Author Topic: 3U Maths Question Thread (Read 1237711 times) Tweet Share

f_tan · « **Reply #2805 on:** September 25, 2017, 11:38:45 pm »

Can anyone help explain this question to me please?

Thank you!

RuiAce · « **Reply #2806 on:** September 25, 2017, 11:40:52 pm »

Quote from: f_tan on September 25, 2017, 11:38:45 pm

Can anyone help explain this question to me please?
(Image removed from quote.)
Thank you!

$\text{One may convert this using an auxiliary angle transform}\\ \text{to get }\boxed{x=13\sin (4t+\theta)}\\ \text{where }\tan \theta = \frac{12}5$
$\text{So differentiating gives }\boxed{v=52\cos (4t+\theta)}\\ \text{implying that the maximum velocity is }52$
$\text{This of course relies on the range of the function}\\ f(t) = 52\cos (4t+\theta)\\ \text{being }-52\le v \le 52$

clarence.harre · « **Reply #2807 on:** September 28, 2017, 08:53:03 am »

Binomial Theroem:

This is question 21 from exercise 10.2 in Maths in Focus Book.

Question:

Write out the expansion of (1 + sinx)^5

My working;
(not really any working to it)
(5C0) + (5C1)sinx + (5C2)(sinx)^2 + (5C3)(sinx)^3 + (5C4)(sinx)^4 + (5C5)(sinx)^5
= 1 + 5sinx + 10(sinx)^2 + 10(sinx)^3 + 5(sinx)^4 + (sinx)^5

Answer:
1 + 5sinx + 10(sinx)^2 + 10(sinx)^3 + 4(sinx)^4 + (sinx)^5
Apparently I'm really close, but the answer throw this mystery 4 in and I'm so tripped out. I always fail the binomial theroem part of tests, so I put my misunderstanding to that. Can anyone please help me out?

Opengangs · « **Reply #2808 on:** September 28, 2017, 09:14:32 am »

Quote from: clarence.harre on September 28, 2017, 08:53:03 am

Binomial Theroem:

This is question 21 from exercise 10.2 in Maths in Focus Book.

Question:

Write out the expansion of (1 + sinx)^5

My working;
(not really any working to it)
(5C0) + (5C1)sinx + (5C2)(sinx)^2 + (5C3)(sinx)^3 + (5C4)(sinx)^4 + (5C5)(sinx)^5
= 1 + 5sinx + 10(sinx)^2 + 10(sinx)^3 + 5(sinx)^4 + (sinx)^5

Answer:
1 + 5sinx + 10(sinx)^2 + 10(sinx)^3 + 4(sinx)^4 + (sinx)^5
Apparently I'm really close, but the answer throw this mystery 4 in and I'm so tripped out. I always fail the binomial theroem part of tests, so I put my misunderstanding to that. Can anyone please help me out?

It should be 5sinx^4, you were correct.

RuiAce · « **Reply #2809 on:** September 28, 2017, 09:26:11 am »

Quote from: clarence.harre on September 28, 2017, 08:53:03 am

Binomial Theroem:

This is question 21 from exercise 10.2 in Maths in Focus Book.

Question:

Write out the expansion of (1 + sinx)^5

My working;
(not really any working to it)
(5C0) + (5C1)sinx + (5C2)(sinx)^2 + (5C3)(sinx)^3 + (5C4)(sinx)^4 + (5C5)(sinx)^5
= 1 + 5sinx + 10(sinx)^2 + 10(sinx)^3 + 5(sinx)^4 + (sinx)^5

Answer:
1 + 5sinx + 10(sinx)^2 + 10(sinx)^3 + 4(sinx)^4 + (sinx)^5
Apparently I'm really close, but the answer throw this mystery 4 in and I'm so tripped out. I always fail the binomial theroem part of tests, so I put my misunderstanding to that. Can anyone please help me out?

You should always consider the fact that MIF is full of mistakes, and when you're in doubt check your answer on WolframAlpha

winstondarmawan · « **Reply #2810 on:** September 28, 2017, 09:53:54 am »

Would appreciate help with the following:
https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/22155294_1336895733102613_415072285_n.jpg?oh=ee857896c722b3a984d8b963d050323d&oe=59CEAAAE
Q3. Thanks in advance.

RuiAce · « **Reply #2811 on:** September 28, 2017, 09:58:09 am »

Quote from: winstondarmawan on September 28, 2017, 09:53:54 am

Would appreciate help with the following:
https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/22155294_1336895733102613_415072285_n.jpg?oh=ee857896c722b3a984d8b963d050323d&oe=59CEAAAE
Q3. Thanks in advance.

This was addressed a while back

Annie657 · « **Reply #2812 on:** September 28, 2017, 10:22:09 am »

Please help me with this question, here is my working but I wasn't really sure what to do and the answer is 1.5. What am I doing wrong?

RuiAce · « **Reply #2813 on:** September 28, 2017, 10:26:46 am »

Quote from: Annie657 on September 28, 2017, 10:22:09 am

Please help me with this question, here is my working but I wasn't really sure what to do and the answer is 1.5. What am I doing wrong?

$a\cos x + b \sin x = R \cos (x \bf{-} \alpha)$
You're meant to flip the sign when you're using the cosine auxiliary angle transformation, which you did not do.
_____________________
$\text{It may help to recall that in fact,}\\ \cos(x-\alpha) = \cos x \cos \alpha + \sin x \sin \alpha\\ \sin(x + \alpha) = \sin x \cos \alpha + \cos x \sin \alpha$
$\text{The sign flips in the compound angle expansion for cos}\\ \text{therefore it flips in the auxiliary angle method as well.}$

winstondarmawan · « **Reply #2814 on:** September 28, 2017, 09:41:31 pm »

Hello! Would appreciate help with the following:
https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/22054248_1337014896424030_1672074058_n.jpg?oh=e182adab17a3164ce623c478b50cb7c6&oe=59CF1AC1
Question 10.
https://scontent-syd2-1.xx.fbcdn.net/v/t35.0-12/22095616_1337208966404623_184734103_o.jpg?oh=3acdf38232ac9db2fa232cb83ee1c701&oe=59CF2B57
Question 4b).
Thanks in advance!

RuiAce · « **Reply #2815 on:** September 28, 2017, 09:55:49 pm »

Quote from: winstondarmawan on September 28, 2017, 09:41:31 pm

Hello! Would appreciate help with the following:
https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/22054248_1337014896424030_1672074058_n.jpg?oh=e182adab17a3164ce623c478b50cb7c6&oe=59CF1AC1
Question 10.
https://scontent-syd2-1.xx.fbcdn.net/v/t35.0-12/22095616_1337208966404623_184734103_o.jpg?oh=3acdf38232ac9db2fa232cb83ee1c701&oe=59CF2B57
Question 4b).
Thanks in advance!

$\text{4b involves a cheeky trick in that you compare the expression}\\ \text{obtained after integrating, to the one from just straight substitution}$
The giveaway to use this method was the fact that all fractions weren't of the form $ \frac1n$, but rather $ 1-\frac1n$. That, and the -1 in the integrated expression doesn't vanish, but instead turns into a plus.
$\begin{align*}\binom{n}{0}+\binom{n}{1}+\binom{n}{2}+\binom{n}{3}+\dots+\binom{n}{n}&=2^n\\ \binom{n}{0}+\frac12\binom{n}{1}+\frac13\binom{n}{2}+\frac14\binom{n}{3}+\dots +\frac1{n+1}\binom{n}{n}&=\frac{2^{n+1}-1}{n+1} \end{align*}$
$\textbf{Upon subtracting,}\\ \begin{align*}\frac12\binom{n}{1}+\frac23\binom{n}{2}+\frac34\binom{n}{3}+\dots + \frac{n}{n+1}\binom{n}{n}&=2^n - \frac{2^{n+1}-1}{n+1}\\ &= \frac{2^n(n+1) - 2(2^n) +1}{n+1}\\ &= \frac{n(2^n) + 2^n - 2(2^n) + 1}{n+1}\\ &= \frac{2^n(n-1)+1}{n+1}\end{align*}$
The other question was addressed quite recently. (Scroll down a bit.)

winstondarmawan · « **Reply #2816 on:** September 29, 2017, 10:13:39 am »

Would appreciate help with the following:
https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/22119112_1337706313021555_425603755_n.png?oh=1d28241a61464e0443ecc82d88de6139&oe=59CFD7C2

RuiAce · « **Reply #2817 on:** September 29, 2017, 10:39:30 am »

Quote from: winstondarmawan on September 29, 2017, 10:13:39 am

Would appreciate help with the following:
https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/22119112_1337706313021555_425603755_n.png?oh=1d28241a61464e0443ecc82d88de6139&oe=59CFD7C2

The question and answer will be moved into the 4U thread some time within the next 12-24 hours as it is Extension 2 material.
$\text{This is basically the approximation }\pi \approx \frac{22}{7}$
For part 1, we may consult with WolframAlpha.

$0 < x < 1 \implies 0 < x^2 < 1\\ \text{and hence }1 < 1+x^2 < 2\\ \text{Reciprocating gives }\frac12 < \frac1{1+x^2} < 1$
$\text{Since }x^4(1-x)^4\ge 0\text{, we may just multiply }x^4(1-x)^4\\ \text{to all three sides of the equation to obtain}\\ \frac12 x^4(1-x)^4 < \frac{x^4(1-x)^4}{1+x^2} < x^4(1-x)^4$
____________________________________________
$\text{This is just the proof of the derivative of }\tan^{-1}y\\ \text{Let }x=\tan^{-1}y \implies y = \tan x \implies \frac{dy}{dx} = \sec^2 x$
$\text{Hence }\frac{dx}{dy} = \frac{1}{\sec^2 x} = \frac{1}{1+\tan^2 x} = \frac{1}{1+y^2}$
____________________________________________
$\int_0^1 \frac{x^4(1-x)^4}{1+x^2}dx = \int_0^1 \left(x^6-4x^5+5x^4-4x^2 4- \frac{4}{1+x^2}\right)dx = \left[\frac{x^7}{7}-\frac{2x^6}{3}+x^5-\frac{4x^3}{3}+4x-4\tan^{-1}x\right]_0^1=\frac{22}7-\pi$
____________________________________________
$\text{We note that}\\ \int_0^1 x^4(1-x)^4dx = \int_0^1 (x^4-4x^5+6x^6-4x^7+x^8)dx=\frac{1}{630}$
$\text{Now, the inequality (from part b)}\\ \frac12 x^4(1-x)^4 < \frac{x^4(1-x)^4}{1+x^2} < x^4(1-x)^4\\ \text{is valid}\textbf{ for all }0 < x < 1$
$\text{Hence we may apply a definite integral over this domain.}\\ \int_0^1\frac12 x^4(1-x)^4 < \int_0^1\frac{x^4(1-x)^4}{1+x^2} <\int_0^1 x^4(1-x)^4$
$\text{This becomes }\frac{1}{1260} < \frac{22}7-\pi < \frac{1}{630}\\ \text{which rearranges to give the desired result.}$

winstondarmawan · « **Reply #2818 on:** September 29, 2017, 04:34:57 pm »

Would appreciate help with the following:
https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/22093754_1337900516335468_1907639337_n.png?oh=88128aeecc6021ceb1432ca005f019c9&oe=59CF971C
For part (iv), my answer is 1.28m and 0.16m whilst the Terry Lee answer is 1.28m and 0.48m.
Thanks in advance.

pikachu975 · « **Reply #2819 on:** September 29, 2017, 05:33:45 pm »

Quote from: winstondarmawan on September 29, 2017, 04:34:57 pm

Would appreciate help with the following:
https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/22093754_1337900516335468_1907639337_n.png?oh=88128aeecc6021ceb1432ca005f019c9&oe=59CF971C
For part (iv), my answer is 1.28m and 0.16m whilst the Terry Lee answer is 1.28m and 0.48m.
Thanks in advance.

Quote from: winstondarmawan on September 29, 2017, 04:34:57 pm

Would appreciate help with the following:
https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/22093754_1337900516335468_1907639337_n.png?oh=88128aeecc6021ceb1432ca005f019c9&oe=59CF971C
For part (iv), my answer is 1.28m and 0.16m whilst the Terry Lee answer is 1.28m and 0.48m.
Thanks in advance.

Terry Lee often makes mistakes, so you might want to check the HSC answers.

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Author Topic: 3U Maths Question Thread (Read 1237711 times) Tweet Share

f_tan

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