IGCSE/GCSE/O & A Level/IB/University Student Forum
Qualification => Subject Doubts => GCE AS & A2 Level => Math => Topic started by: Blizz_rb93 on April 21, 2010, 04:47:44 pm
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Hi everyone, sorry for my inactivity lately. Been studying for my mocks and externals :(
So I have this C1 mock exam tomorrow and I've been solving pastpapers all day, on the January 2010 exam I came across this question that I didn't know how to solve, I checked the mark scheme but it was confusing, can anyone help me?
Here's the question
Question 10 :
f(x) = x2 + 4kx + (3 + 11k), where k is a constant
a) express f(x) in the form (x+p)2 + q, where p and q are constants to be found in terms of K.
Given that the equation f(x) = 0 has no real roots,
b) find the set of possible values of k.
Given that k=1,
c) Sketch the graph of y=f(x), showing the coordinates of any point at which the graph crosses a coordinate axis..
Please anyone help me, any replies will be VERY APPRECIATED!!
Thank you all in advance!!
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do u want the mark scheme?
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Okay, I'm going to try this :) :
a) x^2 + 4kx = (x + 2k)^2 - 4k^2 therefore x^2 + 4kx + (3 + 11k) = (x + 2k)^2 - 4k^2 + 3 + 11k
therefore p = 2k and q = - 4k^2 + 3 + 11k
b) Since f(x) = 0 has no real roots, b^2 - 4ac < 0
a = 1, b = 4k and c = (3 + 11k)
therefore (4k)^2 - 4(1)(3 + 11k) < 0
therefore 16k^2 - 44k - 12 < 0
then divide by 4, so therefore 4k^2 - 11k - 3 < 0
therefore factorise so (4k + 1)(k - 3) < 0
therefore k = -1/4 or k = 3
therefore -1/4 < k < 3
c) k = 1 therefore using (x + 2k)^2 - 4k^2 + 3 + 11k, you know that y = (x + 2)^2 - 4 + 3 + 11 = (x + 2)^2 + 10
when x = 0, y = 14 therefore the curve cuts the y-axis at (0,14)
the curve will not cut the x-axis because f(x) = 0 has no real roots
the lowest point of the curve will be (-2,10)
I hope this helps :)
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thank you for your reply man, but how did you get this part? i dont understand :/
a) x^2 + 4kx = (x + 2k)^2 - 4k^2
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No problem :D
I got that by completing the square.
In the case above, you have x^2 + 4kx and you know that it will end up looking something like (x + ...)^2 + constant. When you complete the square, you have to halve the b value (which in this case is 4k) so that that can go in the bracket. Then you get (x + 2k)^2 although that will give you an extra 4k^2 when you expand the bracket. Therefore you need to minus 4k^2 from this and you end up getting (x + 2k)^2 - 4k^2.
I'm sorry if that's an awful explanation, but I can't really explain it through words properly. If my reasoning makes no sense, you can Google 'completing the square', although I'm sure that you probably know how to do it ;)
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No problem :D
I got that by completing the square.
In the case above, you have x^2 + 4kx and you know that it will end up looking something like (x + ...)^2 + constant. When you complete the square, you have to halve the b value (which in this case is 4k) so that that can go in the bracket. Then you get (x + 2k)^2 although that will give you an extra 4k^2 when you expand the bracket. Therefore you need to minus 4k^2 from this and you end up getting (x + 2k)^2 - 4k^2.
I'm sorry if that's an awful explanation, but I can't really explain it through words properly. If my reasoning makes no sense, you can Google 'completing the square', although I'm sure that you probably know how to do it ;)
sorry for late replies lol, i was eating a while ago and i was using the phone
anyways thank you for your explanation!! you helped me out alot on a)
I already know how to do (b) and right now im gonna read part (c) and see if it helps :)
thank you very much, ill give u rep
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No problem! I'm glad that I could help and thank you for the rep! I only joined these forums yesterday, so I really appreciate it! :)
Good luck with part (c)! :D
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Thank you very much in total :) you helped me out alot :D
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Aww, honestly it was no problem. :)
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Thanks Angel
+rep
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yup thanks :)
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Today I aced the C1 Mock :D :D
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Thanks for the rep :)
Today I aced the C1 Mock :D :D
Ooh, well done! :D
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c2 edexcel
how do we solve this question?
Giving your answers in terms of pie, solve the equation
3 tan2? ? 1 = 0
for ? in the interval ?pie ? ? ? pie
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Home in 1 hour
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whats the "?" in ur question?
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Giving your answers in terms of pie, solve the equation
3 tan2 thita - 1=0
for thita in the interval ?pie ? thita ? pie
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(a) Given that t = log3 x, find expressions in terms of t for
(i) log3 x2,
(ii) log9 x.
(b) Hence, or otherwise, find to 3 significant figures the value of x such that
log3
x2 ? log9 x = 4.
how to do part aii
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Giving your answers in terms of pie, solve the equation
3 tan2 thita - 1=0
for thita in the interval -pie < thita < pie
tan 2 theta= 1/3
2 theta= tan inverse of 1/3
2 theta= 18.4
theta=9.22
period of tan 2theta= +-90 [2theta right...so it will be half the tan theta period...1/2 of 180]
= 9.22, 99.22, -80.78, -170.78
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(a) Given that t = log3 x, find expressions in terms of t for
(i) log3 x2,
(ii) log9 x.
(b) Hence, or otherwise, find to 3 significant figures the value of x such that
log3
x2 ? log9 x = 4.
how to do part aii
for part aii
change the base
and make it 3
so it will be like this (log3 x) /(log3 9)
so log3 x = t
and log3 9 = 2
so the answer is (1/2 t)
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(a) Given that t = log3 x, find expressions in terms of t for
(i) log3 x2,
(ii) log9 x.
(b) Hence, or otherwise, find to 3 significant figures the value of x such that
log3
x2 ? log9 x = 4.
ai)2t
ii)log9 x =lig3 x^(1/2) =1/2 t
retype qb
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for part aii
change the base
and make it 3
so it will be like this (log3 x) /(log3 9)
so log3 x = t
and log3 9 = 2
why cant we use the inverse rule here....like make it 1/log x3 ...............where log x3=1/t ??????
which rule did u use there exactly....... how did u derive the base as 3 in (log3 x) /(log3 9)
so the answer is (1/2 t)
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i used this rule (http://hyperphysics.phy-astr.gsu.edu/Hbase/imgmth/log5.gif)
so log9 x = log x /log 9 , then u put any base u want , i chose 3
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i used this rule (http://hyperphysics.phy-astr.gsu.edu/Hbase/imgmth/log5.gif)
so log9 x = log x /log 9 , then u put any base u want , i chose 3
thankyou :)
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people
could anyone do q7c
plzzz urgentttt
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a=129 r=114/120=0.95
ar^(n-1)<60
120*0.95^n<60
log120 +n log 0.95 < log 60
n?(log 60 -log 120)/log 0.95 =13.52
n=14
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Q4
plz explain in detailsss
thx alotttt
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or you could just keep miultiplying 120 by 0.95 until your answer is less that 60
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or you could just keep miultiplying 120 by 0.95 until your answer is less that 60
why do we do that?
could u be more specific plz?
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Q4
plz explain in detailsss
thx alotttt
Do you need the a) part as well? (i mean the curve)
b) sin-10.35= x-30
x=20.5o+30
=50.5o
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120
120*0.95
120*0.95^2
etc
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Do you need the a) part as well? (i mean the curve)
b) sin-10.35= x-30
x=20.5o+30
=50.5o
no i dont need part (a) thx
but thats not the answer..i got 50.5 but then wat do we do ?
:S ???
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what's the answer?
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what's the answer?
-170.5
50.5
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oh yes
it says all possible values...
sin(180-theta)=sin theta => value of x in this case is great than 180...so rejected
sin(-180-theta)=sin theta
-180-20.5=-200.5
-200.5= x-30
x=170.5
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oh yes
it says all possible values...
sin(180-theta)=sin theta => value of x in this case is great than 180...so rejected
sin(-180-theta)=sin theta
-180-20.5=-200.5
-200.5= x-30
x=170.5
could u please tell me how to do this...?? like are there any rules for the cos , sin and tan curves? any websites or gd resources for
understanding this entire lesson? please!!!!!!!!!!!!
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here http://www.s-cool.co.uk/alevel/maths/trigonometry.html
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here http://www.s-cool.co.uk/alevel/maths/trigonometry.html
i love this site..u've got great resources Nid ;D :P
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i love this site..u've got great resources Nid ;D :P
thank u A.F, thank u ;)
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thx alot :)
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monday m1 c1 prepared? :S