Wrong answer in MarkScheme for Nov 2003 Paper 1?

[Luisfrommexico]

Hello everyone,

So I have finally started to revise for my Maths HL papers today and I found something really weird. On the Maths HL paper 1 examination of Nov of 2003 there is a question - Q 13 - That asks you to solve the following:


Consider the equation (1+ 2k)x² -10x + k -2 = 0, where k belongs to all real numbers. Find the set of values of k for which the equation has real roots.



Now here is the answer given by the IB: _____

Can anyone please attempt this question and tell me what they get because what I got for the answer is not the same for what the mark scheme says, however, my answer does fit the set of values so the discriminant is equal or greater than 0 on all cases of K.

Im just really confused and it is likely that I did something wrong, but when these things happen I attempt the question over and over again until I get the answer that the Ib gives; nevertheless, I havent been able to do so with this one...

PLEASE HELP

Edited April 2, 2013 by Sandwich
Please don't post links to past papers on the forum as they are copyrighted material.

[flinquinnster]

Okay, firstly you're not strictly allowed to post direct links up to copyrighted material on this forum, but let's just pretend I already had a copy of the mark-scheme - which I do.

So, the discriminant is b^2-4ac = (10)^2 - 4(1+2k)(k-2) = 100 - 4(2k^2-3k-2) = 108 - 8k^2 + 12k

For real roots, discriminant must be equal to or greater than 0.

Therefore 108-8k^2 + 12k > 0 (pretend it's a greater than or equal to sign)

-2k^2+3k+27>0

2k^2-3k-27<0

(2k-9)(k-3)<0

If you graph the parabola, you'll find that -3<k<9/2 gives the solutions for the inequality. Any values of k outside the range will result in something >0, which does not fit the inequality we need.

So my conclusion is that the markscheme is right. I actually found that the calculation was quite simple, so maybe you just did something really weird when you were trying to solve the question. Hope I managed to help.