4U Maths Question Thread

[RuiAce]

oh i see what the k does it shifts the graph up and that restricts the the cubic to intersect the x axis more than once. Thank you heaps Rui deeply appreciate it even if you had no internet. The other conics que i asked is from sk patel (6E q8)  textbook, take your time and thank you for helping us out.

(No worries - Pretty much PDFs are a go-to option if I'm trying to answer it without access to a computer with internet, because I can still use the full LaTeX package. The other option is I pull out a whiteboard, which I didn't have at the time either)

I wonder what SK Patel was thinking when he wrote that question; that makes no sense in the literal manner. I've reinterpreted it to be something like this:




I feel as though the wording of the question is what's hard. Not the actual maths involved

[kaustubh.patel]

I'd probs agree with you that the question is very ambiguous (especially after seeing the working) but in mosts part i see it just finding the x intercept with the tangent (i learnt how to derive that before) and the x value of P. Thank you again, I usually do homework late at night any my friend are are probs sleeping so I really appreciate your help.

[arii]

I'm not sure how I'd find the gradient for this question:

Find the equation of the tangent to the curve sqrt(x)+sqrt(y)=sqrt(c) at the point P(a,b) on the curve.

[Sine]

I'm not sure how I'd find the gradient for this question:

Find the equation of the tangent to the curve sqrt(x)+sqrt(y)=sqrt(c) at the point P(a,b) on the curve.

[arii]

Thanks Sine. I managed to successfully figure out the rest.
Anyway, bumped into another question I wasn't sure how to do...

Prove that x^3+y^3=3xy is symmetrical about the line y=x

[RuiAce]

Thanks Sine. I managed to successfully figure out the rest.
Anyway, bumped into another question I wasn't sure how to do...

Prove that x^3+y^3=3xy is symmetrical about the line y=x

You proved something was symmetrical about y=0 (even function) by just showing that the replacement of x with -x didn't change the given equation. You do the same thing here by showing that if you replace x with y (and also y with x), it's symmetrical about the line y=x.

Which is obviously the case, because upon doing this replacement you get \( y^3 + x^3 = 3yx \) which rearranges quickly back into the original equation.

Nor is this a part of the 4U syllabus.

[Dragomistress]

How do you do this?
|2iy|≤4 (Not absolute value)

[RuiAce]

How do you do this?
|2iy|≤4 (Not absolute value)

If that's not an absolute value then I actually don't know what that means. You can't have \(i\) appearing in an equality by itself.

(Modulus and absolute value mean the same thing if that's what you intended to mean)

[Dragomistress]

The question is sketch |z-w|≤4 (w is conjugate of z)

[RuiAce]

The question is sketch |z-w|≤4 (w is conjugate of z)

That is the absolute value then. (As mentioned above, the modulus and the absolute value mean the same thing.)
\begin{align*}|2iy|&\le 4\\ 2y &\le 4\\ y&\le 2 \end{align*}

[Dragomistress]

But apparently, it is -2≤y≤2

[jazzycab]

But apparently, it is -2≤y≤2

The solution to this is \(-2\le y\le 2\) (look at a graph of \(y=x^2-4\) to see why)

[RuiAce]

But apparently, it is -2≤y≤2

Oh my bad. Technically speaking I should have left it as |y|≤2 Which solves to give -2≤y≤2 (it’s  an absolute value inequality).

In general, one can assume that |i|=1 but one can not assume |y|=y. This holds only when y is non negative, which is something we cannot confirm.

[arii]

Find the general solutions of the 1/cos(3x)=1/sin(2x). I got up to the line attached, but I'm not sure if I've lost solutions on the way. (Still haven't figured out how to input fractions and stuff despite Rui's guide)

[RuiAce]

Find the general solutions of the 1/cos(3x)=1/sin(2x). I got up to the line attached, but I'm not sure if I've lost solutions on the way. (Still haven't figured out how to input fractions and stuff despite Rui's guide)

Code: [Select]

[tex]\frac{a}{b}[/tex]
Any further questions should be asked over in that thread
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At no point anywhere else in the proof did you attempt to 'cancel things out', so that was the only instance that was risky.