IGCSE/GCSE/O & A Level/IB/University Student Forum
Qualification => Subject Doubts => IGCSE/ GCSE => Math => Topic started by: moon on October 20, 2009, 12:37:35 pm
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plz, can any body help me in math paper 4 Jun.09 question 11 d (i,ii) ??? ??? ???
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11(d)(i) (n(n+1))/2 + ((n+1)(n+2))/2
=(n^2+n+n^2+2n+n+2)/2
=(2n^2+4n+2)/2
=(2(n^2+2n+1))/2
=n^2+2n+1
So, n^2+2n+1=(n+1)^2
(ii) (n+1)^2=3481
n^2+2n+1=3481
n^2+2n-3480=0
n=(-2+-underroot(2^2-4*1*-3480))/2
n=(-2+-underroot(13924))/2
n=(-2+-118)/2
So, n=58, -60
n=58 (It can't be negative)
Two terms will be n, (n+1)
=58, 59
(58(58+1))/2=1711
(59(59+1))/2=1770
Answer=1711 and 1770
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thanks a lot :) :)
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welcome
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11(d)(i) (n(n+1))/2 + ((n+1)(n+2))/2
=(n^2+n+n^2+2n+n+2)/2
=(2n^2+4n+2)/2
=(2(n^2+2n+1))/2
=n^2+2n+1
So, n^2+2n+1=(n+1)^2
(ii) (n+1)^2=3481
n^2+2n+1=3481
n^2+2n-3480=0
n=(-2+-underroot(2^2-4*1*-3480))/2
n=(-2+-underroot(13924))/2
n=(-2+-118)/2
So, n=58, -60
n=58 (It can't be negative)
Two terms will be n, (n+1)
=58, 59
(58(58+1))/2=1711
(59(59+1))/2=1770
Answer=1711 and 1770
Please explain both parts still didn't get it ...
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Please I need help in JUNE 2009 Q2(d)
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Please explain both parts still didn't get it ...
It's given that the nth term is n(n+1)/k where k comes out to be 2.
so, (n+1)th term will be (n+1)(n+2)/k where k is 2. (v'll add 1 because it's n+1 and not n)
so v'll do this now:
(n(n+1))/2 + ((n+1)(n+2))/2
=(n^2+n+n^2+2n+n+2)/2
=(2n^2+4n+2)/2
=(2(n^2+2n+1))/2
=n^2+2n+1
So, n^2+2n+1=(n+1)^2
Now v have to find 2 consecutive terms with sum of 3481.
v got frm d above ans that the sum of two consecutive nos. is (n+1)^2
now v'll use that and make it equal to 3481 as given below
(n+1)^2=3481
n^2+2n+1=3481
n^2+2n-3480=0
n=(-2+-underroot(2^2-4*1*-3480))/2
n=(-2+-underroot(13924))/2
n=(-2+-118)/2
So, n=58, -60
n=58 (It can't be negative)
Two terms will be n, (n+1)
=58, 59
(58(58+1))/2=1711
(59(59+1))/2=1770
Answer=1711 and 1770
hope u get it now!! :)
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Please I need help in JUNE 2009 Q2(d)
The total count was 148 for 50 rolls.
148=50X2.96
so 10 rolls.....count will be 10x...if x is considered to be the mean
together for sixty rolls
148+10x/60=2.95
therefore x=2.9
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see its quite simple:
mean for 50 rolls is 2.96(found in c part)
mean for 60 rolls is 2.95(given)
so to find the mean for the extra 10 throws....:
[(60*2.95)-(50*2.96)]/10
=>29/10
=>2.9
do u get it bro???
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c u get 2.96 as ur mean i guess in c part outta 50 times
d mean after 50 +10=60 is 2.95
so (148+10x)/60 = 2.95
148+10x = 2.95*60
10x = 177-148
x=29/10
x=2.9
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Please I need help in JUNE 2009 Q2(d)
im gonna use | for the summation (sigma) symbol
we knw that Mean=|fx/|f
therefore the sum of the 60 dice rolls is=Mean*|f
=> 2.95*60=177
the sum of the 50 dice rolls-Mean*|f
=> 2.96 (found in part c)*50=148
therefore the sum of the last 10 rolls=177-148=29
therefore the mean of the 10 rolls=|fx/|f
=>29/10=2.90
okay temme if u dnt get nething?
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why does it have to be june 2009?
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Please I need help in JUNE 2009 Q2(d)
c) Mean score for 50 rolls = ((15*1)+(10*2)+(7*3)+(5*4)+(6*5)+(7*6))/50
= 2.96
d) Mean score for 60 rolls = 2.95
60*2.95 = 177
50*2.96 = 148
So, 177-148 = 29
Mean of 10 rolls = 29/10
= 2.9
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lol 5 answers to that question........haha
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man!!!!!
after so many explanations anyone can understand it!!!
this shows how eager ppl at sf are to help others!!!
Thanks alot cooldude, nid, manbir, wassup and me too!!!
By the way moon....tll us if u didn't understand anything!!u'll have another 10 ppl tryina explain it to u!!
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man!!!!!
after so many explanations anyone can understand it!!!
this shows how eager ppl at sf are to help others!!!
Thanks alot cooldude, nid, manbir, wassup and me too!!!
By the way moon....tll us if u didn't understand anything!!u'll have another 10 ppl tryina explain it to u!!
u may add me too!
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u may add me too!
hawww!!!!but i didn't se uu anywhere on this thread aadi??!!!
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hawww!!!!but i didn't se uu anywhere on this thread aadi??!!!
i meant, u can add me to the 10 ppl.
"haaaww"...hahahaha..!!!..man!!..lolzz...
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ohhhh
ohhkkk.....!!!!!
cool!!!
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At last!!!!! I understood that.but I still think that the puzzle question Q11 d(i)&(ii) was the hardest one in June 2009.Isn't it ??? ??? ONCE AGAIN THANK U VERY MUCH FOR HELPING ME.
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At last!!!!! I understood that.but I still think that the puzzle question Q11 d(i)&(ii) was the hardest one in June 2009.Isn't it ??? ??? ONCE AGAIN THANK U VERY MUCH FOR HELPING ME.
u got q11??
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ya, I got it after long explanation
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Ya Thank u :)
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Can someone help me with Question 5 c i. When i was checking my answers I differentiated the equation (x/2)-(2/x) and got 1/2+2x^-2, which in the question gives me a tangent of 0.75, however this does not seem to be an acceptable answer in the mark scheme.
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which year/session/paper/question no. ??
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Paper 4 June 2009 (the topic of this forum :p)
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Paper 4 June 2009 (the topic of this forum :p)
igcse maths?...
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u need to draw a tangent at the point x=2 on the graph....
and then find the gradient by y2-y1/x2-x1
see the graph here....now just take out the gradient
no need for differentiation!!
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Yep IGCSE 0580 Maths Paper 4 June 2009
Yea I could do it that way, but I just want to check using differentiation because i'm going to need to be able to do that way later. Also sometimes my tangents aren't very accurate so I just want to be able to use differentiation to check it.
Thanks
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ohhkkk....ur welcome
but i cant help u with differentiation....aadi or shreycool may help u...i dont do add. maths!!
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Yep IGCSE 0580 Maths Paper 4 June 2009
Yea I could do it that way, but I just want to check using differentiation because i'm going to need to be able to do that way later. Also sometimes my tangents aren't very accurate so I just want to be able to use differentiation to check it.
Thanks
dy/dx (x/2 - 2/x) = 1/2 + 2/x2 => substituing x=2 --> m = 1
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hey aadi cud u explain me differentiation more briefly...i wud want to learn to check my answer if possible...pm me or better come on gmail...
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its difficult to learn differentiation in such a short span of..there are different methods for different type
for e.g here...u have to apply the u/v rule..
so yea..maybe later..
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Thanks but i don't understand why it is x^2, isn't it x^-2 because the original power is -1 and we reduce it by 1 we get -2
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ya ure rite!!
but x^-2 becomes 1/x^2!!!
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n ono...its not that simple
its the u/v method
here
d/dx (u/v) = (vdu-udv)/v2
break the expressin into two halves
1. x/2
2. 2/x
first..
u = x
v = 2
du = 1
dv = 0
->(vdu-udv)/v2
= 2/4
second..
u = 2
v =x
du = 0
dv = 1
(vdu-udv)/v2
=> (0-2)x2 = -2/x2
so the final differentiation = 1/2 - (-2/x2) = 1/2 + 2/x2
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its difficult to learn differentiation in such a short span of..there are different methods for different type
for e.g here...u have to apply the u/v rule..
so yea..maybe later..
ya ya...kk..
u teach me after the ig'z!!
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OMG I'm such an idiot. This is exactly the kind of thing i'm going to do tomorrow. I can do the work but I just make so many errors. After substituting x=2 into the equation i forgot to multiply by 2 before adding a half.
Thank you all so much and sorry about the hassle
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OMG I'm such an idiot. This is exactly the kind of thing i'm going to do tomorrow. I can do the work but I just make so many errors. After substituting x=2 into the equation i forgot to multiply by 2 before adding a half.
Thank you all so much and sorry about the hassle
thats k..
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baapre aadi yuo tuped so much just to explain him differentiation!!
i guess adi toxic is now scared!!
By the way there is one trick by differentiation you can find the equation of the normal and tangnt in maths !!
i ahve used it many times!!
although this is not correct method!!
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I need help in NOV 2004 Q7 Please.It is diffiicult.
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I need help in NOV 2004 Q7 Please.It is diffiicult.
the whole q?
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please I need a3, bii and c2 ???
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I need help in NOV 2004 Q7 Please.It is diffiicult.
7 a)i) y = px + qx + r
y = 4^2+(-2*4)+(-3)
y = 5
ii) On K and L y=0
so, px + qx + r=0
x-2x-3=0
Solve this n u'll get x=-1, 3
K(-1,0) L(3,0)
iii) On M x=0
so, y=r
y=-3
M(0,-3)
b)i) Upside down
ii) Eqn, will be x^2=y, so just put some values n get the shape of graph
c)i) c=0 because it passes through d origin
ii) (3,0) (4,8)
put these in the eqn.
3^2a + 3b = 0
4^2a + 4b = 8
solve these n u'll get a = 2, b = –6
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please I need a3, bii and c2 ???
for aiii
best wud be to draw the graph, and see where the minimum point lies. write its cordinates
for confirmation u can use differentiation
dy/dx = 2x - 2
2x - 2 =0
2x = 2
x = 1 substituting y = 4
dont bother if u dont understand this
for bii
u know then the value are applied in the equation
u r left with only y = x2 which is a parabola and passes through the origin (0,0)
for cii
it says it passes through the origin
which means, c =0
u r left with
y= ax^2 + bx
substitute the cordinates (3, 0) and (4, 8).
to get
0 = 9a + 3b
and
8 = 16a + 4b
from the second we can get
a = 8-4b / 16
substitute this in the first to get
72 - 36b
-------- = -3b
16
72 - 36b = -48b
solve to get
b = -6 and a = 2
@wassup - for aiii - it is not they-axis, its a line of symmetry
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there u go:
a)iii) K(-1,0) and L(3,0)
M lies on the line of symmetry so the the point on the x-axis along this line of symmetry will be the mid-point of the line KL
Mid point = ((x2+x1)/2 , (y2+y1)/2)
Mid point = (1,0)
As the point M lies on the same line of symmetry thus its x-coordinate is 1
by using the value of x(as 1) and the values of p, q and r given find y
y= -4
b)ii)The graph will be the same figurebut its base will not go below the origin(0,0)
c)ii)
put the values for each coordinate in the equation forming 2 equations nd then solve them simultaneously
for (3,0)
0=a(3)2+b(3) +0
=>9a+3b=0
for (4,8)
8=a(4)2+b(4)+o
=>16a+4b=8
solve them simultaneously to get a=2, and b=-6
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for aiii
best wud be to draw the graph, and see where the minimum point lies. write its cordinates
for confirmation u can use differentiation
dy/dx = 2x - 2
2x - 2 =0
2x = 2
x = 1 substituting y = 4
dont bother if u dont understand this
for bii
u know then the value are applied in the equation
u r left with only y = x2 which is a parabola and passes through the origin (0,0)
for cii
it says it passes through the origin
which means, c =0
u r left with
y= ax^2 + bx
substitute the cordinates (3, 0) and (4, 8).
to get
0 = 9a + 3b
and
8 = 16a + 4b
from the second we can get
a = 8-4b / 16
substitute this in the first to get
72 - 36b
-------- = -3b
16
72 - 36b = -48b
solve to get
b = -6 and a = 2
@wassup - for aiii - it is not they-axis, its a line of symmetry
aadi won't it be tooo time consuming to draw the graph and do the differentiation(a part)....just see wat i did....i think dts the right method!!
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aadi won't it be tooo time consuming to draw the graph and do the differentiation(a part)....just see wat i did....i think dts the right method!!
yes. that would be better. i considered that and thus didnt comment. u wanted to hear this. lol
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iii) On M x=0
so, y=r
y=-3
M(0,-3)
This answer isn't like that in the mark scheme it should b (1, -4)
Thanks u all for ur help & I think that it is not that much difficult.
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This answer isn't like that in the mark scheme it should b (1, -4)
Thanks u all for ur help & I think that it is not that much difficult.
sorry i thot it was d y axis.
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yes. that would be better. i considered that and thus didnt comment. u wanted to hear this. lol
hey nthng like that dude...just wanted to let u know....
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abbe kitan jhagde ho tum dono!!!!