Author Topic: VCE Methods Question Thread! (Read 6009056 times) Tweet Share

knightrider · « **Reply #7815 on:** January 09, 2015, 12:52:34 pm »

Quote from: Conic on January 09, 2015, 02:13:10 am

I think that should be given full marks. In Q4 of the 2014 exam a function had units in cm, and a question asked for an answer to the nearest mm, so if you want to be absolutely sure you should check the assessment report for that exam when it comes out.

Thanks was it exam 1 or 2?

knightrider · « **Reply #7816 on:** January 09, 2015, 12:57:02 pm »

How would you do this question?

The domain of the function $y=\sqrt{\frac{ x-3}{x-6}$ is

Hancock · « **Reply #7817 on:** January 09, 2015, 01:34:52 pm »

Quote from: knightrider on January 09, 2015, 12:57:02 pm

How would you do this question?

The domain of the function $y=\sqrt{\frac{ x-3}{x-6}$ is

Denominator cannot equal 0. Therefore x =/= 6.
Stuff under the square root has to be positive ->
Case 1 (both numerator and denominator are positive);
Case 2 (both numerator and denominator are negative);

Case 1:
x - 3 >= 0 ------> x >= 3
x - 6 > 0 ------> x > 6
Combining gives: x > 6

Case 2:
x - 3 =< 0 ------> x =< 3
x - 6 < 0 ------> x < 6
Combining gives x =< 3

(Remember that x - 6 =/= 0)

So we have x > 6, x =< 3 and x =/= 6. Since x > 6 already takes into account x =/= 6, we only need two inequalities.

x > 6 and x <= 3

Conic · « **Reply #7818 on:** January 09, 2015, 03:11:13 pm »

Quote from: catherine_mimi on January 09, 2015, 12:50:48 pm

Hello, I have a question (the attached picture)

I am having some trouble with the last question. dii)
I have figured that:
(x-a)^n-1[x(n+m)-(nb+ma)]>0 in order for f'(x)>0
This means that the two factors can be both positive or negative factors.

If both are positive:
x>a and x>(nb+ma)/(n+m)

If both are negative x<a and x<(nb+ma)/(n+m)

Apparently (nb+ma)/(n+m)>a
and that is why there are two answers:
x>(nb+ma)/(n+m) amd x<a

I don't understand why (nb+ma)/(n+m)>a
Can somebody please explain?

We know that $\beta > \alpha$ and that $m,n>0$ . So

$\beta > \alpha \implies n \beta > n \alpha \quad\quad\quad\quad\quad\quad\:\:(\textrm{as }n>0)$

$\implies n \beta + m \alpha > n \alpha + m \alpha \quad (\textrm{adding }m\alpha\textrm{ to both sides})$

$\implies n \beta + m \alpha > (m+n)\alpha$

$\implies \frac{n \beta + m \alpha}{m+n} >\alpha.$

Quote from: knightrider on January 09, 2015, 12:52:34 pm

Thanks was it exam 1 or 2?

Exam 2.

knightrider · « **Reply #7819 on:** January 09, 2015, 03:42:52 pm »

Quote from: Hancock on January 09, 2015, 01:34:52 pm

Denominator cannot equal 0. Therefore x =/= 6.
Stuff under the square root has to be positive ->
Case 1 (both numerator and denominator are positive);
Case 2 (both numerator and denominator are negative);

Case 1:
x - 3 >= 0 ------> x >= 3
x - 6 > 0 ------> x > 6
Combining gives: x > 6

Case 2:
x - 3 =< 0 ------> x =< 3
x - 6 < 0 ------> x < 6
Combining gives x =< 3

(Remember that x - 6 =/= 0)

So we have x > 6, x =< 3 and x =/= 6. Since x > 6 already takes into account x =/= 6, we only need two inequalities.

x > 6 and x <= 3

Thanks Hancock

How does the combining process work?

keltingmeith · « **Reply #7820 on:** January 09, 2015, 04:09:19 pm »

Quote from: knightrider on January 09, 2015, 03:42:52 pm

Thanks Hancock
How does the combining process work?

Combining is simply taking the intersection of the two intervals - i.e, what they have in common.

So, when is x both >6 and >=3? When x>6.
When is x both <6 and <=3? When x<6.

knightrider · « **Reply #7821 on:** January 09, 2015, 04:25:44 pm »

Quote from: EulerFan101 on January 09, 2015, 04:09:19 pm

Combining is simply taking the intersection of the two intervals - i.e, what they have in common.

So, when is x both >6 and >=3? When x>6.
When is x both <6 and <=3? When x<6.

Thanks eulerfan

for your second one isn't it meant to be
when $x\leq 3$

keltingmeith · « **Reply #7822 on:** January 09, 2015, 04:29:37 pm »

Quote from: knightrider on January 09, 2015, 04:25:44 pm

Thanks eulerfan for your second one isn't it meant to be
when $x\leq 3$

... Erm, pretty sure that's what I wrote...?

Just not with TeX because laziness, hahah.

knightrider · « **Reply #7823 on:** January 09, 2015, 04:35:22 pm »

Quote from: EulerFan101 on January 09, 2015, 04:29:37 pm

... Erm, pretty sure that's what I wrote...? Just not with TeX because laziness, hahah.

Oh okay thanks

in your quote it said for the second one
"When is x both <6 and <=3? When x<6." thats why i was just checking.

keltingmeith · « **Reply #7824 on:** January 09, 2015, 04:46:23 pm »

Quote from: knightrider on January 09, 2015, 04:35:22 pm

Oh okay thanks

in your quote it said for the second one
"When is x both <6 and <=3? When x<6." thats why i was just checking.

OH WHOOPS. Hahah, my bad, apparently I'm too tired to be on the forums.

Yep, thanks for catching me! Should <=3, not <6.

knightrider · « **Reply #7825 on:** January 09, 2015, 05:50:34 pm »

How would you do this question?
The function

$(f:x\leq 0\rightarrow R)$ where $f(x)=4-x$ has the range

brightsky · « **Reply #7826 on:** January 09, 2015, 06:03:00 pm »

Sketch f(x) = 4 - x over the domain (-infinity, 0]. You'll see that the range is [4, infinity).

Alternatively, do a little algebra:

x ≤ 0
-x ≥ 0
4 - x ≥ 4

Since y = 4 - x, it follows that:

y ≥ 4

Hence, the range is, once again, [4, infinity).

SE_JM · « **Reply #7827 on:** January 09, 2015, 06:06:02 pm »

Hello~ I have a question i'm having trouble with. I was a bit reluctant to ask more questions (coz, no one answered me on my last question

)

BUT, the world is teeming with questions i don't know and AN has one of the most patient helpers

So, here goes my question:
A rocket is shot vertically upward with an initial velocity of 400 feet per second. Its height s after t seconds is s=400t-16t². How fast is the distance changing from the rocket to an observer on the ground 1800 feet away from the launching site, when the rocket is still rising and is 2400 feet above the ground?

The correct answer is: increasing at a rate of 64 feet per second

Thank you!

knightrider · « **Reply #7828 on:** January 09, 2015, 06:32:53 pm »

Quote from: brightsky on January 09, 2015, 06:03:00 pm

Sketch f(x) = 4 - x over the domain (-infinity, 0]. You'll see that the range is [4, infinity).

Alternatively, do a little algebra:

x ≤ 0
-x ≥ 0
4 - x ≥ 4

Since y = 4 - x, it follows that:

y ≥ 4

Hence, the range is, once again, [4, infinity).

Thanks brightksky

how did you get this part from ?
$4 - x \geq 4$

keltingmeith · « **Reply #7829 on:** January 09, 2015, 06:35:41 pm »

Quote from: SE_JM on January 09, 2015, 06:06:02 pm

Hello~ I have a question i'm having trouble with. I was a bit reluctant to ask more questions (coz, no one answered me on my last question )

OOPS. Swear I got to everyone, sorry. >.< I will answer you before the day is done, I swear! I am not allowed to go on leave tomorrow until you have been helped!

(srs tho, never be afraid to ask questions. We're very scary people, but we're like that to everyone, so life lessons in bravery +1)

Quote from: SE_JM on January 09, 2015, 06:06:02 pm

BUT, the world is teeming with questions i don't know and AN has one of the most patient helpers

So, here goes my question:
A rocket is shot vertically upward with an initial velocity of 400 feet per second. Its height s after t seconds is s=400t-16t². How fast is the distance changing from the rocket to an observer on the ground 1800 feet away from the launching site, when the rocket is still rising and is 2400 feet above the ground?

The correct answer is: increasing at a rate of 64 feet per second

Thank you!

Sounds like a rate question! Let's see what we can get:

1. Identify the rate we want. It's with respect to time (since we want to know how fast), and we want to know distance from the observer. So, we'll call its distance from the observer x, and we want to find dx/dt.

2. Diff x(t). If we do not have x(t), set up a rate equation so we can find it. We know s, so we could use $\frac{dx}{dt}=\frac{dx}{ds}\cdot\frac{ds}{dt}$

3. Diff x(s) and s(t).

Now, finding x(s) might seem impossible at first - BUT, we know the horizontal distance from the rocket, and the vertical, so we can set up a triangle to help us out! Using Pythag, we have $x^2=s^2+1800^2$ . Solving for x, we get $x=\sqrt{s^2+1800^2}$ . NOTE, we ignore the negative value, because x MUST be positive (it is a distance, after all).

So, $\frac{dx}{ds}=\frac{s}{\sqrt{s^2+1800^2}}$ and $\frac{ds}{dt}=400-32t$ . Putting this together, we get:

$\frac{dx}{dt}=\frac{dx}{ds}\cdot\frac{ds}{dt} \\ \frac{dx}{dt}=\frac{s}{\sqrt{s^2+1800^2}}\cdot\left(400-32t)$

Now we just need to plug and play! We want to know the rate when s=2400. This does pose a problem, since we have a t in the equation - so, we put s=2400 into the equation for t, and solve for t:

$2400=400t-16t^2 \\ t^2-25t+150=0 \\ (t-10)(t-15)=0 \\ t=10,15$

BUT, we want to know about when the rocket is still rising - at t=15, the rocket is now falling, so we take t=10. Now, we can use our formula:

$\frac{dx}{dt}=\frac{2400}{\sqrt{2400^2+1800^2}}\cdot\left(400-32(10)\right)) \\ \frac{dx}{dt}=\frac{2400}{3000}\cdot\left(80\right) \\ \frac{dx}{dt}=64$

Which is what they claim to be right~
Of course, you could've substituted s(t) in instead of using the related rates method as I did. I just felt that my way was a bit cleaner~

Quote from: knightrider on January 09, 2015, 06:32:53 pm

Thanks brightksky
how did you get this part from ?
$4 - x \geq 4$

From -x>=0, add 4 to both sides.

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