Author Topic: 3U Maths Question Thread (Read 1520970 times) Tweet Share

RuiAce · « **Reply #1650 on:** March 10, 2017, 11:29:01 pm »

Quote from: shreya_ajoshi on March 10, 2017, 11:26:29 pm

Thanksssss. Yep I need help with C
I also have another question:
A basketball team of 15 players is to be culled into a starting squad of 5 players. In how many ways can this be done so that:
A) the shortest player is not included
B) the tallest player is included
C) the three eldest people are included and thr three youngest people are not included

I'm confused about it because they've hardly given us any info :/

$\text{Reject the shortest player, so we only have 14 players to consider}\\ \text{We still need 5, so we have }\binom{14}{5}$
_____________
$\text{Forcibly include the tallest player}\\ \text{Then, we have 14 players left, and we only need to choose 4 others}\\ \therefore \binom{14}4$
_____________
$\text{Forcibly include the 3 eldest players}\\ \text{This leaves us with 12 players left to choose 3 from}$
$\text{But then reject the two youngest from our sample space}\\ \text{So we only have 10 players left to choose 3 from}\\ \therefore \binom{10}3$

hanaacdr · « **Reply #1651 on:** March 11, 2017, 04:54:46 am »

Quote from: Shadowxo on March 10, 2017, 08:25:59 am

Way I'd do it is, by kind of expanding it out and figuring what I need,
$a \leq b \\ a+ab \leq b+ab \\ a(1+b) \leq b(1+a) \\ \frac{a}{1+a} \leq \frac{b}{1+b}$

Alternatively, you could do it like ellipse,
$a+1 \leq 1+b$
$\frac{1}{1+a}\geq \frac{1}{1+b}$
$\frac{1}{1+a}-\frac{1+a}{1+a} \geq \frac{1}{1+b} - \frac{1+b}{1+b}$
$-\frac{a}{1+a}\geq -\frac{b}{1+b}$ $\frac{a}{1+a}\leq \frac{b}{1+b}$
(fixed some errors)
And your working for ii seems to be right

I really like the way you approached this question
thank you so much!
Hanaa

hanaacdr · « **Reply #1652 on:** March 11, 2017, 05:01:11 am »

Quote from: RuiAce on March 09, 2017, 09:54:05 pm

Hint for first part: Consider the following theorems.

1. Alternate segment theorem
2. Three right angles in that diagram, two of which are angle AMQ and angle ALQ

would <QNB be a right angle?

RuiAce · « **Reply #1653 on:** March 11, 2017, 07:50:06 am »

Quote from: hanaacdr on March 11, 2017, 05:01:11 am

would <QNB be a right angle?

$\text{Yes.}$
$\text{Read the question again. They state that }QN \perp PB$

hanaacdr · « **Reply #1654 on:** March 11, 2017, 10:51:32 am »

Quote from: RuiAce on March 11, 2017, 07:50:06 am

$\text{Yes.}$
$\text{Read the question again. They state that }QN \perp PB$

true, thank you

VydekiE · « **Reply #1655 on:** March 11, 2017, 11:15:22 am »

Hi, it would be great if I could get some help on these inverse trig questions
1) Evaluate sin[cos^-18/10 + tan^-1(-6/8)]
2) Prove that sin^-1x=cos^-1square root of 1-x^2 for x is greater than or equal to 0 and x is less than or equal to 1
Thank you!!

RuiAce · « **Reply #1656 on:** March 11, 2017, 11:32:39 am »

Quote from: VydekiE on March 11, 2017, 11:15:22 am

Hi, it would be great if I could get some help on these inverse trig questions
1) Evaluate sin[cos^-18/10 + tan^-1(-6/8)]
2) Prove that sin^-1x=cos^-1square root of 1-x^2 for x is greater than or equal to 0 and x is less than or equal to 1
Thank you!!

$\text{Expand out the compound angle first for 1)}\\ \begin{align*}&\quad\sin \left(\cos^{-1}\frac{8}{10}+\tan^{-1}\left(-\frac68\right)\right)\\&=\sin \left(\cos^{-1}\frac{8}{10}-\tan^{-1}\frac68\right)\\ &=\sin \left(\cos^{-1}\frac8{10}\right)\cos \left(\tan^{-1}\frac68\right)-\cos \left(\frac{8}{10}\right)\sin \left(\tan^{-1}68\right)\end{align*}$
$\text{Then the process is something you are expected to know.}\\ \text{For the first one, let }\theta = \tan^{-1}\frac68\\ \text{so that }\tan \theta = \frac{6}{8}=\frac{\text{opposite}}{\text{adjacent}}\\ \text{Draw a right-angled triangle to figure out what }\sin \theta\text{ and }\cos \theta\text{ are}.$
$\text{For the second one, let }\theta = \cos^{-1}\sqrt{1-x^2}\\ \text{so }\cos \theta = \frac{\sqrt{1-x^2}}{1}=\frac{\text{adjacent}}{\text{hypotenuse}}$

Dragomistress · « **Reply #1657 on:** March 11, 2017, 11:56:01 am »

May someone help me with this.

Thanks

jamonwindeyer · « **Reply #1658 on:** March 12, 2017, 01:09:37 pm »

Quote from: JKFlacka on March 11, 2017, 11:56:01 am

May someone help me with this.

Thanks

Hey! So you need to first find the equation of the line passing through those two points, using the two point formula:

$\frac{y-y_1}{x-x_1}=\frac{y_2-y_1}{x_2-x_1}\\\frac{y}{x+2}=\frac{-7}{5}\\\therefore 7x+5y+14=0$

To prove that (-7,7) lies on the line, we just substitute those values to see if it is true:

$7(-7)+5(7)+14=0\\0=0\\$

For the second part, you are using the rule for division in a given ratio, but in reverse. Find the m and n values that the simultaneous equations, the first being:

$\frac{nx_1+mx^2}{m+n}=-7\\\therefore\frac{-2n+3m}{m+n}=-7$

And similar for $y$ coordinates: You'll get simultaneous equations that can be solved to find the ratio you need

yasmineturner · « **Reply #1659 on:** March 14, 2017, 08:13:00 pm »

Hey

I am really lost with the following projectile motion questions and I was wondering if someone could please help me:

1. From a stationary helicopter, a steel ball is projected vertically upwards with a velocity of 4.9m/s. If a ball reaches the ground in 5 seconds, how high was the helicopter when the ball was projected.

2. An object falling in a vertical line passes a window 3m high in 1/6 of a second. Find the distance above the top of the window from which the object was let fall.

3. Steve tosses an apple to Adam who is sitting near him. Adam catches the apple at exactly the same height Steve released it. Suppose that the initial speed of the apple is V=5m/s and the initial angle, alpha, of elevation is given by tan(alpha)=2. Find x(dot), x, y(dot on top of y), and y by integrating x(double dot)=0 and y(double dot)=-10, taking the origin at Steve’s hands.

The thing I do not understand about question three is that it introduces tan(alpha)=2, whereas usually, questions state the angle of elevation instead.

Thankyou so much

RuiAce · « **Reply #1660 on:** March 14, 2017, 08:23:14 pm »

Quote from: yasmineturner on March 14, 2017, 08:13:00 pm

Hey I am really lost with the following projectile motion questions and I was wondering if someone could please help me:

1. From a stationary helicopter, a steel ball is projected vertically upwards with a velocity of 4.9m/s. If a ball reaches the ground in 5 seconds, how high was the helicopter when the ball was projected.

2. An object falling in a vertical line passes a window 3m high in 1/6 of a second. Find the distance above the top of the window from which the object was let fall.

3. Steve tosses an apple to Adam who is sitting near him. Adam catches the apple at exactly the same height Steve released it. Suppose that the initial speed of the apple is V=5m/s and the initial angle, alpha, of elevation is given by tan(alpha)=2. Find x(dot), x, y(dot on top of y), and y by integrating x(double dot)=0 and y(double dot)=-10, taking the origin at Steve’s hands.

The thing I do not understand about question three is that it introduces tan(alpha)=2, whereas usually, questions state the angle of elevation instead.

Thankyou so much

$\text{3. }\alpha\text{ is the angle of elevation.}\\ \text{Since we are given that }\tan \alpha = 2 = \frac{opp.}{adj.}\\ \text{(and obviously an angle of elevation is between }0^\circ\text{ and }90^\circ$
$\text{Draw out a right-angled triangle}.\\\text{Since }\tan \alpha =\frac{2}{1}\\ \text{the opposite side will have length }2\\ \text{and the adjacent side will have length }1\\ \text{By Pythagoras' theorem, this means that the hypotenuse has length }\sqrt{5}$
$\text{Hence reading off the triangle}\\ \cos\alpha= \frac{1}{\sqrt{5}}, \, \sin \alpha=\frac{2}{\sqrt{5}}$
$\text{Following the usual process}\\ \ddot{x} = 0 \implies \dot{x} = C_1\\ \text{Since when }t=0, \dot{x}=V\cos \alpha = 5 \times \frac{1}{\sqrt{5}}=\sqrt{5}\\ C_1=\sqrt{5}\\ \text{Hence }\dot{x}=\sqrt{5}$
$\ddot{y}=-10\implies \dot{y}=-10t+C_2\\ \text{Since when }t=0, \dot{y}=V\sin \alpha = 5\times \frac{2}{\sqrt{5}}=2\sqrt5\\ C_2=2\sqrt{5}\\ \text{Hence }\dot{y}=-10t+2\sqrt{5}$
$\text{You should be able to finish it off from here.}$

grace.estelle · « **Reply #1661 on:** March 14, 2017, 10:18:38 pm »

Hi, can someone please explain how to do question iii)?
Thanks!

RuiAce · « **Reply #1662 on:** March 14, 2017, 10:22:17 pm »

Quote from: grace.estelle on March 14, 2017, 10:18:38 pm

Hi, can someone please explain how to do question iii)?
Thanks!

C is correct because it is the definition of continuity. The limit as you approach from the left, must always be equal to the limit as you approach from the left, AND equal to f(a) itself to satisfy continuity.

A is wrong because it just means we have a stationary point
B and D are useless. It's saying y = dy/dx, which is irrelevant to continuity.

(Side note: Continuity is, in fact, a prerequisite for differentiability)
_________________________________
$\text{It may be worthwhile to demonstrate a function that is DISCONTINUOUS}\\ \text{to show what contradicts continuity.}$
$\text{Consider the derivative of }|x|\text{, which is given by}\\ f(x)=\frac{d}{dx}|x|=\begin{cases}1& x>0\\ -1& x<0\end{cases}\\ \text{Then }\lim_{x\to 0^+}f(x) = 1\\ \text{But }\lim_{x\to 0^-}f(x) = -1\\ \text{Since these are unequal, there is discontinuity}\\ \text{Graphically, the graph jumps up from }y=-1\text{ to }y=1$
$\text{In fact, }f(0)\text{ doesn't even exists.}\\ \text{This further affirms that it's discontinuous}$
$\text{For the sake of the HSC maths courses}\\ \text{Usually a discontinuous function is a piecewise function}$

grace.estelle · « **Reply #1663 on:** March 14, 2017, 10:30:28 pm »

Quote from: RuiAce on March 14, 2017, 10:22:17 pm

C is correct because it is the definition of continuity. The limit as you approach from the left, must always be equal to the limit as you approach from the left, AND equal to f(a) itself to satisfy continuity.

A is wrong because it just means we have a stationary point
B and D are useless. It's saying y = dy/dx, which is irrelevant to continuity.

(Side note: Continuity is, in fact, a prerequisite for differentiability)

Thank you so much!

grace.estelle · « **Reply #1664 on:** March 14, 2017, 10:46:23 pm »

please help!! I don't know how to approach this

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