Mathematics: Post your doubts here!

[Faaiz Haque]

http://papers.xtremepapers.com/CIE/...AS Level/Mathematics (9709)/9709_s07_qp_1.pdf

Question 10 iii please

 

[Jelleh Belleh]

Thought blocker can you reply please , read above

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]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w11_qp_12.pdf
Q6 how to do pls explain

Suppose you extend a line starting from d to ab il call this peter
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 )
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]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w11_qp_12.pdf
Q6 how to do pls explain

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

 

[Relon]

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.

 

[Haya Ahmed]

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/...S Level/Mathematics (9709)/9709_w12_qp_13.pdf

Q6 (ii)
Q8 (ii) and (iii)

 

[itallion stallion]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s11_qp_31.pdf
please help me Question 10
Mark scheme link
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s11_ms_31.pdf

Sorry for delay,if u can't understand any step tell me I will do it in detail.

 

[itallion stallion]

Can someone plz help me with this one
http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w13_qp_33.pdf
Q6 part 2
Thanks a bunch!

 

[ZaqZainab]

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)

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]

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

You explained well than me, so I deleted my post

 

[ZaqZainab]

You explained well than me, so I deleted my post

you didn't have to
btw c was not 5

 

[Thought blocker]

you didn't have to
btw c was not 5

I directly input x as -1 and I got c as 5 and in ms too it is 5 :/ I guess you are correct, but help me y my method is false ?

 

[ZaqZainab]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s07_qp_1.pdf

Question 10 iii please

using the dy/dx you found in (i)
find the gradient of the curve at x=-2
by substituting -2 as x
gradient of the curve = 4
gradient of normal -1/4
Gradient of the normal =(y+2/x+2=-1/4
It says intersect the x axis so y=0
you will get x=-10

 

[ZaqZainab]

I directly input x as -1 and I got c as 5 and in ms too it is 5 :/ I guess you are correct, but help me y my method is false ?

lol i didnt see the ms
oh c isnt 5 they mean c+'intergrated part'=5
5 is y :/
when you integrate dy/dx and subsitute -1 you don't get 0 there
wa

 

[Thought blocker]

lol i didnt see the ms
oh c isnt 5 they mean c+'intergrated part'=5
5 is y :/
when you integrate dy/dx and subsitute -1 you don't get 0 there
wa

oh, hell with me

 

[ZaqZainab]

oh, hell with me

I am guessing you did not integrate

 

[Thought blocker]

I am guessing you did not integrate

yes I forgot, I directly input -1

 

[princeali97]

Why does F(x)=(2x^2) -8x+5 for the domain 0<equal x <equal 5 does not have an inverse? What are the conditions for an inverse?

 

[Faaiz Haque]

using the dy/dx you found in (i)
find the gradient of the curve at x=-2
by substituting -2 as x
gradient of the curve = 4
gradient of normal -1/4
Gradient of the normal =(y+2/x+2=-1/4
It says intersect the x axis so y=0
you will get x=-10

thanks

 

[ZaqZainab]

Why does F(x)=(2x^2) -8x+5 for the domain 0<equal x <equal 5 does not have an inverse? What are the conditions for an inverse?

to have an inverse it should be a one to one function