Math SL paper 1

[vx4594]

How did you all think it went? Personally fine until I got to the longer questions where 8 was very simple but 9 & 10 a bit challenging, but I don't remember it all perfectly either.

[DiegoA]

How did everyone found Paper 1 and Paper 2 in Maths SL? It's been more than 24 hours for both papers

Personally, I found them more difficult than past papers.

[Guest joaninhagcl]

how did it go for everyone??

[jeanlean]

Hey guys!

How did everyone find Paper 1 and Paper 2?

[tusonliu]

Paper 1 was difficult compared to past papers.

What was the answer for the question #7 which asked about the value of K when discriminant was higher than 0??? I´ve gotten only one result: 2. Was that because 2 was an assymptote and K could be any number?

Paper 2 difficult as well.

There was this question about... LOL cant remember.... something about functions of Section A. I think it was question 7.

[Kiandra S.]

k has two values when the discriminant of the quadratic function is greater than 0.

[g_ar]

i found paper 2 harder than alot of past papers, there was alot of unfamiliar questions....i just hope they make the grade boundaries lower

[khimberleigh]

Paper 1 was difficult compared to past papers.

What was the answer for the question #7 which asked about the value of K when discriminant was higher than 0??? I´ve gotten only one result: 2. Was that because 2 was an assymptote and K could be any number?

I got K is greater than 2

I found paper 1 pretty good, but paper 2 much harder than past papers.

[vx4594]

If I recall correctly it was + or - 2.. don't remember perfectly

Paper 1 was difficult compared to past papers.

What was the answer for the question #7 which asked about the value of K when discriminant was higher than 0??? I´ve gotten only one result: 2. Was that because 2 was an assymptote and K could be any number?

Paper 2 difficult as well.

There was this question about... LOL cant remember.... something about functions of Section A. I think it was question 7.

[DiegoA]

k has two values when the discriminant of the quadratic function is greater than 0.

The discriminant of the function had to be = 0. I got that it could be any number except +2.

It was something like (k-2)(k-2)=0

[g_ar]

k has two values when the discriminant of the quadratic function is greater than 0.

The discriminant of the function had to be = 0. I got that it could be any number except +2.

It was something like (k-2)(k-2)=0

no, the discriminant had be >0, because the function has two distinct real roots.....so the answer is k>2

Edited November 15, 2013 by g_ar

[Samster]

I honestly found that Paper 1 harder than the May 2013 one and all the other ones that I had tried. I had no clue how to do most of question 10!

I'm hoping that it is scaled down significantly though!

Edited November 15, 2013 by Samster

[Kiandra S.]

k has two values when the discriminant of the quadratic function is greater than 0.

The discriminant of the function had to be = 0. I got that it could be any number except +2.

It was something like (k-2)(k-2)=0

no, the discriminant had to be k>0, because the function has two distinct real roots.....so the answer is k>2

Yes discriminant greater than zero, it it is = 0 then the two root have the same value.

[vx4594]

I saw in some past paper with a discriminant question very similar that you solved for 0 once you had the discriminant expression and I got + - 2, which is what they do on the same exact question on another past paper..

[g_ar]

I saw in some past paper with a discriminant question very similar that you solved for 0 once you had the discriminant expression and I got + - 2, which is what they do on the same exact question on another past paper..

the discriminant only equals zero when the function has a repeated or double root.

discriminant is >0 if there are two distinct real roots

and discriminant <0 if there are no real roots.

[gabslion]

k has two values when the discriminant of the quadratic function is greater than 0.

The discriminant of the function had to be = 0. I got that it could be any number except +2.

It was something like (k-2)(k-2)=0

If it has two values, then the discriminant has to be more than 0. Therefore, you have to find a range with +- 2 for the values

[DiegoA]

k has two values when the discriminant of the quadratic function is greater than 0.

The discriminant of the function had to be = 0. I got that it could be any number except +2.

It was something like (k-2)(k-2)=0

If it has two values, then the discriminant has to be more than 0. Therefore, you have to find a range with +- 2 for the values

The question said that the quadratic function had two different real roots. Therefore, the discriminant had to be >0. We all agree up to that point.

Now, the original function was something along the lines of kx^2+(k+2)x+2.

Which can be re-written as Ax^2+Bx+C.

Therefore, the "B" value was (k+2). Given that the discriminant of a function is b^2-4ac, the discriminant resulted in another function, which had to be >0.

The discriminant gave you the following: (k^2+4k+4)-(8k) which is k^2-4k+4. THAT function had to be OVER 0 for the ORIGINAL function to have two real roots.

In numbers, (k-2)(k-2)>0

For the above equation to be true, k must be ANY number EXCEPT +2. Try it yourselves. That equation is true for any number except for +2 (yeah, you can even try with -2).

(I used capitals for the sake of you understanding what I'm trying to express because its difficult to explain math by writting it).

That being said, that was what I did on my exam, and if I am wrong in any point, please say so.

-Diego

[sarah:)]

k has two values when the discriminant of the quadratic function is greater than 0.

The discriminant of the function had to be = 0. I got that it could be any number except +2.

It was something like (k-2)(k-2)=0

If it has two values, then the discriminant has to be more than 0. Therefore, you have to find a range with +- 2 for the values

The question said that the quadratic function had two different real roots. Therefore, the discriminant had to be >0. We all agree up to that point.

Now, the original function was something along the lines of kx^2+(k+2)x+2.

Which can be re-written as Ax^2+Bx+C.

Therefore, the "B" value was (k+2). Given that the discriminant of a function is b^2-4ac, the discriminant resulted in another function, which had to be >0.

The discriminant gave you the following: (k^2+4k+4)-(8k) which is k^2-4k+4. THAT function had to be OVER 0 for the ORIGINAL function to have two real roots.

In numbers, (k-2)(k-2)>0

For the above equation to be true, k must be ANY number EXCEPT +2. Try it yourselves. That equation is true for any number except for +2 (yeah, you can even try with -2).

(I used capitals for the sake of you understanding what I'm trying to express because its difficult to explain math by writting it).

That being said, that was what I did on my exam, and if I am wrong in any point, please say so.

-Diego

Yes, discriminant had to be greater than 0.

I understand that it's not +2 and also understand that it's any value greater than 2. However, how is it also less than two, like how'd you work out that part... just trial and error?

[DiegoA]

Yes, discriminant had to be greater than 0.

I understand that it's not +2 and also understand that it's any value greater than 2. However, how is it also less than two, like how'd you work out that part... just trial and error?

Because you know its not +2 you just have to try with any value bigger than +2, and any value lower than +2. I tried with 0 and 5, and noticed that with both numbers the function>0. Therefore, any value bigger or smaller than +2 satisfied the equation

[Alicia Maldonado]

Are you all talking about TZ0?