1. ## Re: International Baccalaureate Maths Marathon

I came across a Maths SL question today. Wasn't sure if this answer was correct from this certain Maths SL textbook.

$lim_{x \rightarrow \infty} \ \frac{3x^2+2}{x-3}$

I always thought that dominate powers held true. The back of the back has the answer as "D.N.E (increases without bound)"

2. ## Re: International Baccalaureate Maths Marathon

Originally Posted by davidgoes4wce
I came across a Maths SL question today. Wasn't sure if this answer was correct from this certain Maths SL textbook.

$lim_{x \rightarrow \infty} \ \frac{3x^2+2}{x-3}$

I always thought that dominate powers held true. The back of the back has the answer as "D.N.E (increases without bound)"
They are basically saying the limit is infinity (the numerator has higher degree than denominator).

3. ## Re: International Baccalaureate Maths Marathon

My answer would have been $+ \infty$

4. ## Re: International Baccalaureate Maths Marathon

The interval notation seems a bit different to what I have read in the past. (This is from an IB source as well)

Would it be right for the 2nd row to write it along the lines of:

$x \in (a,b)$

I noticed them using the brackets ][

$The book writes it as x \in ]a \ b[$

5. ## Re: International Baccalaureate Maths Marathon

Originally Posted by davidgoes4wce

The interval notation seems a bit different to what I have read in the past. (This is from an IB source as well)

Would it be right for the 2nd row to write it along the lines of:

$x \in (a,b)$

I noticed them using the brackets ][

$The book writes it as x \in ]a \ b[$
Yes, you can use (a, b). You can read about the two notations here: https://en.wikipedia.org/wiki/Interv...ding_endpoints .

6. ## Re: International Baccalaureate Maths Marathon

Originally Posted by davidgoes4wce
My answer would have been $+ \infty$
Keep in mind that infinity is not a number.

7. ## Re: International Baccalaureate Maths Marathon

I think there is a typo in this question for Q 13 (b) . I cant get why its $p^2 \leq 3pq$

8. ## Re: International Baccalaureate Maths Marathon

Originally Posted by davidgoes4wce

I think there is a typo in this question for Q 13 (b) . I cant get why its $p^2 \leq 3pq$
$\noindent It's a typo in the definition of the function, it should just be$

$f(x) = x^{3} + px^{2} + qx,$

$\noindent i.e. leading coefficient should be 1.$

9. ## Re: International Baccalaureate Maths Marathon

Originally Posted by InteGrand
$\noindent It's a typo in the definition of the function, it should just be$

$f(x) = x^{3} + px^{2} + qx,$

$\noindent i.e. leading coefficient should be 1.$
If I do that the derivative becomes

$f'(x)=3x^2+2px+q \geq 0$

10. ## Re: International Baccalaureate Maths Marathon

Originally Posted by davidgoes4wce
If I do that the derivative becomes

$f'(x)=3x^2+2px+q \geq 0$
They mean the derivative is non-negative for all (real) x. We can thus use discriminants to obtain the desired inequality.

11. ## Re: International Baccalaureate Maths Marathon

Originally Posted by InteGrand
They mean the derivative is non-negative for all (real) x. We can thus use discriminants to obtain the desired inequality.
$\triangle=b^2-4ac$

$(2p)^2-4(3)(q) \geq 0$

$4p^2-12q \geq 0$

$4p^2 \geq 12q$

$p^2 \geq 3q$

This is the opposite direction of the inequality required in the proof. There has been a type in the whole question.

12. ## Re: International Baccalaureate Maths Marathon

Originally Posted by davidgoes4wce
$\triangle=b^2-4ac$

$(2p)^2-4(3)(q) /geq 0$

$4p^2-12q /geq 0$

$4p^2 \geq 12q$

$p^2 \geq 3q$

This is the opposite direction of the inequality required in the proof. There has been a type in the whole question.
$\noindent The question is correct. The discriminant inequality should be the other way round, i.e. \Delta \leq 0 if the quadratic is non-negative for all real x.$

13. ## Re: International Baccalaureate Maths Marathon

By the way this is a question from Newington, seems like the staff don't know how to teach the subject properly there.

By setting the discriminant to less than equal to zero, we are trying to prove that there are no solutions for the real values of x, using the quadratics roots formula. Right?

14. ## Re: International Baccalaureate Maths Marathon

Originally Posted by davidgoes4wce
By the way this is a question from Newington, seems like the staff don't know how to teach the subject properly there.

By setting the discriminant to less than equal to zero, we are trying to prove that there are no solutions for the real values of x, using the quadratics roots formula. Right?
Why do you think they don't know how to teach the subject properly there? Is it something their worked answers said?

The reason we can set the discriminant to be less than or equal to 0 is that we are told the quadratic is non-negative for all real x ("non-negative definite"). This implies (since it's a quadratic with real coefficients) that its discriminant is non-positive.

15. ## Re: International Baccalaureate Maths Marathon

I'll take your word for it Integrand, I am still confused as to what non-negative means in this case.

I thought the discriminant for sure was $\triangle \geq 0$ reading the question first up.

16. ## Re: International Baccalaureate Maths Marathon

Originally Posted by davidgoes4wce
I'll take your word for it Integrand, I am still confused as to what non-negative means in this case.

I thought the discriminant for sure was $\triangle \geq 0$ reading the question first up.
$\noindent Let's recall some discriminant theory. Let Q(x) = ax^{2} + bx + c be a real quadratic (i.e. the coefficients a,b,c are real numbers) and a \neq 0. Let \Delta = b^{2} - 4ac be the discriminant. Then we have$

$\bullet \, \Delta > 0 \iff Q(x) \text{ has two distinct real roots } \iff Q(x) \text{ is strictly positive in some places and strictly negative in other places}$

$\bullet \,\Delta = 0 \iff Q(x) \text{ has a double real root } \iff Q(x) \text{ is either always non-negative or non-positive (depending on the sign of }a) \text{ and attains the value 0 at some real }x$

$\bullet \, \Delta < 0 \iff Q(x) \text{ has no real root } \iff Q(x) \text{ is either always (strictly) negative or positive (depending on the sign of }a).$

$\noindent Since in our question, we were told the quadratic f'(x) was non-negative always, one of the last two bullet points above applies, so we have either \Delta = 0 or \Delta < 0, i.e. we have \Delta \leq 0.$

And sorry I typoed before, I wrote non-negative for the discriminant but meant to say non-positive. I have corrected this now.

17. ## Re: International Baccalaureate Maths Marathon

Originally Posted by InteGrand
Why do you think they don't know how to teach the subject properly there? Is it something their worked answers said?

.
Take back my comment about the school. Don't know much about the school personally except I have bumped across a couple of students from here. I hear a lot of complaints about students about how well a teacher can teach or can't teach a particular subject. (from all schools and universities). I say to any student or parent, never put the blame on the school/teachers or tutors. They only play about 1/5th of the role in the outcome of a grade.

I think from memory one of the authors from the Cambridge maths textbook for HSC came from there. Julie something.......

Also I have driven past Stanmore and that school fairly often.

18. ## Re: IB Maths Marathon

Q8b (i) (a+2i)(1+2i)=4-bi

I think there was another mistake in the Cambridge textbook when going through complex numbers, if someone could confirm that would be great.

I got a=8, b=-18

a=8, b=-24

19. ## Re: IB Maths Marathon

You are correct. Easy verification by substitution then expansion and grouping of reals and imaginary components which I know you know, so why did you have doubts?

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