Hey there! Sign in to join this conversationNew here? Join for free
Turn on thread page Beta
    Offline

    3
    ReputationRep:
    (Original post by cheetahs56)
    I can see 59/72, what do you think grade boundaries will be?

    Posted from TSR Mobile
    Same. I feel like crap. I so so hope you can lose 15 and get an A

    Posted from TSR Mobile
    Offline

    1
    ReputationRep:
    I got k = -5,
    A = -2, -27
    and PQ as 4.5 or 9/2
    Offline

    2
    ReputationRep:
    How did you do question 2? The hidden quadratic?

    Posted from TSR Mobile
    Offline

    0
    ReputationRep:
    Ok, I thought this was a really easy exam, but that's my own opinion- don't bite me for it!

    I brought my answers out of the exam on scrap paper, so I'll do my best to translate. I'm pretty confident they're all right, if I'm wrong correct me, but I went through with teachers and other students afterwards and these seem to be correct answers. If anybody did get similar, shout up!

    1)i) 12 rt.5
    ii) 4 rt.5
    iii) 5 rt.5

    2) x^3 = 1/8 and -1 ... therefore x = 1/2 and -1

    3) f'(x) = -(12x^-3) +2
    f''(x) = 36x^-4

    4)i) 3(x+3/2)^2 + 13/4 [Because 3(3/2)^2 = 3(9/4) = (27/4) .. so then take that from the ten to give 13/4]
    ii) Vertex is (-3/2, 13/4)
    iii) b^2 - 4ac = 9^2 - 120 = -39

    5)i) Basically the same graph as 1/x^2... Curves symmetrical and never touch any of the axes
    ii) Stretch in Scale Factor 1/2 in y (vertical) direction

    6)i) x^2 + (y+4)^2 = 40 Centre is (0, -4) Radius is rt.40 = 2 rt.10
    ii) B co-ordinate is (-2, -10)

    7)i) x < -1/8
    ii) 0 </= x </= 5 ( </= means greater than or equal to! solved this one by drawing out the curve.. )

    8) m of perpendicular is 1/3, so m of line is -3.. put into formula to get 3x + y -1/2 = 0 .. x2 to get into integers, 6x + 2y - 1 = 0

    9)i) Positive quadratic curve (smiley face, not sad face). Intercepts y at (0,-6) and x at (-3/2,0) and (2,0)
    ii) Vertex at x=1/4 so function is decreasing for x < 1/4
    iii) Submit into formula for curve, use quadratic formula to solve, find that points P and Q are (-2,4) and (2.5,4) so distance is 4.5

    10)i) Solve to find that k = -5
    dy/dx = -3x^2 - 6x + 4 - k
    0 = -3(-3)^2 - 6(-3) + 4 - k
    k = -27 + 18 + 4 = -5

    ii) d2y/dx2 = -6x - 6 ... submit -3 in to get 12, 12>0 therefore it's a minimum point
    iii) This one was really long winded for 5 marks, basically you had to put the formulas together, solve the quadratic to find the two points on the cubic curve that satisfied a gradient of 9, you then put those two points (0 and -2) into the y=9x-9, and see which works. 0 doesn't work, but -2 does, so A = (-2, -27)

    Hope this helps! Please point out any mistakes you think I've made, would be awesome if someone could scan a copy of the question sheet too
    Offline

    19
    ReputationRep:
    (Original post by eggfriedrice)
    Same, I was tempted to ask for scrap paper to do the test again and see if I get the same answers, but instead I just decided to check through 3 times. Luckily I spotted and corrected a few mistakes.
    Oh wow, their application thing opened early then. I haven't even touched my application in a few weeks lol, I literally have half a box to fill and send off.
    yours sounds long hahaha, for mine I had two huge engineering questions and you pick one or both n u had to write a whole page for and the rest of it was on your university choices and grades and references.
    Offline

    1
    ReputationRep:
    (Original post by roar96)
    How did you do question 2? The hidden quadratic?

    Posted from TSR Mobile
    substitute y for x^3 then solve as normal i think you got -1/8 and -1 then take the 3rd root of these
    Offline

    0
    ReputationRep:
    (Original post by Anubis2)
    'Fraid you're wrong, the last is (-3/2, 13/4).
    yeah I know, I remember now I did put (-3/2,13/4)

    for y it was something like 10-17/4 that's where I got the 17/4 bit from in my head.
    Offline

    0
    ReputationRep:
    (Original post by Anubis2)
    You do not have to write the answer in any specific part of the answer booklet, only next to the correct question number. Congrats for getting (-2,-27) as that is right.
    Thanks, but what I meant was that did the last question to find the tangent consist of about a third of the last inner page and the entire back outside page in the answer booklet? Because I only used the bottom space in the back inner page for the tangent question and I got a feeling I put the answer in the wrong question box. Sorry, I phrased the question badly.
    Offline

    19
    ReputationRep:
    (Original post by roar96)
    How did you do question 2? The hidden quadratic?

    Posted from TSR Mobile
    let y = x^3 therefore y^2 = x^6 then substitute and factorise.
    Offline

    12
    ReputationRep:
    (Original post by eggfriedrice)
    Could have sworn on my paper it said "coordinates". My eyes must be playing tricks on me. -_-
    (Original post by a10)
    it did say co-ordinates but only one of the pair satisfies the equation..

    co-ordinates usually mean 2 points (x and y) if they wanted one value its usually referred to as ordinate.

    Did the exam go well for you?
    (Original post by Zakee)
    It did say coordinates. However, remember a coordinate is simply one value. (i.e) find out the x coordinate in the equation (4,11), the coordinate is 4. The coordinates however are (4,11).
    To be precise it specified "the coordinates of the point". Point, singular. Hence only one answer.

    The coordinates weren't (4,11) - they were (-2,-27). That was an example, right?
    Offline

    2
    ReputationRep:
    (Original post by John Dogg)
    substitute y for x^3 then solve as normal i think you got -1/8 and -1 then take the 3rd root of these
    But i swear that gave you a cubic to solve...

    Posted from TSR Mobile
    Offline

    1
    ReputationRep:
    Hi guys,
    I accidentally put k=5 for 10i even though I did the right workings (can't add up). I also then used this for part iii with the tangent 9x-9 but It wouldn't factories as I got the wrong quadratic. How many marks will I lose?
    thanks
    Offline

    0
    ReputationRep:
    (Original post by roar96)
    But i swear that gave you a cubic to solve...

    Posted from TSR Mobile
    No It didn't I am afraid that is the correct method!
    Offline

    0
    ReputationRep:
    (Original post by Dugald)
    what are the rules for A* at A2. say I happen to get a low A in this paper, will I still be able to get an A*?
    to get an A* you have to have 80% of all ums and average 90% in C3 and C4
    Offline

    0
    ReputationRep:
    (Original post by roar96)
    But i swear that gave you a cubic to solve...

    Posted from TSR Mobile
    the original equation was something like 8x6 + 7x3.. so you would end up with a quadratic when you substitute y=x3
    Offline

    12
    ReputationRep:
    I don't think they can ask you to solve cubics in core 1, thats restricted to c2 with factor theorem. Unless they already give you the factors.

    For those questions they're always in a form that can easily turn into a quadratic.
    Offline

    12
    ReputationRep:
    Given that Mr M is posting answers at some point tonight, discussing solutions seems a little fruitless.

    However, if anyone wants to post a question fully I can post my answer to it. Does anyone remember the two inequalities (questions not answers), or the 7-mark Question 8?
    Offline

    0
    ReputationRep:
    (Original post by metabalo)
    Thanks, but what I meant was that did the last question to find the tangent consist of about a third of the last inner page and the entire back outside page in the answer booklet? Because I only used the bottom space in the back inner page for the tangent question and I got a feeling I put the answer in the wrong question box. Sorry, I phrased the question badly.
    If you put the answer in the wrong box, don't worry. My teacher's an exam marker and said that it's just to help set it out. You'll get the marks for the working, even if it's in another question's box
    Offline

    1
    ReputationRep:
    what did guys put for the question where you had to state what happens to graph as it transforms from 2/x^2 to 1/x^2. I wrote that it stretches parallel in the x axis by 0.5
    Offline

    2
    ReputationRep:
    (Original post by a10)
    let y = x^3 therefore y^2 = x^6 then substitute and factorise.
    Ahhh... (Y^3)^2 .... Darn... I was quite sure it would have to be cubed.... How stupid of me
    Posted from TSR Mobile
 
 
 
Reply
Submit reply
Turn on thread page Beta
Updated: May 15, 2013
Poll
“Yanny” or “Laurel”

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Write a reply...
Reply
Hide
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.