3U Maths Question Thread

[Mahan]

Hey Mahan!! Thanks so much for tending to my questions so promptly they make a lot of sense 

I was wondering if there had been a formatting malfunction in the LHS bar statement towards the top of the solution and if there was anyway to fix it - if not that's ok. Thanks again, i'll definitely have more questions soon 

I fixed it. 

[Kle123]

This question stems from a 4u exam question but ill just try to illustrate the problem on geogebra. My question: if you were just given the gradients (i.e. no scaled diagram) how would you know if angle ABC is obtuse (which it clearly is)?

[Kle123]

Can someone pls explain how you get x=-5 in this. i got x=4 but not -5 as well.

|2x+5| + |x| = 17

Thx very much! 

maybe because -5 isnt a solution? (you can test it out by subbing it in if you want) LHS=10 =/= RHS. x=-22/3 is the other solution.

[MEL7401]

Yep its not a solution...
sorry looked at wrong answer 
but figured it out now

[Shadowxo]

This question stems from a 4u exam question but ill just try to illustrate the problem on geogebra. My question: if you were just given the gradients (i.e. no scaled diagram) how would you know if angle ABC is obtuse (which it clearly is)?

Hi (from a VCE perspective, so not sure if it's valid for HSC)
First way would be by intuition, you can tell by just a rough sketch.
Mathematically, you could find the difference in the angles they make with the positive direction of the x axis. m=tan(x), so the angle for the 1/2 gradient would be arctan(1/2), and the angle for the -1 gradient would be 180+arctan(-1), as arctan(-1) is negative. Find the difference between them, 180+arctan(-1)-arctan(1/2), and that's your angle, 108º.
Also, you know m1*m2=-1 if two lines are perpendicular (90º), and you could use that.

[RuiAce]

This question stems from a 4u exam question but ill just try to illustrate the problem on geogebra. My question: if you were just given the gradients (i.e. no scaled diagram) how would you know if angle ABC is obtuse (which it clearly is)?

As far as the HSC goes you wouldn't have this sort of situation unless you found the points A and C to begin with, OR this is a complex locus question. Best post the previous parts.

[MEL7401]

I need help!
I got how x= 22/15 but not 4/3

Thanks

[RuiAce]

I need help!
I got how x= 22/15 but not 4/3

Thanks

Can you please post your working? There are many guidelines as to how you are expected to approach this question. If you only got the x<22/15 part then it is likely you did not follow one of them.

[MEL7401]

Dont worry about other question... thx anyway

but could someone pls help with
3x+1/x-4 > to 1/3

Thx heaps!

[jamonwindeyer]

Dont worry about other question... thx anyway

but could someone pls help with
3x+1/x-4 > to 1/3

Thx heaps!

Just as above, there are a few ways to do this question. I'll show you the cleanest, but you might not have seen it yet: If you haven't, post your working and I'll show it done the way you attempted it!



The issue here of course is that multiplying by \(x-4\) could change the direction of the inequality. We can fix this by multiplying instead by \((x-4)^2\), which is guaranteed to be positive



Note that there are shortcuts you could have taken there, but I think this is clearer. Once you are here, you need to draw a sketch of the parabola \(y=8x^2-25x-28\), using the intercepts you find in the factored form. Then use that sketch to see where the parabola is above or on the \(x\)-axis - And those regions are your answers. Here, you'll get \(x\ge4, x\le\frac{-7}{8}\)

If you've seen this method before, hopefully it makes sense. If not, show an example/attempt of the working you would have tried! I'll make mine match

[legorgo18]

Hey can i get some help with this question (I know its really easy dont judge pls xd but i just forgot everything about locus)

A is a point where the circle with eq x^2 + y^2 = 16 cuts the x-axis. Find the locus of the midpoints of all chords of this circle that contain the point A.

Cheers

[MEL7401]

Just as above, there are a few ways to do this question. I'll show you the cleanest, but you might not have seen it yet: If you haven't, post your working and I'll show it done the way you attempted it!



The issue here of course is that multiplying by \(x-4\) could change the direction of the inequality. We can fix this by multiplying instead by \((x-4)^2\), which is guaranteed to be positive



Note that there are shortcuts you could have taken there, but I think this is clearer. Once you are here, you need to draw a sketch of the parabola \(y=8x^2-25x-28\), using the intercepts you find in the factored form. Then use that sketch to see where the parabola is above or on the \(x\)-axis - And those regions are your answers. Here, you'll get \(x\ge4, x\le\frac{-7}{8}\)

If you've seen this method before, hopefully it makes sense. If not, show an example/attempt of the working you would have tried! I'll make mine match

Hey thx Jamon!
Yep- i usually use that method too...
just wasn't working out tho 
but i understand why now!!

[Dragomistress]

May someone please help me with this question.

Thanks!

[MEL7401]

When studying 4 a 3U exam- do u recommend doing chapter reviews or past papers???..... when ur running out of time!!!

[kiwiberry]

When studying 4 a 3U exam- do u recommend doing chapter reviews or past papers???..... when ur running out of time!!!

Definitely past papers if you're running out of time!! Past papers the best way to prepare for any maths exam anyway, because they're the only thing that will give you an idea of what the actual exam will be like.