AQA maths core 1, 13th May 2015

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Reply 480

aoxa

Original post by SunDun111

What was the original equation of the curve can you remember?

y= x^4 - 3x^2 -2

I think?

I got -10 as well

Original post by aoxa

You got -10 by using the dy/dx method.

If you used the method of y-y/x-x to get the gradient, you got -7 or something. I don't think you were supposed to se that method though.

You can't use y-y/x-x. That is for a straight line. The straight line may have intersected at A but that doesn't make the gradient of the curve the same at that very point. You are finding the tangent to the curve and therefore the gradient must be the same as the curve, not the line. That isn't the right method and you'd probably score only 1 mark for attempting to calculate the equation of a line.

(edited 8 years ago)

Reply 483

username1412036

Finished the markscheme. Thanks everyone!

Reply 484

iJAKE

I got -10 for that one but I know I dropped a definite 5 marks on the first 2 parts of question 6:'(

Reply 485

iJAKE

Original post by Sekaib

Yeah I get 162/5 for the area behind the curve and the line.
I got the curve as 108/5 and the trapezium at 54 but I split the trapezium into a triangle and a rectangle but I couldn't remember the formula lol.

I got exactly the same! It came out as a negative number so i just flipped it and think i got the shaded region's area of 32.4

Reply 486

aoxa

Original post by -jordan-

You can't use y-y/x-x. That is for a straight line. The straight line may have intersected at A but that doesn't make the gradient of the curve the same. You are finding the tangent to the curve and therefore the gradient must be the same as the curve, not the line. That isn't the right method and you'd probably score only 1 mark for attempting to calculate the equation of a line.

I did not use the y-y/x-x method - I was saying this is probably what the teacher used to get the answer of -7. To get this you used the y-y/x-x on the straight line AB, and the worked out the reciprocal of ABs gradient, this gives the perpendicular to AB. The problem is that this method does not necessarily give the tangent to the curve.

Reply 487

student0042

Original post by brokenlevel23

Finished the markscheme. Thanks everyone!

You just need to change bi to 5x ... I put 4x by accident in my original comment.

Reply 488

username1412036

Original post by student0042

You just need to change bi to 5x ... I put 4x by accident in my original comment.

Oh yeah lol

Reply 489

tanyapotter

if i couldn't factorise and got the critical values of the last inequality wrong, would i gain a mark for drawing a sign diagram and having p < x < q ? or do i lose all marks completely

Reply 490

JackC3001

Original post by brokenlevel23

FINAL MARKSCHEME:

Question 1
(a) -3/5 [2]
(b)(i)
perp. gradient = 5/3
y-?=5/3(x-?)
5x-3y+1=0 [3]
(b)(ii) A(9,-4) [3]

Question 2
7 + √15 [5]

Question 3
(a) y=-10x-4 [4]
(b)(i) 108/5 [5]
(b)(ii). 162/5 [3]

Question 4
(a) (x+1)^2 + (y-3)^2 = 50 [2]
(b)(i). C(-1,3) [1]
(b)(ii) 5√2 [2]
(c)
k^2-6x-16
(k-8)(k+2)
k=8, k=-2 [2]
(d) min. distance = 7 [2]

Question 5
(a) p = 3/2, q = -¼ [3]
(b)(i) (-3/2, -¼) [2]
(b)(ii) line of symmetry is x=-3/2 [1]
(c) y = x^2-x+4 [3]

Question 6
**IT WAS AN OPEN TOP CYLINDER DAMNIT**
(a)(i) h = 24/r -r/2 [3]
(a)(ii) Sub πr^2 into part (i) to get V = 24πr -π/2 r^3 [3]
(b)(i) 24π -3π/2 r^2 [2]
(b)(ii)
r=4 when dy/dx=0
r=4 => d^2y/dx^2(=-12π)<0, therefore maximum [4]

Question 7
(a) http://i.imgur.com/rVJ169A.png [3]
(b)(i) R = 36 [2]
(b)(ii) R= 0 therefore root [2]
(b)(iii)(x-2)(x^2-5x+10) [2]
(b)(iv) x=2 [3]

Question 8
(a) (Show) x^2 + 3(k-2)x -13-k=0 [1]
(b) 9k^2-32k-16<0 [3]
(c) -4/9< k<4 [4]

Very nice, glad i could help - just put n = 7 for question 2 as this is what the question actually specified. It gave you + √15 in the question.

Also, I just re did the question for 7biii - it is most definitely (x+2) and therefore 7biv is x=-2

(edited 8 years ago)

Reply 491

tanyapotter

Original post by JackC3001

Very nice, glad i could help - just put n = 7 for question 2 as this is what the question actually specified. It gave you + √15 in the question.

was it not x=-2 for the only root?

Reply 492

Excuse Me!

Original post by brokenlevel23

FINAL MARKSCHEME:

Question 7
(a) http://i.imgur.com/rVJ169A.png [3]
(b)(i) R = 36 [2]
(b)(ii) R= 0 therefore root [2]
(b)(iii)(x-2)(x^2-5x+10) [2]
(b)(iv) x=2 [3]

Pretty certain that the graph 7a went through the point (0,3)