May 2014 Math Sl Paper 1 TZ1

[Casstafiore]

Did anyone get an answer for the part c of question 10? What value of k did you guys get?

[Dan123456]

it was 4 i believe

[Dan123456]

Did u guys find it easy? I actually found it slightly difficult and I'm hoping that the mark boundaries will be low.

Also, what do u guys think will be the mark boundaries for the math exploration?

[UCL14May]

it was 4 i believe

Is that the answer for the last question of Question 9? The vector question? I got 4 seconds for the last question of Question 9...

[kobe24]

k was equal to 1. This is because the total area of a square with side length k is going to be k^2. But the question told us that the area was k cm. Consequently k^2 = k. You solve that quadratic and you get k=0 and k=1. k=0 cant be a solution therefore k=1.

[Guest YOLOEH]

k was equal to 4. The mention of an 'indefinite' number of triangles meant that the sum of the areas would require the use of the infinite geometric series formula. As each triangle was isosceles and the length of the side of the square was k cm, the area for the first triangle (or first term in the sequence) would be given by A= (1/2)*(k/2)^2 or A= (k^2)/8. The common ratio, r, would be 1/2 as the area of each subsequent triangle would be half of the area preceding it. As we were told that the sum of the geometric series was k cm^2, we could equate that to the geometric series expression, thereby giving: k = [(k^2)/8] / [1-(1/2)]. This could then be simplified to k = (k^2)/4 and then rearranged to 4k = k^2. Divide both sides by k and you are left with k=4.