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Mathematics: Post your doubts here!

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I have almost tried this several times but then I give up..Help me
 

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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 :cry:

thanx in advance
 
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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)

thanx in advance
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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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)

thanx in advance
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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Reactions: Dug
Messages
948
Reaction score
5,542
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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)

thanx in advance
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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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 :)
thanx so much(y)
 
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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 :)
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 minus
 
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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 :)
oh so q9(i) is on intergration?...i havent yet learnt intergration..sry 4 the trouble:oops:..n sry it was supposed 2 be q9(iii), not q9(ii)..so if u cud plz ans dat:D
 
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oh so q9(i) is on intergration?...i havent yet learnt intergration..sry 4 the trouble:oops:..n sry it was supposed 2 be q9(iii), not q9(ii)..so if u cud plz ans dat:D
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 :)
 
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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.
hope u understood :)
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)
 
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