I was wondering whether the 2U Maths work done in Year 11 will be tested in the HSC exams. OR, will it it just come back in the form of harder HSC topics, and be somewhat of a foundation to complete questions in these topics.

For all mathematics courses, preliminary content IS examinable in the HSC and makes up 20% of the marks.

Mate2425 · « **Reply #3832 on:** October 19, 2018, 10:34:43 pm »

Hi guys for this question HSC 2013 Q14d, the working out does not resonate a complete understanding, could you please explain it in another way please.
http://educationstandards.nsw.edu.au/wps/wcm/connect/2cdf8537-92dc-4328-ab39-74a1f1dd489f/maths-hsc-exam-2013.pdf?MOD=AJPERES&CACHEID=ROOTWORKSPACE-2cdf8537-92dc-4328-ab39-74a1f1dd489f-lGhNsb6

Many thanks, in advance!

jamesrandom · « **Reply #3833 on:** October 19, 2018, 10:37:07 pm »

Can someone help me with the attached question.

S200 · « **Reply #3834 on:** October 19, 2018, 11:05:18 pm »

Quote from: jamesrandom on October 19, 2018, 10:37:07 pm

Can someone help me with the attached question.

Find the derivative, sub in the point to get the gradient..
Then use the gradient-point formula ($y-y_1 =m(x-x_1)$), where $m$ is the gradient, to get the full equation.

RuiAce · « **Reply #3835 on:** October 20, 2018, 12:04:57 am »

Quote from: Mate2425 on October 19, 2018, 10:34:43 pm

Hi guys for this question HSC 2013 Q14d, the working out does not resonate a complete understanding, could you please explain it in another way please.
http://educationstandards.nsw.edu.au/wps/wcm/connect/2cdf8537-92dc-4328-ab39-74a1f1dd489f/maths-hsc-exam-2013.pdf?MOD=AJPERES&CACHEID=ROOTWORKSPACE-2cdf8537-92dc-4328-ab39-74a1f1dd489f-lGhNsb6

Many thanks, in advance!

$\text{We first note that}\\ \begin{align*}\int_{-1}^2 f(x)\,dx &= 3\\ \int_{-3}^{-1} f(x)\,dx &= 0\\ \int_2^3 f(x)\,dx &= 0 \end{align*}$
The last two come from noticing symmetry - if we have the same area appearing above and below the x-axis, they cancel each other out. Whereas the first one occurs because the region between $x=-1$ and $x=2$ is just a rectangle with length 3 and breadth 1.
$\text{So upon adding, }\boxed{\int_{-3}^3 f(x)\,dx = 3}.$
$\text{Note that for any positive real number }b,\\ \int_{-3-b}^{3+b}f(x)\,dx \text{ now combines the area established above,}\\ \text{as well as the two areas, one being between }x=-3-b\text{ and }x=3,\\ \text{and the other between }x=3\text{ and }x=3+b.$

$\text{Since }\int_3^{3+b}f(x)\,dx\text{ represents the area of a rectangle}\\ \text{below the }x\text{-axis, length }b\text{ and breadth }1,\\ \int_3^{3+b}f(x)\,dx = -b.$
$\text{Similarly }\int_{-3-b}^3 f(x)\,dx = -b.\\ \text{Therefore combining all three integrals together,}\\ \boxed{\int_{-3-b}^{3+b}f(x)\,dx = 3 - 2b}.$
$\text{So if we want }\int_{-3-b}^{3+b}f(x)\,dx = 0,\text{ then we'd have}\\ 3-2b = 0 \implies \boxed{b = \frac{3}{2}}.\\ \text{But comparing with the original question, what we really want is }a\text{, which is related}\\ \text{by }\boxed{a = 3+b}.\text{ Hence }a = 3 + \frac32 = \frac92.$

Sr-1425 · « **Reply #3836 on:** October 20, 2018, 07:52:52 am »

Hi, I'm struggling to find a value for t in which ln 0.2t = 5t.
Any help would be much appreciated.

fun_jirachi · « **Reply #3837 on:** October 20, 2018, 08:33:20 am »

You have from rearranging the question that 4.8t=0. Therefore, t=0.

clovvy · « **Reply #3838 on:** October 20, 2018, 08:37:30 am »

Quote from: fun_jirachi on October 20, 2018, 08:33:20 am

You have from rearranging the question that 4.8t=0. Therefore, t=0.

it says ln(0.2t)=5t... I tried regular rearranging and I am not getting anywhere so I check wolfram and it seems that it is way beyond HSC unless I misread the question or the person mistyped

fun_jirachi · « **Reply #3839 on:** October 20, 2018, 08:43:11 am »

Oh! Must not have read that correctly, my bad.

Well thinking about it graphically, you have y=ln0.2x and y=5x. The graphs of both those functions never cross, so there is no solution, which is why you're struggling to find it. Sometimes you've gotta look at it unconventionally and think about the graphical method, which many people seem to forget

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