Guys plz help me in no6(ii).thx in advance.http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w11_qp_33.pdf.please
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Mathematics: Post your doubts here!
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http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w12_qp_33.pdf
Q.1
I am not able to solve it. Can someone help?
Q.1
I am not able to solve it. Can someone help?
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what is the answer for it?
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Check it here... Do u know how to do it?
http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w12_ms_33.pdf
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See if you can get it!!Check it here... Do u know how to do it?
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w12_ms_33.pdf
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ln(x+5)=1+lnx
ln(x+5)=lne+lnx
ln(x+5)=ln(ex)
x+5=ex
ex-x=5
x(e-1)=5
x=5/e-1
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Can sum1 plz help me with the foll qs:
2007 oct-nov p1:
Q4(ii),(iii)
Q9(i),(ii)
Q11(v)
2008 may-jun p1:
Q8(i)
2008 oct-nov p1:
Q7(i)
Q10(iii),(iv)
plz guys im begging u 2 help me
thanx in advance
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ok May/june 08
8i) f(x) = 4x - 2k and g(x) = 9/(2-x)
fg(x) = x
4(9/2-x) - 2k = x
36/(2-x) - 2k = x, multiply everything by 2-x
36 - 2k(2-x) = x(2-x)
36 - 4k + 2kx = 2x - x^2
x^2 + 2kx - 2x +36 - 4k = 0
x^2 +x(2k - 2) + 36 - 4k = 0
since they have equal roots b^2 - 4ac = 0
so,
(2k-2)^2 - 4(1)(36-4k) = 0
i'm sure now u can find the values of k
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0ct-nov 08
7i) first you have to find the perimeter of both the square and circle then add both and equate to 80cm.( because 80cm is the total length)
so,
Perimeter for square = 4x , for circle = 2(pi)r hence,
4x + 2pi*r = 80 divide everything by 2 and make r subject of formula
==> r = (40 -2x)/pi
now find area, area of square = x^2 , area of circle = pi*r^2
so, A = x^2 + pi*r^2, if u substitute r into this equation u will find ur answer
10 (iii) here u use the completion of square method:
h(x) = 6x - x^2
= -(x^2 - 6x)
= -[x^2 - 6x + (-3)^2 - (-3)^2]
= -[(x - 3)^2 -9]
= 9 - (x - 3)^2
(iv) here u have to find inverse of h:
h(x) = 9 - (x-3)^2
let y = 9- (x-3)^2
(x-3)^2 = 9 -y, taking square root both sides
x - 3 = +/ - sq.root of(9-y)
x= 3 +/- sq.root of(9-y) hence
inverse of h = 3 +/- sq.root(9 -x)
hope u got it
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0ct/nov 07
m not so sure of q4
9i) here since they are asking u to find equation of the curve u have to integrate dy/dx
if u integrate dy/dx u'll get:
y= 4x - (x^2 / 2) + c
u have a point (2,9), substitute this point and find the value of c:
===> 9 = 4(2) - (2^2 / 2) + c
c = 3
then substitute c in ur equation and u'll have ur answer :
y = 4x - (x^2 / 2) + 3
ii) here u will have to find the gradient of normal first then u can find the equation, to find grad. of normal first find grad of tangent (i.e dy/dx) at x=2
so,
dy/dx at x=2 will be = 2
now to find grad of norm use m1*m2 = -1
u can let m1= 2 (grad of tangent) so ur m2 = -1/2
now since u have the gradient of normal and a point (2,9) u can find the equation using:
y - y1 = m(x -x1) and u'll get
y = 10 - x/2
hope u understood
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thanx so muchok May/june 08
8i) f(x) = 4x - 2k and g(x) = 9/(2-x)
fg(x) = x
4(9/2-x) - 2k = x
36/(2-x) - 2k = x, multiply everything by 2-x
36 - 2k(2-x) = x(2-x)
36 - 4k + 2kx = 2x - x^2
x^2 + 2kx - 2x +36 - 4k = 0
x^2 +x(2k - 2) + 36 - 4k = 0
since they have equal roots b^2 - 4ac = 0
so,
(2k-2)^2 - 4(1)(36-4k) = 0
i'm sure now u can find the values of k
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http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w12_qp_33.pdf
I don't get how to do Q7, Q8 ii, Q9 ii or 10... Basically last half of the paper... This paper kills...
I don't get how to do Q7, Q8 ii, Q9 ii or 10... Basically last half of the paper... This paper kills...
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Thanx.. i got Q7(i), but cud u plz xplain how exactly u used the completing the squares method in 10(iii)..n 10(iv) the ans in ms is 3+root(9-x), theres no minus0ct-nov 08
7i) first you have to find the perimeter of both the square and circle then add both and equate to 80cm.( because 80cm is the total length)
so,
Perimeter for square = 4x , for circle = 2(pi)r hence,
4x + 2pi*r = 80 divide everything by 2 and make r subject of formula
==> r = (40 -2x)/pi
now find area, area of square = x^2 , area of circle = pi*r^2
so, A = x^2 + pi*r^2, if u substitute r into this equation u will find ur answer
10 (iii) here u use the completion of square method:
h(x) = 6x - x^2
= -(x^2 - 6x)
= -[x^2 - 6x + (-3)^2 - (-3)^2]
= -[(x - 3)^2 -9]
= 9 - (x - 3)^2
(iv) here u have to find inverse of h:
h(x) = 9 - (x-3)^2
let y = 9- (x-3)^2
(x-3)^2 = 9 -y, taking square root both sides
x - 3 = +/ - sq.root of(9-y)
x= 3 +/- sq.root of(9-y) hence
inverse of h = 3 +/- sq.root(9 -x)
hope u got it
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oh so q9(i) is on intergration?...i havent yet learnt intergration..sry 4 the trouble0ct/nov 07
m not so sure of q4
9i) here since they are asking u to find equation of the curve u have to integrate dy/dx
if u integrate dy/dx u'll get:
y= 4x - (x^2 / 2) + c
u have a point (2,9), substitute this point and find the value of c:
===> 9 = 4(2) - (2^2 / 2) + c
c = 3
then substitute c in ur equation and u'll have ur answer :
y = 4x - (x^2 / 2) + 3
ii) here u will have to find the gradient of normal first then u can find the equation, to find grad. of normal first find grad of tangent (i.e dy/dx) at x=2
so,
dy/dx at x=2 will be = 2
now to find grad of norm use m1*m2 = -1
u can let m1= 2 (grad of tangent) so ur m2 = -1/2
now since u have the gradient of normal and a point (2,9) u can find the equation using:
y - y1 = m(x -x1) and u'll get
y = 10 - x/2
hope u understood
..n sry it was supposed 2 be q9(iii), not q9(ii)..so if u cud plz ans dat
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not a prob...sure just waitoh so q9(i) is on intergration?...i havent yet learnt intergration..sry 4 the trouble
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ok to find co-ordinates of Q u have to solve the equation of the curve and the equation of the normal simultaneously. if u do that one of ur co-ordinates will be for P and the other one for Q.oh so q9(i) is on intergration?...i havent yet learnt intergration..sry 4 the trouble
hope u understood
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if u dont mind cud u plz do it..coz thats wat i did but wasnt getting the ans n that other question in the other post q10(iii)ok to find co-ordinates of Q u have to solve the equation of the curve and the equation of the normal simultaneously. if u do that one of ur co-ordinates will be for P and the other one for Q.
hope u understood