Author Topic: 3U Maths Question Thread (Read 1606809 times) Tweet Share

kaustubh.patel · « **Reply #3165 on:** January 28, 2018, 02:02:48 am »

Hey guys please help with this question
I'm pretty sure this may be a 4u question but it's in a 3u paper (tara angelican 1999)

Opengangs · « **Reply #3166 on:** January 28, 2018, 07:37:30 am »

Dragomistress · « **Reply #3167 on:** January 28, 2018, 05:10:42 pm »

Hey, how do you complete this?

006896 · « **Reply #3168 on:** January 28, 2018, 06:39:50 pm »

Hi hi!
Help please.
Thank you!

RuiAce · « **Reply #3169 on:** January 28, 2018, 06:54:34 pm »

Quote from: Dragomistress on January 28, 2018, 05:10:42 pm

Hey, how do you complete this?

Note that this question is not highlighted in purple and is thus of 2U level.
$\text{Assuming per-annum compounding, since nothing was specified.}\\ \text{Note that deposits are made at the }\textbf{end}\text{ of every year.}$
$\text{Let }A_n\text{ be the amount at the }\textbf{end}\text{ of the }n\text{-th year.}\\ \text{For the first ten years, following the usual recursive approach we have}\\ \begin{align*}A_1 &= 5000\\ A_2 &= A_1(1.06) + 5000\\ &= 5000(1.06) + 5000\\ A_3 &= A_2(1.06) + 5000\\ &= 5000 (1.06)^2 +5000 (1.06) + 5000\\ &\vdots\\A_{10}&= 5000(1.06)^{9}+ \dots + 5000(1.06) + 5000\\ &= 5000\underbrace{(1 + 1.06 + \dots + 1.06^9)}_{\text{geom. series, }r=1.06}\\ &= 5000\left(\frac{1.06^{10}-1}{1.06-1}\right)\\ &= 65903.97471\dots\end{align*}$
I will use a rounded value of 65903.97 from here onwards, although preferably do not round until the very end.
$\text{The rate is then increased from 6 percent to 10 percent.}\\ \text{So from here,}\\ \begin{align*}A_{11} &=A_{10}(1.1) + 5000\\ &= 65903.97(1.1) + 5000\\ A_{12}&=A_{11}(1.1) + 5000\\ &= 65903.97(1.1)^2 + 5000(1.1) + 5000\\ &\vdots\\ A_{25}&=65903.97(1.1)^{15} + 5000(1.1)^{14}+\dots + 5000(1.1) + 5000\\ &= 65903.97(1.1)^{15} + 5000\underbrace{(1+1.1+\dots+1.1^{14})}_{\text{geom. series, }r=1.1}\end{align*}\\ \text{The rest should be easy.}$

RuiAce · « **Reply #3170 on:** January 28, 2018, 07:03:50 pm »

Quote from: 006896 on January 28, 2018, 06:39:50 pm

Hi hi!
Help please.
Thank you!

$\text{In the usual convention, assume that }\\ a\text{ is opposite }A\text{ and etc. in }\triangle ABC.$
Note that this should've been clearly stated, but the convention is common enough for it not to be.
$\text{Via the (rearranged) cosine rule, we have}\\ \begin{align*}\cos A &= \frac{b^2+c^2-a^2}{2bc}\\ \cos B&= \frac{a^2+c^2-b^2}{2ac}\end{align*}$
$\text{Since }\boxed{a\cos A = b\cos B},\text{ upon subbing in we have}\\ a\left(\frac{b^2+c^2-a^2}{2bc}\right) = b\left(\frac{a^2+c^2-b^2}{2ac}\right)\\ \text{Multiplying both sides by }2c\text{ reduces this to}\\ \boxed{\frac{a}{b}(b^2+c^2-a^2) = \frac{b}{a}(a^2+c^2-b^2)}$
_____________________________________________________________
$\text{From here, just cross multiply and tackle the algebra.}\\ \begin{align*}a^2(b^2+c^2-a^2)&=b^2(a^2+c^2-b^2)\\ a^2b^2+a^2c^2-a^4&=a^2b^2+b^2c^2-b^4\\ (a^2c^2-b^2c^2) - (a^4-b^4)&=0\\ c^2(a^2-b^2) - \underbrace{(a^2-b^2)(a^2+b^2)}_{\text{diff. of two sq.}}&=0\end{align*}$
$\text{Factorising }(a^2-b^2)\text{ then gives}\\ \left[c^2- (a^2+b^2)\right](a^2-b^2)=0$
$\text{So either }a^2-b^2=0\text{, which would imply }a=b\\ \text{and thus the triangle is isosceles,}$
$\text{or alternatively }c^2-(a^2+b^2)=0\text{ which would imply }c^2=a^2+b^2\\ \text{so by the converse of Pythagoras' theorem, the}\\ \text{triangle is right-angled.}$

fireives1967 · « **Reply #3171 on:** January 30, 2018, 01:13:34 pm »

How would you do c, d and e? I remember learning it the term before, but I forgot the method.

I remember having to do pythag with the numbers in front, but I forgot the method in finding everything else

Highly appreciated!

RuiAce · « **Reply #3172 on:** January 30, 2018, 01:38:27 pm »

Quote from: fireives1967 on January 30, 2018, 01:13:34 pm

How would you do c, d and e? I remember learning it the term before, but I forgot the method.

I remember having to do pythag with the numbers in front, but I forgot the method in finding everything else

Highly appreciated!

$\text{To start each one off:}$
$\text{For c), you will need the auxiliary angle method.}\\ \text{This turns the equation to }\boxed{2 \cos \left(x-\frac\pi3\right) = 2}\\ \text{with our domain }0\le x \le 2\pi\text{ becoming }\boxed{-\frac\pi3\le x-\frac\pi3 \le \frac{5\pi}{3}}$
Alternatively, you could use the auxiliary angle method differently to obtain $2 \sin \left(x+\frac\pi6\right) = 2 $ with the domain changing to $\frac\pi6\le x+\frac\pi6 \le \frac{13\pi}{6} $.
$\text{For d), the Pythagorean identity gives us a quadratic in terms of }\sin x.\\ 1-\sin^2 x - \sin x = -1 \implies \boxed{\sin^2x +\sin x - 2 = 0}$
$\text{For e), the triple angle expansion is:}\\ \begin{align*}\sin 3x &= \sin (2x + x)\\ &= \sin 2x \cos x + \cos 2x \sin x\\ &= 2\sin x\cos^2 x + (1-2\sin^2x)\sin x\\ &= 2\sin x (1-\sin^2x) + \sin x(1-2\sin^2x)\\ &= \dots\\ &= 3\sin x - 4\sin^3 x\end{align*}$
If you have any further trouble, you should post any progress you have made thus far, including any relevant train of thought.

loz0829 · « **Reply #3173 on:** January 30, 2018, 10:32:12 pm »

Hi, I'm not sure how to do question 8. Help me please

Thank you

RuiAce · « **Reply #3174 on:** January 31, 2018, 12:58:16 am »

Quote from: loz0829 on January 30, 2018, 10:32:12 pm

Hi, I'm not sure how to do question 8. Help me please
Thank you

$\text{It can be shown that}\\ \begin{align*}x&=Vt\cos\theta\\ y&=Vt\sin\theta-\frac{1}{2}gt^2\end{align*}\\ \text{This is left as your exercise.}$
$\text{The Cartesian equation of motion can be found by rearranging }\boxed{t=\frac{x}{V\cos \theta}}\\ \text{and upon subbing in and using trig identities, we have}\\ \boxed{y = -\frac{gx^2}{2V^2}\sec^2\theta+ x\tan\theta}$
$\text{The range can be found by observing that for the time of flight, }y=0.\\ \text{Following the usual process, this gives a time of flight of }t=\frac{2V\sin\theta}{g}\\ \text{and hence a range of }\boxed{R = \frac{2V^2\sin\theta\cos\theta}{g}}$
_______________________________________
$\text{Because the range is 45, we have}\\ 45 = \frac{2V^2\sin\theta\cos\theta}{g} \implies \boxed{\frac{g}{2V^2} = \frac{\sin\theta\cos\theta}{45}}$
$\text{Subbing this in to our equation of motion gives}\\ \boxed{y = -\frac{x^2\sin\theta\cos\theta}{45}\sec^2\theta + x\tan\theta}\\ \text{Recalling relevant trigonometric identities, this reduces to}\\ \boxed{y = -\frac{x^2}{45}\tan\theta + x\tan\theta}$
_______________________________________
$\text{From here the question is easy.}\\ \text{As the ball passes through }(11.25,22.5),\\ \begin{align*}22.5 &= -\frac{11.25^2}{45}\tan\theta + 11.25\tan\theta\\ \frac{22.5}{8.4375}&=\tan\theta\end{align*}\\ \text{You should be able to finish the question from here.}$
Remember, to find $V$, you'll need to use $g=10$.

Lefkiiii6 · « **Reply #3175 on:** January 31, 2018, 03:58:42 pm »

Find the greatest coefficient of (2x-1)⁵.

Could somebody please show me the process of doing this binomial question?

jamonwindeyer · « **Reply #3176 on:** January 31, 2018, 05:45:49 pm »

Quote from: Lefkiiii6 on January 31, 2018, 03:58:42 pm

Find the greatest coefficient of (2x-1)⁵.

Could somebody please show me the process of doing this binomial question?

Hey! Here, since the power is only five, I think it's probably easiest to just write out the expansion in full!

$\left(2x-1\right)^5=\binom{5}{0}\left(2x\right)^5\left(-1\right)^0+\binom{5}{1}\left(2x\right)^4\left(-1\right)^1+\binom{5}{2}\left(2x\right)^3\left(-1\right)^2\\+\binom{5}{3}\left(2x\right)^2\left(-1\right)^3+\binom{5}{4}\left(2x\right)^1\left(-1\right)^4+\binom{5}{5}\left(2x\right)^0\left(-1\right)^5$

Evaluate all of these and you'll find:

$\left(2x-1\right)^5=32x^5-80x^4+80x^3-40x^2+10x-1$

Of these, 80 is the largest

note that this isn't the general method for this question type, but with only six terms in the expansion, I'd say it is the easiest!

3.14159265359 · « **Reply #3177 on:** January 31, 2018, 06:39:34 pm »

Quote from: jamonwindeyer on January 31, 2018, 05:45:49 pm

Hey! Here, since the power is only five, I think it's probably easiest to just write out the expansion in full!

$\left(2x-1\right)^5=\binom{5}{0}\left(2x\right)^5\left(-1\right)^0+\binom{5}{1}\left(2x\right)^4\left(-1\right)^1+\binom{5}{2}\left(2x\right)^3\left(-1\right)^2\\+\binom{5}{3}\left(2x\right)^2\left(-1\right)^3+\binom{5}{4}\left(2x\right)^1\left(-1\right)^4+\binom{5}{5}\left(2x\right)^0\left(-1\right)^5$

Evaluate all of these and you'll find:

$\left(2x-1\right)^5=32x^5-80x^4+80x^3-40x^2+10x-1$

Of these, 80 is the largest note that this isn't the general method for this question type, but with only six terms in the expansion, I'd say it is the easiest!

hey, I have a question regarding your working out. say the equation from the expansion was

20x^3 - 80x^2 +15x -3

here, will the greatest coefficient still be 80 (or -80???) or will it be 20?

ty!

jamonwindeyer · « **Reply #3178 on:** January 31, 2018, 06:53:39 pm »

Quote from: Mimi5601 on January 31, 2018, 06:39:34 pm

hey, I have a question regarding your working out. say the equation from the expansion was

20x^3 - 80x^2 +15x -3

here, will the greatest coefficient still be 80 (or -80???) or will it be 20?

ty!

I think its a question of semantics there, I'd say it would technically be 20 but you could definitely argue that -80 is greater in magnitude - Questions in the HSC wouldn't have that ambiguity

RuiAce · « **Reply #3179 on:** January 31, 2018, 06:54:20 pm »

Quote from: Mimi5601 on January 31, 2018, 06:39:34 pm

hey, I have a question regarding your working out. say the equation from the expansion was

20x^3 - 80x^2 +15x -3

here, will the greatest coefficient still be 80 (or -80???) or will it be 20?

ty!

Will add; in the HSC, in the 1% chance that they ask such a question, it'll almost always be the coefficient greatest in magnitude.

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Author Topic: 3U Maths Question Thread (Read 1606809 times) Tweet Share

kaustubh.patel

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