Mathematics: Post your doubts here!

[Dug]

View attachment 22700

Please solve this!!

Thank you

Separate variables and integrate:
⌡1/y dy = ⌡6x/(x² + 4) dx
ln|y| = 3ln(x²+4) + c

Put values:
ln(32) = 3ln(4) + c
ln(32) = ln(64) + c
c = ln(32) - ln(64)
c = ln(1/2)

ln|y| = 3ln|x² + 4| + ln(1/2)
ln|y| = ln(x² + 4)³ + ln (1/2)
ln|y| = ln[(x² + 4)³ (1/2)]
y = (1/2)(x² + 4)³

 

[tanmaydube]

Thanks a lot Dug!!!

 

[Mayedah]

It depends upon which 2 shapes u r using..
there r many different combinations that can be used..
Which 2 trinagles did u use?

Thnx i got it

 

[tanmaydube]

Help again, stuck at the first part!!!

 

[tanmaydube]

Cannot do B (ii)

 

[parthrocks]

can somebody help me with this
special request to @daredevils

 

[anishh]

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

 

[19islandprincess96]

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?

 

[parthrocks]

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

 

[parthrocks]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w12_qp_33.pdf
Q.1
I am not able to solve it. Can someone help?

what is the answer for it?

 

[19islandprincess96]

what is the answer for it?

Check it here... Do u know how to do it?
http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w12_ms_33.pdf

 

[parthrocks]

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

See if you can get it!!

 

[19islandprincess96]

See if you can get it!!

Now that you have done it, it seems so simple and I feel stupid!
Thanks a ton!

 

[anishh]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w12_qp_33.pdf
Q.1
I am not able to solve it. Can someone help?

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

 

[Iffat]

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

 

[Kumkum]

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

 

[asexamskillme111]

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s10_qp_43.pdf

Can someone do Q7? I think I've asked a question like this before. I don't get why only box A is involved in the calculation for part ii. I also don't get how to go along solving for part iii. Thank you.

 

[Kumkum]

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

 

[Kumkum]

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

 

[Iffat]

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