Exam Questions (1 Viewer)

[Shazer2]

I was wondering if I could get some help with a few questions from the exam I sat today.











Thanks.

 

[nexus_lad_69]

1) 2^x x 2^2(x+1) = 2^-1

x + 2(x+1) = -1

 

[Shazer2]

Is there any chance you can use LaTeX? I find it hard to understand plain text. If you find writing LaTeX difficult, you can get an extension for chrome called "Daum equation editor" to make it easier.

Thank you!

 

[deswa1]

Is there any chance you can use LaTeX? I find it hard to understand plain text. If you find writing LaTeX difficult, you can get an extension for chrome called "Daum equation editor" to make it easier.

Thank you!

If you give me fifteen mins- I'll write up all the answers by hand and scan them- hopefully others can latex some whilst I'm doing them. Also, 4 has more than one answer

 

[Carrotsticks]

Question 1



But all of this is equal to 1/2, which is the same thing as 2^{-1}

So equating powers of 2:

3x+2 = -1

3x = -3

x = -1

Question 2

Domain: All real x because there are no possible discontinuities or undefined regions.

Range: y >= 0 because it is a quadratic with the vertex on the x axis (specifically at x=-3).

Question 3

Sub the point in and show that LHS = RHS.

2(1) + 3(4) - 14 = 0 = RHS

Question 4

If you draw a sketch, you can clearly see that D will be on the bottom right of the parallelogram.

To get from B to C, we add 2 to the x coordinate and subtract 2 from the y coordinate.

So to get from A to D, we do the same. So D is (2,-2)

D can also be seen to be on the top right, where you apply the same mentality to see that to get from A to C, we add 3 to the x coordinate and 5 to the y coordinate. So similarly to get from B to D, we do the same ie: D is (4,12)

Question 5

 

[Shazer2]

If you give me fifteen mins- I'll write up all the answers by hand and scan them- hopefully others can latex some whilst I'm doing them. Also, 4 has more than one answer

No problem! There was a lot more that I had trouble with in the exam (didn't complete whole last page, 15 marks down the drain).

 

[Shazer2]

Question 1



But all of this is equal to 1/2, which is the same thing as 2^{-1}

So equating powers of 2:

3x+2 = -1

3x = -3

x = -1

Question 2

Domain: All real x because there are no possible discontinuities or undefined regions.

Range: y >= 0 because it is a quadratic with the vertex on the x axis (specifically at x=-3).

Question 3

Sub the point in and show that LHS = RHS.

2(1) + 3(4) - 14 = 0 = RHS

Question 4

If you draw a sketch, you can clearly see that D will be on the bottom right of the parallelogram.

To get from B to C, we add 2 to the x coordinate and subtract 2 from the y coordinate.

So to get from A to D, we do the same. So D is (2,-2)

D can also be seen to be on the top right, where you apply the same mentality to see that to get from A to C, we add 3 to the x coordinate and 5 to the y coordinate. So similarly to get from B to D, we do the same ie: D is (4,12)

Question 5

Alright, I got the start of question 1 right in the exam, I got to but then stopped. Will I get any marks for that? Also, I believe I got the domain/range question correct, but I didn't show any working I just stated the domain and range. Will I get ANY marks there?

I should have known how to do Q3, but clearly didn't. Question 4, I'm still confused by, I did draw a sketch and realised it would be at the bottom right. Question 5, I see now, couldn't think how to do so in exam.

Thanks Carrotsticks.

 

[deswa1]

@Carrot- how the eff do you latex/type that fast? That was amazing. Oh and I think for 4) there is another point (-2,2) considering where AC is parallel to BD

 

[Carrotsticks]

Alright, I got the start of question 1 right in the exam, I got to but then stopped. Will I get any marks for that? Also, I believe I got the domain/range question correct, but I didn't show any working I just stated the domain and range. Will I get ANY marks there?

I should have known how to do Q3, but clearly didn't. Question 4, I'm still confused by, I did draw a sketch and realised it would be at the bottom right. Question 5, I see now, couldn't think how to do so in exam.

Thanks Carrotsticks.

You would get maybe 1 mark for doing that, but it really depends on how many marks were allocated to it overall. If it was worth 3, then you would definitely get 1 mark. If worth 2, maybe if you're lucky. If worth 1, then no.

Different schools mark differently. I know for sure my school would have docked off a mark if I didn't include a sketch of the curve, showing the domain/range. Other schools let you get away with just stating it for more simple questions like that. Domain/Range can get quite complicated later on when doing Inverse Trig, so some working out is needed then. For this, perhaps not.

Q4 can be bottom right, or top right. Sketch it, and you will see.

 

[Shazer2]

@deswa1, if you run Chrome, you should look into the Daum equation editor extension. It allows you to click and type to generate LaTeX code. It's very fast and efficient.

 

[J2good4u]

was that the independent 2 hour exam. My brother also sat it today he said it was alright... how did you go?

 

[Carrotsticks]

@Carrot- how the eff do you latex/type that fast? That was amazing. Oh and I think for 4) there is another point (-2,2) considering where AC is parallel to BD

Yep you are correct, I missed that case =)

 

[Shazer2]

was that the independent 2 hour exam. My brother also sat it today he said it was alright... how did you go?

I'm not sure what independent is? The exam, yeah, I went very bad.

 

[Shazer2]

You would get maybe 1 mark for doing that, but it really depends on how many marks were allocated to it overall. If it was worth 3, then you would definitely get 1 mark. If worth 2, maybe if you're lucky. If worth 1, then no.

Different schools mark differently. I know for sure my school would have docked off a mark if I didn't include a sketch of the curve, showing the domain/range. Other schools let you get away with just stating it for more simple questions like that. Domain/Range can get quite complicated later on when doing Inverse Trig, so some working out is needed then. For this, perhaps not.

Q4 can be bottom right, or top right. Sketch it, and you will see.

I'm not quite sure what the marking for it was like. For the sketch, I sketched something that looked like this:

But I don't understand how you know to "add 2 to x and subtract 2 from y" or whatever?

 

[J2good4u]

That's alright, its only year 11, did you have problems with anything else?

 

[Shazer2]

That's alright, its only year 11, did you have problems with anything else?

Yeah the whole back page was trigonometry and I didn't really answer any of it. I also have trouble with proofs for triangles.

 

[deswa1]

These are the three possible positions of D that make it a parallelogram (3 is stuffed up because it doesn't really fit on your diagram but the point should be clear):

 

[J2good4u]

Those are generally tough, especially the two unit ones, if you want I help you out with it.

 

[Shazer2]

Yeah the diagram I drew in the exam was the 1st one, deswa1. I'm wondering, how do you actually FIND that point? Is there a certain way to do it? I'm unsure.

 

[deswa1]

Yeah the diagram I drew in the exam was the 1st one, deswa1. I'm wondering, how do you actually FIND that point? Is there a certain way to do it? I'm unsure.

There's a long way and a shortcut:

Long way:
1. Note that BC is parallel to AD. Find the gradient of BC and this must therefore be the gradient of AD. Find the equation of the line AD.
2. Note also that DC is parallel to AB. Find the gradient of AB and this must therefore be the gradient of CD. Find the equation of the line CD.
3. Now D lies on both AD and CD, solve the two equations simultaneously and you'll find the point D.

Short way: Recognise that BC is parallel to AD. Also note that to get from B to C, you increase the x co-ordinate by 2 and decrease the y co-ordinate by 2. Do the same to AD (because the two lines are parallel). So A is the point (0,0). By increasing/decreasing the x and y co-ordinate respectively by 2, you get D to be the point (2,-2). Do the same for the other points