Mathematics: Post your doubts here!

[Shreeram]

Thanks!! though you've used add/sub formulas , which aren't a part of my sylllabus (i'm not appearing for P3 this year) ,so could you pls explain further :
how does tanπ -tanx / 1+tanπ tanx = -tanx/1 ?
thanks in advance!

you would find that the list of formula sheet. Then applying it would give u the answer btw its for p1 only.

 

[PhyZac]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s11_qp_33.pdf
Q 7 part (iii) please PhyZac Dug syed1995

I dint know how to do this, but searched through google and learned a useful method.

First we know the roots are
-√3 + i mod 2 arg 5/6pi
-√3 - i mod 2 arg -5/6pi

So according to De Moivre's theorem

therefore,

z^6 = r^6 (cos 6t + i sin6t)
2^6 (cos6(5/6pi) + isin 6(5/6pi))
64 (-1 + i (0))
-64

same thing with other root!

link.
http://www.cliffsnotes.com/study_guide/De-Moivres-Theorem.topicArticleId-11658,articleId-11634.html

 

[immie.rose]

you would find that the list of formula sheet. Then applying it would give u the answer btw its for p1 only.

oh yeahhh i found those in the formula sheet! thanks again!

 

[Shreeram]

Can anyone help me to give a detailed explanation for june 12 p 42 q7 thanks

 

[Shreeram]

Hopefully this helps

Can u help me to give a detailed explanation for june 12 p 42 q7 thanks

 

[Scafalon40]

I dint know how to do this, but searched through google and learned a useful method.

First we know the roots are
-√3 + i mod 2 arg 5/6pi
-√3 - i mod 2 arg -5/6pi

So accordingto De Moivre's theorem

therefore,

z^6 = r^6 (cos 6t + i sin6t)
2^6 (cos6(5/6pi) + isin 6(5/6pi))
64 (-1 + i (0))
-64

same thing with other root!

link.
http://www.cliffsnotes.com/study_guide/De-Moivres-Theorem.topicArticleId-11658,articleId-11634.html

 

[sma786]

Hi, can someone please solve Question 5 http://papers.xtremepapers.com/CIE/...AS Level/Mathematics (9709)/9709_s07_qp_1.pdf

 

[iKhaled]

Hi, can someone please solve Question 5 http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s07_qp_1.pdf

here you go!

(i) divide that thingy into 2 triangles from the middle and u will have triangle AOX with angle 1/6pi.. so AX= 12tan1/6pi = 4√3

(ii) to get the area of the shaded region we need to find the area of the sector then subtract it from the whole thingy which is the 2 triangles..area of the sector will be= 1/2(12)^2(1/3pi)
area of sector = 24pi and the area of the 2 triangles will be = 1/2(4√3)(12) x 2 = 48√3

area of shaded region = 48√3 - 24pi

thats it, i hope you got it!

 

[PhyZac]

Hi, can someone please solve Question 5 http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s07_qp_1.pdf

Okay, tangent means making 90 degrees with a point on circle.

so AX make 90 degree B A and BX make 90 at B and thus, line OX will bisect then angle 1/3pi (means cuts in half) giving 1/6pi

tan (1/6pi) = AX/12 (opposite over adjacent (radius))
1/√3 = AX/12
AX = 12/√3

(ii) Now the area can be found by finding area of the two triangles AOX and BOX (they are same) and find the sector area...and subtract
area of AOX = 12/√3 x 12 x 1/2 (1/2bh = area of triangle)
=24√3
then into 2 because 2 triangle =48√3

Now the sector 1/2 r^2 theta
1/2 x 12^2 x 1/3pi = 24pi

48√3 - 36pi =

 

[applepie1996]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w07_qp_3.pdf
question 10 part 2
anyone ?
PhyZac

 

[Nurul Huda sabid]

i m doing Cambridge advanced level pure mathematics 2 & 3 by hugh neil douglas quading....... the problem is i lost the answers of chapter 15-Rational function......plz help me!

 

[Scafalon40]

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s12_qp_31.pdf
Q3 part (ii) b please!
PhyZac

 

[VelaneDeBeaute]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s12_qp_31.pdf
Q3 part (ii) b please!
PhyZac

From the previous part, you have obtained a simplified form of the expression as p(x) = (x-2)(x+2)(x-4)
Since the question now calls for evaluating p(x squared), you just substitute the x in the expression for x squared.
Your expression would become somewhat like this
(I'll be using x2 for x squared)
p(x2) = (x2 -2)(x2 +2)(x2 -4)
Now, you equate them to zero, each of them. You'll obtain six roots, two of which will be imaginary.

 

[Scafalon40]

From the previous part, you have obtained a simplified form of the expression as p(x) = (x-2)(x+2)(x-4)
Since the question now calls for evaluating p(x squared), you just substitute the x in the expression for x squared.
Your expression would become somewhat like this
(I'll be using x2 for x squared)
p(x2) = (x2 -2)(x2 +2)(x2 -4)
Now, you equate them to zero, each of them. You'll obtain six roots, two of which will be imaginary.

Thanks! : )

 

[PhyZac]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s12_qp_31.pdf
Q3 part (ii) b please!
PhyZac

Okay, now u have to get a first which is
(x-2)(x-2)(x+4)

Now when p(x^2)=0
therefore
(x^2-2)(x^2-2)(x^2+4)

Use the quadratic formula.
for first root (and second since they are same)
0+/-√0-(4)(1)(-2) / 2
= +/-2√2/2
=+/-√2

and third root
0+/-√0-(4)(1)(4) / 2
+/-4i/2
+/- 2i

 

[sma786]

What equation did we take in Q10, part 2 ?
http://papers.xtremepapers.com/CIE/...AS Level/Mathematics (9709)/9709_s07_qp_1.pdf

 

[PhyZac]

What equation did we take in Q10, part 2 ?
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s07_qp_1.pdf

for stationary point, take the dy/dx equation and equate to 0.
for nature of stationary point use d2y/dx2 and is -ve number comes means max point if +ve number comes means minimum.

 

[Kumkum]

AOA
can someone plz help me with 10c
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w09_qp_12.pdf

 

[syed1995]

AOA
can someone plz help me with 10c
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w09_qp_12.pdf

We take tan inverses of the gradient and subtract them.

tan^-1(m1) - tan^-1(m2) should give you the angle between the lines..

tan^-1(2) - tan^-1(1/2)
gives 36.9 Degrees Angle.

 

[Kumkum]

We take tan inverses of the gradient and subtract them.

tan^-1(m1) - tan^-1(m2) should give you the angle between the lines..

tan^-1(2) - tan^-1(1/2)
gives 36.9 Degrees Angle.

Thank you