cherylim23's methods question thread :)

[|ll|lll|]

Yeah, but you could also say that when Dis = 0, there are two real solutions. Just that solutions are the same...

[iNerd]

Thought three complex solutions would be weird (thanks for the clarification)... On the solution, yep, one real distinct solution for an inflection cubic, but remember that are really three real solutions (just one -one real one- is repeated three times) <--- wording is important. I never thought of that (the 'distinct stuff') until I read kyzoo's tips... Don't think its taught too well in textbooks

Yes, think about it this way. When the discriminant is zero, there are actually two solutions, but two repeated solutions that's all.

Hence arises the confusion when the question states, for what values of (insert variable) would the equation have two solutions.
I don't know whether to let discriminant > 0 or discriminant >= 0

IMO the question should state two distinct solutions...otherwise let discriminant > O

[|ll|lll|]

Also, for a cubic polynomial, when there are two real and equal roots, the last root must definitely be real and not complex yes? Is there any proof for this?

[m@tty]

If all co-efficients are real, then yes, that is correct.

Say the two real roots of a polynomial f with real co-efficients are at x=a, and a third root is x=b+ci.







But since the coefficients must all be real, c must equal 0.

So the third root is also "real" rather than "complex".

(I will remind you here that all real numbers are in fact complex numbers as . ie. 2=2+0i)

If however, there were imaginary coefficients, the third root could be "complex".

This is more like an algebraic proof - but there are much more intuitive ways of understanding it. Think of it graphically, or if you are familiar with the conjugate root theorem for polynomials with real coefficients, then it is clearly seen that there must be a third real root.

[|ll|lll|]

^ Thanks m@tty

As for the attached question, I don't get why the answer is E and not A. They are both technically right...

[pi]

As for the attached question, I don't get why the answer is E and not A. They are both technically right...

They are both right, especially as no horizontal scaling is given... See m@tty's post, I made a stupid error...

[m@tty]

Ah, no. It is asymptotic to y=0 - therefore only E.

If A was , then it would be another possibility with the given info. But it is not.

[iNerd]

Ah, no. It is asymptotic to y=0 - therefore only E.

If A was , then it would be another possibility with the given info. But it is not.

Haven't done exponentials in ages but does the +1 in A indicate that the asymptote is at y = 1 ?

[pi]

Ah, no. It is asymptotic to y=0 - therefore only E.

If A was , then it would be another possibility with the given info. But it is not.

Haven't done exponentials in ages but does the +1 in A indicate that the asymptote is at y = 1 ?

Yep, that right. lol, can't believe I got that wrong... (better wrong now than in an exam though). Thanks m@tty, that was a really stupid error on my part.

[m@tty]

Haha, it's ok. Maths is actually really easy to make errors in..

Ah, no. It is asymptotic to y=0 - therefore only E.

If A was , then it would be another possibility with the given info. But it is not.

Haven't done exponentials in ages but does the +1 in A indicate that the asymptote is at y = 1 ?

Yep.

You don't need to remember it. Think, as x becomes very large() hence .

[|ll|lll|]

I should facepalm myself because it took me so long and I never figured that out. If it was A, the asymptote would be at y=1 I assume.

Thanks again, matty.

/Edit: and thanks to everyone else

[|ll|lll|]

For this particular question attached, there may be more than one approach to part c)

How would any of you guys do it?

I did it differently from the worked solutions...


/Edit: Also, comparing these two graphs: and
What's the transformation from f(x) to g(x)?

Thanks for reading

[pi]

I would have found the amplitude (and hence minimum) and then subbed in t=9 to prove it was a minimum. Then (just to be sure), would have done the gradient table thing to show that it is in fact a minimum.

[|ll|lll|]

Thanks Rohitpi. I wouldn't have thought of the gradient table though.

This is what the worked solutions showed - I reckon it's a longer method

/Edit: Actually, technically speaking, during that point of the year, say that question came out on a SAC, I don't think a gradient table would be required because we haven't learnt to differentiate circular functions 8-)

[pi]

Thanks Rohitpi. I wouldn't have thought of the gradient table though.

This is what the worked solutions showed - I reckon it's a longer method

I think my (or your) method is the reverse of that... Only difference is that ours would probably save a couple of minutes. I think with a gradient table, both would be marked equally.