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Mathematics: Post your doubts here!

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Its quite simple for the question 7 part 2 just do it like
3/(1 + 2x)^2 < 1/3 and then solve for x in the next step you will get like this

(1 + 2x)^2 (>)9
then
once when you solve it using the (a+b)^2 formulae you will get it like 4x^2 + 4x +1>9
you will get like
4x^2 +4x -8 and then solve it quadratically......ok get it like x > 1, x < –2 ok gt it?
 
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Maths
Q 5 may june 2011 p11
Plz solve the second part for me

Q7 part ii

Q11 part iv

Pleaase solve...


5ii is really simple .. you just gotta substitute y=sin^2(Theta).

The equation 2sin^4(Theta)+Sin^2(Theta)-1=0 becomes 2y^2 + y -1 =0

2y^2 + y -1 =0

2y^2 +2y -y -1 = 0

2y(y+1)-1(y+1)=0
(2y-1)(y+1)=0
y=1/2 or y=-1

sin^2(Theta)=1/2 or -1
Sin(Theta)=√1/2 or √-1 (which isn't possible.)
Theta= Sin-1(√1/2)
Theta=45

Thetha = 45 Degrees and 135 Degrees (Since Sin is only positive in first and 2nd quadrant)
 
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Q 5 may june 2011 p11
Plz solve the second part for me

Q7 part ii

Q11 part iv

Pleaase solve...

Q11 iv .. what's hard in it?

find inverse of f(x) and put g(x) into that inverse equation.

y= 2x +1
(y-1)/2=x

f-1(x)=(x-1)/2

f-1(g(x))= (x^2-2-1)/2
=(x^2-3)/2
 
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can someone please tell me what is "List of formulae (MF9) " stated on P1 under Additional materials ????
 
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Assalamo'alaikum, I need help with this question:

Solve the equation
ln(x + 5) = 1 + lnx,
giving your answer in terms of e.

i'm home studying and no one around can help me right now.
 
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Its quite simple for the question 7 part 2 just do it like
3/(1 + 2x)^2 < 1/3 and then solve for x in the next step you will get like this

(1 + 2x)^2 (>)9
then
once when you solve it using the (a+b)^2 formulae you will get it like 4x^2 + 4x +1>9
you will get like
4x^2 +4x -8 and then solve it quadratically......ok get it like x > 1, x < –2 ok gt it?


This question was repeated many times > and i didnt now it ....
so any help plz ? <>> http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s11_qp_41.pdf

q7 >>> last part ... (iii) Find the time from the instant the string becomes slack until it becomes taut again
 
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Question 7 (ii)


There is only one way to solve this.

You need to remember that for a line to be tangent it needs to be like

tangent.gif


Meaning that the line y=mx+c needs to intersect the curve at exactly one which .. (that point being the minimum/maximum point on the curve) Meaning both the roots need to be same.

that is only true in b^2-4ac=0 .. in b^2-4ac=0 .. there is only 1 root.

so to solve this question

y = m(x – 2)
y = mx - 2m ===> 1
y = x^2 − 4x + 5 ===> 2

Substitute 1 in 2..

mx -2m = x^2 -4x +5 // shifting it to one side gives us..

x^2-4x + 5 - mx + 2m=0 // arranging them in order of powers gives us

x^2 -(4+m)x +5 + 2m=0

a= 1
b=4+m
c=5+2m

using b^2-4ac =0 we get..

(4+m)^2-4(1)(5+2m)=0 // now opening the brackets
16+8m+ m^2 -20-8m=0 // +8 , -8 get cancelled
16+m^2-20=0
m^2-4=0
m^2=4
m=√4
m= 2 or m = -2

now we need to substitute m into the equation to get the co-ordinates of x.

x^2 -(4+m)x +5 + 2m=0
x^2 - (4+2)x +5 +2(2)=0 ------- or ----- x^2 -(4-2)x +5 +2(-2)=0
x^2-6x+5+4=0 ------- or ---------- x^2 -2x +5 -4
x^2-6x+9=0 ------------ or --------- x^2 -2x +1 =0
x^2-3x-3x+9 ------------ or --------- x^2-x-x+1
x(x-3)-3(x-3) ------------ or --------- x(x-1)-1(x-1)
(x-3)(x-3) ------------ or --------- (x-1)(x-1)

x=3 or 1 ..

Now finally equate the values of x into y=x^2 − 4x + 5 to get the y-coordinates of the co-ordinates..


I know it seems a little too lengthy but usually they stop at values of m and don't ask the co-ordinates itself.. it's the first time i have seen them ask the co-ordinate as well.

As far as derivate goes.. that's not possible and will yield wrong answer since we can only get 1 value of m via that method and you won't get full marks.. the only correct way to solve this question is via b^2-4ac=0
 
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http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w11_qp_61.pdf
In qs 5i , the qs says 94% of the letters are within 12g of mean, so isnt it supposed to be (20+12<x<20-12)?? my answer's not matching with the mark scheme......:cry:
pleaseee help Dug syed1995 Ashique
also tell me how to find the p is 0.97

That is one question which I still haven't understood.. I myself need help in that question :p

Tkp , Dug , PhyZac you guys got any idea how this one would be solved?
 
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