Mathematics: Post your doubts here!

Jelleh Belleh

Jelleh Belleh

Messages
164
Reaction score
193
Points
53
Faaiz Haque said:
Thought blocker can you reply please , read above
Click to expand...
I can show you how to solve the beginning part:-

1-tan^2x / 1+tan^2x
= ( (1/1) - (sin^2x/cos^2x) ) / ( (1/1) + (sin^2x/cos^2x) )
multiply both (1/1) by cos^2x in order to get common denominator
=> ( (cos^2x - sin^2x)/cos^2x ) / ( (cos^2x + sin^2x)/cos^2x)
both cos^2x denominators cancel out
continue

I hope it's clear.
 
Rutzaba

Rutzaba

Messages
4,493
Reaction score
15,418
Points
523
omaaaar said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w11_qp_12.pdf
Q6 how to do pls explain
Click to expand...
Suppose you extend a line starting from d to ab il call this peter :p
suppose you extend yet another line from e to ab il call it david
now luk at the diagram u will find peter=david (they are twins :p)
now if peter = david then if we find peter we will get david...
to find peter we need the fbi kidding lol
we have a theta=pi/3 and a hypotenuse 6
so sin= opp/hyp
sin pi/3= peter/6
peter= 6 sinpi/3
6*(root 3)/2 = peter
peter= 3 root 3
since this is also equals to david
david again can be found by sin=opp/hyp
just in this case the hyp is 10 and the theta theta
sintheta= opp/hyp
sintheta=david/10
sintheta=3 root 3/10
theta= sin inverse of (3 root 3)/10
rememb me in ur prayers ^_^
 
Rutzaba

Rutzaba

Messages
4,493
Reaction score
15,418
Points
523
omaaaar said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w11_qp_12.pdf
Q6 how to do pls explain
Click to expand...
for part b...
take peter for instance the point at wch peter connects line ab lt me name is p
and the point at ab where david touches it let it be q
now the distance ab= ap+pq+qb
ap and qp we can by trig
qp is equals to de
so by
cos pi/3= adj/ hyp
cos pi/3= ap/ 6
ap=3
cos theta=adj/hyp
cos( sin inverse of (3 root 3)/10) (pichle janam mai theta tha)= qb/10
qb=8.54
now the length of ab= radius of small circle plus radius of the big circle
ab=6+10
ab=16
16-8.54-3=4.46
qp=de=4.46
now the perimeter = arcdx plus arc ex plus line de
arc dx= s= r theta
s= 6*pi /3
arc ex= s = r theta
10*cos( sin inverse of (3 root 3)/10)
s=8.54
8.54+3+4.46= 16cm approx... i rounded too much
 
Last edited:
Relon

Relon

Messages
64
Reaction score
45
Points
28
Can someone please answer this questions about As Math ( P1 & M1 )
i am a private candidate and I am a bit confused about the Answer Booklet/Paper,
how many pages it will be?
does it have a limited pages ?
can i solve questions randomly e.g ( solve last question first )
In mechanics do i have to draw a diagram?

Any help would be appreciated.
 
itallion stallion

itallion stallion

Messages
271
Reaction score
429
Points
73

Attachments

  • image.jpg
    image.jpg
    1.6 MB · Views: 6
ZaqZainab

ZaqZainab

Messages
1,764
Reaction score
3,472
Points
273
Haya Ahmed said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w12_qp_11.pdf

Q10 part (ii) and (iii) explain please :)


http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w12_qp_13.pdf

Q6 (ii)
Q8 (ii) and (iii)
Click to expand...
10 (ii)g(x)=4x^2 -24x +11 with domain x ≤ 1
Substitute 1
4-24+11=-9
but how do we know if its ≤ or ≥
The domain says x is less than 1 so x can be 0
Substitute 0
11
Which is more than -9 so f(x)≥ -9
iii) you do completing the square
4x^2 -24x +11 ----> 4(x-3)^2 -25 =y
make x the subject
3+- 0.5 root of (y+25)/4= x
we have to remove the +- to do that look at the domain if it says x≥ hen we use + if it says x≤ then we use - 3- o.5 root of (x+25) = g(x) inverse
the range of g(x) is the domain of g(x) inverse
x≥ -9
6(ii) fg(x)=1= 0.5(cosx) + pie/6
make cosx the subject
cox=(1-(pie/6) )*2 =0.952
x= cos inverse of 0.952 make sure your calculator is in radians mode
x=0.31 you can also get 5.97 but that does'nt come in the range
8(ii) you have show that it has a stationary point at x=-1
dy/dy = 2(3x+4)^3/2 - 6x - 8
substitute -1
2(-3+4)^3/2 +6 -8 =0
hence proved dy/ dx at x=-1 is 0 that is its stationary point
and now determine its nature find the derivative of derivative
which is= 3(3x+4)^1/2 (3) -6
substitute -1
= 9-6= +3 which is a positive value and the point it Minimum
(iii) Integrate dy/dx
you get
y=((4(3x+4)^5/2 )/15) -(3x^2)-(8x)+c
substitute y=5 and x=-1 you get c as -4/15
 
Thought blocker

Thought blocker

XPRS Moderator
Messages
8,545
Reaction score
34,867
Points
718
ZaqZainab said:
10 (ii)g(x)=4x^2 -24x +11 with domain x ≤ 1
Substitute 1
4-24+11=-9
but how do we know if its ≤ or ≥
The domain says x is less than 1 so x can be 0
Substitute 0
11
Which is more than -9 so f(x)≥ -9
iii) you do completing the square
4x^2 -24x +11 ----> 4(x-3)^2 -25 =y
make x the subject
3+- 0.5 root of (y+25)/4= x
we have to remove the +- to do that look at the domain if it says x≥ hen we use + if it says x≤ then we use - 3- o.5 root of (x+25) = g(x) inverse
the range of g(x) is the domain of g(x) inverse
x≥ -9
6(ii) fg(x)=1= 0.5(cosx) + pie/6
make cosx the subject
cox=(1-(pie/6) )*2 =0.952
x= cos inverse of 0.952 make sure your calculator is in radians mode
x=0.31 you can also get 5.97 but that does'nt come in the range
8(ii) you have show that it has a stationary point at x=-1
dy/dy = 2(3x+4)^3/2 - 6x - 8
substitute -1
2(-3+4)^3/2 +6 -8 =0
hence proved dy/ dx at x=-1 is 0 that is its stationary point
and now determine its nature find the derivative of derivative
which is= 3(3x+4)^1/2 (3) -6
substitute -1
= 9-6= +3 which is a positive value and the point it Minimum
(iii) Integrate dy/dx
you get
y=((4(3x+4)^5/2 )/15) -(3x^2)-(8x)+c
substitute y=5 and x=-1 you get c as -4/15
Click to expand...
You explained well than me, so I deleted my post :p
 
ZaqZainab

ZaqZainab

Messages
1,764
Reaction score
3,472
Points
273
You must log in or register to reply here.
Top