4U Maths Question Thread

[shekhar.patel]

In the Vector question, I am getting a different answer to the book's. My answer and book's answer is attached. Would you please help me where I went wrong or if the book answer is wrong.

[fun_jirachi]

Hi there, I was trying to just say that the conjugate only affects the imaginary part for example, x= a+ib then x bar = a- ib.
But yes, I see your way of solving the question. Many thanks for the help 

This makes things clearer - but as with your original statement 'since the conjugate roots only applies to imaginary components we can say a, b and c are real' - the jump from the first part to the second part is very awkward; the logic that takes you from A to B is very flawed. I can't quite see how or why that was assumed?Perhaps explaining your line of thinking could help us better your understanding Though it's true in this case, you can't assume this every time, and you certainly shouldn't be assuming things in the first place - it's much safer to use one of the methods described above

In the Vector question, I am getting a different answer to the book's. My answer and book's answer is attached. Would you please help me where I went wrong or if the book answer is wrong.

The book's answer is wrong. In future, a more helpful check (when given the value of each of the complex numbers!) is to actually compute what z-w is: in this case it's \(\left(1-\frac{1}{\sqrt{2}}\right) + \left(\sqrt{3}+\frac{1}{\sqrt{2}}\right)i\) - which is clearly in the first quadrant, not on the imaginary axis as the book's answer suggests. In actuality, the book's answer and your answer is identical - the book likely hasn't taken care in drawing a diagram properly, as well as positioning the origin of the ray at the actual origin.

Hope this helps

[shekhar.patel]

This makes things clearer - but as with your original statement 'since the conjugate roots only applies to imaginary components we can say a, b and c are real' - the jump from the first part to the second part is very awkward; the logic that takes you from A to B is very flawed. I can't quite see how or why that was assumed?Perhaps explaining your line of thinking could help us better your understanding Though it's true in this case, you can't assume this every time, and you certainly shouldn't be assuming things in the first place - it's much safer to use one of the methods described above

The book's answer is wrong. In future, a more helpful check (when given the value of each of the complex numbers!) is to actually compute what z-w is: in this case it's \(\left(1-\frac{1}{\sqrt{2}}\right) + \left(\sqrt{3}+\frac{1}{\sqrt{2}}\right)i\) - which is clearly in the first quadrant, not on the imaginary axis as the book's answer suggests. In actuality, the book's answer and your answer is identical - the book likely hasn't taken care in drawing a diagram properly, as well as positioning the origin of the ray at the actual origin.

Hope this helps

Hi, yes the reasoning behind checking the answer is helpful. Thanks

[shekhar.patel]

I need help in this question, I tried to rotate vector by +- i but I am getting the wrong answer.

[fun_jirachi]

Hey there!

Note that while \(\vec{AB}\) represents the same complex number as \(\vec{OC}\), you can't just use that to find \(\vec{OB}\), since \(\vec{OB}\) clearly does not equal \(\vec{OC}\) (one is the diagonal of the square, the other is a side!). What you've actually done is rotate  \(\vec{OA}\) clockwise 90 degrees to give \(\vec{OC}\). 

There are two ways to do this:
a) Consider that \(\vec{OB}=\vec{OA}+\vec{AB}\) - you've got both values; the first given in the question and the second you've just calculated as equal to  \(\vec{OC}\). Add them up and you'll have \(\vec{OB}\).

b) Consider that \(\vec{AB}\) is a rotation anti-clockwise of \(\vec{AO}\) by 90 degrees. Recalling that  \(\vec{AO}\) = \(-\vec{OA}\), we find that \(\vec{AO}=-2-i\). Rotating to find \(\vec{AB}\), the computation then becomes identical to the one above via addition of vectors.

There technically is also a 'cheat' way of doing it by inspection; it's a relatively simple square and some people might pick up that B is 1+3i straightaway, but there's no working out involved; for the most part this is only helpful for checking your answer.

Hope this helps

[shekhar.patel]

Hey there!

Note that while \(\vec{AB}\) represents the same complex number as \(\vec{OC}\), you can't just use that to find \(\vec{OB}\), since \(\vec{OB}\) clearly does not equal \(\vec{OC}\) (one is the diagonal of the square, the other is a side!). What you've actually done is rotate  \(\vec{OA}\) clockwise 90 degrees to give \(\vec{OC}\). 

There are two ways to do this:
a) Consider that \(\vec{OB}=\vec{OA}+\vec{AB}\) - you've got both values; the first given in the question and the second you've just calculated as equal to  \(\vec{OC}\). Add them up and you'll have \(\vec{OB}\).

b) Consider that \(\vec{AB}\) is a rotation anti-clockwise of \(\vec{AO}\) by 90 degrees. Recalling that  \(\vec{AO}\) = \(-\vec{OA}\), we find that \(\vec{AO}=-2-i\). Rotating to find \(\vec{AB}\), the computation then becomes identical to the one above via addition of vectors.

There technically is also a 'cheat' way of doing it by inspection; it's a relatively simple square and some people might pick up that B is 1+3i straightaway, but there's no working out involved; for the most part this is only helpful for checking your answer.

Hope this helps

Hi there. Yes I see the way of doing it, I think a) is much easier for me to follow, but I don't get how OB=OA + AB, and the fact that we added OA after rotating is not quite clear. Would you please explain me why that is. Thanks

[shekhar.patel]

I just need some clarification here about why the following cannot be one of the possible positions.
Thanks

[fun_jirachi]

Hi there. Yes I see the way of doing it, I think a) is much easier for me to follow, but I don't get how OB=OA + AB, and the fact that we added OA after rotating is not quite clear. Would you please explain me why that is. Thanks

\(\vec{OB}=\vec{OA}+\vec{AB}\) because in general, for any complex numbers X, Y and Z, \(\vec{XZ}=\vec{XY}+\vec{XZ}\) - consider the following diagram:

Click for the diagram!

Also, consider the square we have as well, as per the diagram below:

Click for the next diagram!

When we rotate \(\vec{OA}\), we have \(\vec{OC}\), which you'll note is identical to \(\vec{AB}\). From the above, we must then add to have \(\vec{OB}\). Note that here \(\vec{OB}\) is the diagonal of the square; you'll need to add the two sides to get the value of the complex number.

I just need some clarification here about why the following cannot be one of the possible positions.
Thanks

Before I begin assisting you with this question, have you tried using the Argand diagram to help you in working out the answer? It's very helpful, and may make you realise why the answers are correct

Hope this helps

[shekhar.patel]

\(\vec{OB}=\vec{OA}+\vec{AB}\) because in general, for any complex numbers X, Y and Z, \(\vec{XZ}=\vec{XY}+\vec{XZ}\) - consider the following diagram:

Click for the diagram!

Also, consider the square we have as well, as per the diagram below:

Click for the next diagram!

When we rotate \(\vec{OA}\), we have \(\vec{OC}\), which you'll note is identical to \(\vec{AB}\). From the above, we must then add to have \(\vec{OB}\). Note that here \(\vec{OB}\) is the diagonal of the square; you'll need to add the two sides to get the value of the complex number.

This is clear now. I understand it. Thanks

[Vxncent288]

Hi,
I was wondering what resources I could use to study for the new 3U and 4U courses? Are there certain websites I can find past HSC papers? I'm a bit worried about studying for mathematics with the new syllabuses having come in so any advice/recommendations would be greatly appreciated!

[shekhar.patel]

Hi,
I was wondering what resources I could use to study for the new 3U and 4U courses? Are there certain websites I can find past HSC papers? I'm a bit worried about studying for mathematics with the new syllabuses having come in so any advice/recommendations would be greatly appreciated!

I mean there are only so many changes in the new syllabus. You can use the new Cambridge 3 unit and 4 unit textbooks. They contain really great questions, and relevant to the new syllabus. If you are doing past hsc papers ( find them on THSC) and you feel that you haven’t quite learnt the content to do the question, check the NESA syllabus and see if that topics is listed there. If it’s not just chill. Having a good understanding and knowing your syllabus is definitely helpful. Maybe print out the major topics and outcomes of the syllabus and stick it on your study desk.
Other than that the resources that are available are your teachers, tutor, textbooks, syllabus, past papers and your creativity.
Hope that helps 

[shekhar.patel]

Hi, I am not sure how to do this question, any hints to start off will be much appreciated .
Thanks

[RuiAce]

Hi, I am not sure how to do this question, any hints to start off will be much appreciated .
Thanks

For \(y^2 = f(x)\) curves, the curve is always symmetric about the \(x\)-axis. Hence the area under the \(x\)-axis should equal the area above it.

The upper curve is \( y = x^{\frac{3}{2}} \). (Recall that square rooting is equivalent to taking powers of a half.)

So the area under the upper curve will be \( \int_0^3 x^{\frac32}\,dx \). The combined area will be double this.

[shekhar.patel]

Ok, I see. Got the answer thanks.

[shekhar.patel]

In this question we need to find the area of the curve bounded by the abscissas. So we need to make x the subject. I have many operations but I can't find a way of making x the subject.