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    For 2013 Q1A (lmao) why is it not a>2 ????
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    (Original post by Mystery.)
    For 2013 Q1A (lmao) why is it not a>2 ????
    Because you use the discrimination equation. Since you want different roots a cant equal 0 because the answer to the discrimination equation will be zero. A can be less than or bigger than 0 since it's being squared so either way the answer will be positive
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    I've slowly started to get the hang of the multiple choice with the help of the RZC PowerPoint (highly recommend) however I'm still not grasping what first steps to take with the longer questions and was wondering if anybody could recommend any resources to help with the longer questions
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    1H and J from 2013????
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    (Original post by Mystery.)
    1H and J from 2013????
    1H

    Square both sides of LHS equation and you get a circle equation (be careful though, don't sketch for x<0 or y<0). the other is a straight line. Draw a quick sketch and work out area using area of sector and area of triangle equations etc

    1J

    Theres not a hope in hell Ill be able to explain my method for this one and i didnt bother trying to understand the mark scheme answer (i dont understand it either). Sorry but theres at least 1H for you. Maybe someone will come along soon and help you with 1J
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    (Original post by theaverage)
    I don't understand how they go about the long division in 2013 question 1, G?

    Posted from TSR Mobile
    I can help you with that. PM me.
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    A mad robot sets off towards the North East on a journey from the origin. It travels in stages by moving forward and then rotating on the spot. It follows these pseudo-code instructions:

    SUB JOURNEY

    DISTANCE = 1000

    WHILE (DISTANCE > 0.001)
    MOVE DISTANCE
    STOP
    ROTATE(90, DEGREES, CLOCKWISE)
    DISTANCE = DISTANCE / 2
    END WHILE

    EXPLODE

    END SUB

    Where does the robot explode?

    Someone want to explain the solution...?

    It starts off like below:
    The forward motion, if it occurs, following the nth turn will be of distance 1000×2−n.This will occur for each n for which

    1000×2−n>0.001
    Taking logs gives

    where does the 2 come from?
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    (Original post by Mystery.)
    x
    Sorry quick question just before I read this. Is this a question 5 from either 2014 or 2015? (i havent done those papers yet and dont want spoilers so just wanna check )
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    (Original post by DylanJ42)
    Sorry quick question just before I read this. Is this a question 5 from either 2014 or 2015? (i havent done those papers yet and dont want spoilers so just wanna check )
    No,its from nrich haha
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    (Original post by Mystery.)
    1H and J from 2013????
    also I decided to read through the mark scheme answer for 1J so ill try talk you through this;
    Spoiler:
    Show
    http://imgur.com/a/2Bqhz (a link to a quick graph sketch etc, itll help me explain it)

    So first part of the mark scheme, since  \displaystyle 0 \leq x &lt; n then  \displaystyle 2^x can take the values  \displaystyle 1 \leq 2^x &lt; 2^n . Since  \displaystyle [2^x] cant actually equal  \displaystyle 2^n (bc its a < sign) and  \displaystyle [2^x] must be an integer then  \displaystyle 2^x can take the integer values of  \displaystyle [1 , 2^n - 1] inclusive (i think that notation is right, i hope that makes sense).


    the next line with  \displaystyle [2^x] = k for blah blah blah... is straight forward enough, just set k equal to a few numbers and youll see that what they say there is true. ( or maybe youll see it better from the sketch i done)

    Okay now for the main part, notice how they have just split the graph into a lot of rectangles (ive colour coded the first few for you). By saying that the base is  \displaystyle log_2(k+1) - log_2k and the height is  \displaystyle k

    From there its just some rearranging and tidying up

    A lot for 4 marks there actually :s
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    (Original post by DylanJ42)
    also I decided to read through the mark scheme answer for 1J so ill try talk you through this;
    Spoiler:
    Show
    http://imgur.com/a/2Bqhz (a link to a quick graph sketch etc, itll help me explain it)

    So first part of the mark scheme, since  \displaystyle 0 \leq x &lt; n then  \displaystyle 2^x can take the values  \displaystyle 1 \leq 2^x &lt; 2^n . Since  \displaystyle [2^x] cant actually equal  \displaystyle 2^n (bc its a < sign) and  \displaystyle [2^x] must be an integer then  \displaystyle 2^x can take the integer values of  \displaystyle [1 , 2^n - 1] inclusive (i think that notation is right, i hope that makes sense).


    the next line with  \displaystyle [2^x] = k for blah blah blah... is straight forward enough, just set k equal to a few numbers and youll see that what they say there is true. ( or maybe youll see it better from the sketch i done)

    Okay now for the main part, notice how they have just split the graph into a lot of rectangles (ive colour coded the first few for you). By saying that the base is  \displaystyle log_2(k+1) - log_2k and the height is  \displaystyle k

    From there its just some rearranging and tidying up

    A lot for 4 marks there actually :s
    Ah thankyou so much. I'll read through it later. Currently doing some late night studying on sequences and series ;laugh;
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    (Original post by Mystery.)
    Ah thankyou so much. I'll read through it later. Currently doing some late night studying on sequences and series ;laugh;
    no problem I was about to go to bed until I seen a post in the MAT thread, I thought id read it and here I am :laugh:

    also for your nrich problem is 1000 x (2- n) from the mark scheme? its just im working out the distance travelled after the nth turn to be  \displaystyle \frac{1000}{2^n}
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    (Original post by DylanJ42)
    no problem I was about to go to bed until I seen a post in the MAT thread, I thought id read it and here I am :laugh:

    also for your nrich problem is 1000 x (2- n) from the mark scheme? its just im working out the distance travelled after the nth turn to be  \displaystyle \frac{1000}{2^n}
    Yeah, I copied it from there.
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    (Original post by Mystery.)
    Yeah, I copied it from there.
    hmm you see if the robot hasnt taken any turns n= 0 then the distance is 1000.

    If its taken its first turn the distance will be 500

    after the 2nd turn itll be 250 etc

    this fits; distance after nth turn  \displaystyle = \frac{1000}{2^n} , whereas  \displaystyle 1000 \times (2-n) doesnt work, even for n=0 as the robot cant travel 2000 units.
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    Name:  Screen Shot 2016-10-21 at 11.13.48.png
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    Can someone please explain to me how to do this? It is from 2010. Thank you
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    (Original post by molly221)
    Name:  Screen Shot 2016-10-21 at 11.13.48.png
Views: 104
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    Can someone please explain to me how to do this? It is from 2010. Thank you
    Check option B. Note that option B actually equals to 1.5.

    C is less than 1 because its argument is smaller than the base. C eliminated.

    5^(1.5) is equal to 5 times sqrt(5). Because sqrt (5) > 2, hence 5^(1.5) is larger than 10. Option D is smaller than 1.5.

    By a similar approach we have option A is larger than 1.5. Hence, the answer is A.
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    (Original post by LaserRanger)
    Check option B. Note that option B actually equals to 1.5.

    C is less than 1 because its argument is smaller than the base. C eliminated.

    5^(1.5) is equal to 5 times sqrt(5). Because sqrt (5) > 2, hence 5^(1.5) is larger than 10. Option D is smaller than 1.5.

    By a similar approach we have option A is larger than 1.5. Hence, the answer is A.
    Thanks, helped a lot.
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    (Original post by danielhx)
    also for Q4 2014, i dont understand the last part how they can just swap beta and alpha just like that
    (Due to convenience I just type alpha as a and beta as b)
    Note that originally
    0<a<b<pi/2
    Now
    0<b<a<pi/2

    Do you see the whole question actually remains the same except this switch? Hence all steps you have done from i) to v) can be directly applied. You can assume this as a change of the name of variables, ie. somehow you are just giving a new name to a and b respectively.
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    (Original post by KloppOClock)
    For question 3iv on 2009 I answered it differently but am wondering if it is still correct.

     \displaystyle \frac{3n-1}{(4n-1)} \leq An/(B+n)



1 is negligible for large values of n. Cancel the n's. Ignore B for large values of n as B is a constant.

Left with;

0.75 \leq A

    Would this still get the marks?
    I don't think so. Check the question, for 3 iv), n can be small numbers as long as it is larger than 1.
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    (Original post by danielhx)
    can someone help me understand 1G of 2014? I understand that the term required is nC4 i dont see how the coefficient is 4(nC4)
    In my opinion, a clearer solution (if you don't feel dizzy with two summations at the same time) would be applying binomial expansion in the trinomial for two times. You will get two binomial coefficients which multiply to get 4 times nC4.
 
 
 
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