Mathematics: Post your doubts here!

[unique840]

One more help please guys 9709 May/june 2009 P1 Q4 help please

which part?

 

[Sajjad Hossain]

salam ....
can u plz tell me tat hw 2 expand a binomial having three terms? m really stuck hea !!! plz help me out

It wud be nice if u wud say which years question it is or just the question and what the answer is supposed to be

 

[Sajjad Hossain]

which part?

Both :'(

 

[unique840]

Both :'(

part1) a represents amplitude i.e 6
b represents number of cycles i.e 2
c represents the value of y from where the graph is starting i.e 3

 

[unique840]

Both :'(

part2) y = 6sin(2x) + 3
at y = 0
0 = 6sin(2x) +3
6sin(2x) = -3
sin(2x) = -1/2
2x = sin inverse of (1/2) keeping the calc in radian mode
the value of sin is pie/6
it is -1/2 so sin -ve is in 3rd and 4th quadrant
so 2x = 7pie/6 or 11pie/6
x = 7pie/12 or 11pie/12
smallest value of x will be 7pie/12

 

[BeanZ 08]

It wud be nice if u wud say which years question it is or just the question and what the answer is supposed to be

lawwlz... Its frm P1( The course text book,nt d past paper) and it is as followz:
Expand (1+x+2x(sq))(sq) check your answer with a numerical sustitution.
(sq) i used hea 2 denote square cz i dn noe hw 2 ri8 it using key board?

 

[ffaadyy]

salam ....
can u plz tell me tat hw 2 expand a binomial having three terms? m really stuck hea !!! plz help me out

Please post that question.

 

[ffaadyy]

lawwlz... Its frm P1( The course text book,nt d past paper) and it is as followz:
Expand (1+x+2x(sq))(sq) check your answer with a numerical sustitution.
(sq) i used hea 2 denote square cz i dn noe hw 2 ri8 it using key board?

(1+x+2x^2)^2

Suppose 'x+2x^2' to be one term.

[1 + (x+2x^2)]^2

Now expand the above equation.

[1 + (x+2x^2)]^2
1 + (x+2x^2)(2) + [(x+2x^2)^2(2)(1)]/2
1 + 2x + 4x^2 + (x+2x^2)^2
1 + 2x + 4x^2 + x^2 + 4x^3 + 4x^4
1 + 2x + 5x^2 + 4x^3 + 4x^4

Therefore, the final answer is '1 + 2x + 5x^2 + 4x^3 + 4x^4'.

 

[Gémeaux]

Cud anyone please help me with q.3 paper 41 may june 2011? =(
http://www.xtremepapers.com/CIE/International A And AS Level/9709 - Mathematics/9709_s11_qp_41.pdf

 

[ffaadyy]

Cud anyone please help me with q.3 paper 41 may june 2011? =(
http://www.xtremepapers.com/CIE/International A And AS Level/9709 - Mathematics/9709_s11_qp_41.pdf

Refer to the above diagram and solve the horizontal and the vertical components of forces.

Horizontal Component:

T cos θ + T sin θ = 15.5
T cos θ = 15.5 - T sin θ

Vertical Component:

T cos θ + 8 = T sin θ
15.5 - T sin θ + 8 = T sin θ
23,5 = 2 T sin θ
11.75 = T sin θ

Therefore, T sin θ = 11.75~12

Put back this value of 'T sin θ' in the equation 'T cos θ = 15.5 - T sin θ'.

T cos θ = 15.5 - T sin θ
T cos θ = 15.5 - 12
T cos θ = 3.5.

Therefore, T cos θ = 3.5,

To find the value of θ, use the formula 'tan θ = opposite/adjacent'.

tan θ = opposite/adjacent
tan θ = (T sin θ) / (T cos θ)
tan θ = 12/3,5
θ = 73.73o

 

[Gémeaux]

Horizontal Component:
T cos θ + T sin θ = 15.5
T cos θ = 15.5 - T sin θ
Vertical Component:
T cos θ + 8 = T sin θ
15.5 - T sin θ + 8 = T sin θ
23,5 = 2 T sin θ
11.75 = T sin θ
Therefore, T sin θ = 11.75~12
Put back this value of 'T sin θ' in the equation 'T cos θ = 15.5 - T sin θ'.
T cos θ = 15.5 - T sin θ
T cos θ = 15.5 - 12
T cos θ = 3.5.
Therefore, T cos θ = 3.5,
To find the value of θ, use the formula 'tan θ = opposite/adjacent'.
tan θ = opposite/adjacent
tan θ = (T sin θ) / (T cos θ)
tan θ = 12/3,5
θ = 73.73o

thankyou so much

 

[XPFMember]

Assalamoalaikum wr wb!

Some more links added to the first post, and the old ones which weren't working have been fixed...

Check this out! - Nice website, with video tutorials!

My P1 Notes! - Only few chapters available at the moment!

Maths Notes by destined007

Some Notes for P1 and P3 - shared by hamidali391

A LEVEL MATHS TOPIC WISE NOTES

MATHS A LEVEL LECTURES

compiled pastpapers p3 and p4 - by haseebriaz

Permutations and Combinations (my explanation)- P6

Permutations and Combinations - P6

Vectors - P3

Complex No. max/min IzI and arg(z) - P3

Range of a function. - P1

 

[Hussnain]

Thank you very much

 

[Hussnain]

i m replacing theta with x
1/(2cosx +i - i2sinx +2 -i)
+i -i cancels out
1/(2cosx -i2sinx +2)
multiply and divide the whole fraction with the conjugate
[1/(2cosx + 2 - i2sinx)] * [(2cosx + 2 + i2sinx) / (2cosx + 2 + i2sinx)]
in the denominator: ( (2cosx + 2 + (2cosx + 2 + i2sinx) ) - i2sinx) * (2cosx + 2 + i2sinx)
taking (2cosx + 2) as a. and (i2sinx) as b. we apply the formula (a+b)(a-b) = a^2 - b^2; so
(2cosx + 2 + i2sinx) / {[(2cosx+2)^2] - [(i2sinx)^2]}
(2cosx + 2 + i2sinx) / {[4(cos^2)x + 8cosx + 4] + 4(sin^2)x}
-(i^2) is changed to +1 in the above step
4(cos^2)x + 4(sin^2)x is changed by identity [(cos^2)x +(sin^2)x = 1] to 4
(2cosx + 2 + i2sinx) / (8cosx + 8)
we jus have to prove the real part. so separate the real and imaginary part
(2cosx + 2) / (8cosx + 8)
(cosx + 1) / 4(cosx + 1)
cosx + 1 cancels out
1/4 is left which is constant for all values of theta

THANK YOU VERY MUCH

 

[Hussnain]

HELP NEEDED.
M/J/10.
P3
Q 4 (ii)
REPLY FAST.
I HAVE MY EXAM TOMORROW

 

[Hussnain]

This is how we'll do this question:

1/(z+2-i)

We are given that ' z = 2 cos θ + i ( 1 - 2 sin θ )' so we'll put in this value of 'z' in the above equation.

1/(z+2-i)
1/[2 cos θ + i ( 1 - 2 sin θ )+2-i]

Arrange the real numbers and the imaginary numbers.

1/[(2 cos θ + 2) + i ( 1 - 2 sin θ -1 )]
1/[(2 cos θ + 2) - i ( 2 sin θ )]
THANKS ALOT

Next, multiply the equation '1/[(2 cos θ + 2) - i ( 2 sin θ )]' by the conjugate of '[(2 cos θ + 2) - i ( 2 sin θ )]'

{1/[(2 cos θ + 2) - i ( 2 sin θ )]} x [(2 cos θ + 2) - i ( 2 sin θ )] / [(2 cos θ + 2) + i ( 2 sin θ )]
[2 cos θ + 2 + i (2 sin θ)] / [(2 cos θ + 2)^2 + (4 sin^2 θ)]

As the question has asked us to deal with the real part only, we'll remove 'i (2 sin θ)' which is imaginary.

(2 cos θ + 2) / [(2 cos θ + 2)^2 + (4 sin^2 θ)]
(2 cos θ + 2) / (4 cos^2 θ + 8 cos θ + 4 + 4 sin^2 θ)
(2 cos θ + 2) / (4 cos^2 θ + 4 sin^2 θ + 8 cos θ + 4)
(2 cos θ + 2) / [4 (cos^2 θ + sin^2 θ) + 8 cos θ + 4]

Recall the identity 'cos^2 θ + sin^2 θ=1'.

(2 cos θ + 2) / [4 (cos^2 θ + sin^2 θ) + 8 cos θ + 4]
(2 cos θ + 2) / [4 (1) + 8 cos θ + 4]
(2 cos θ + 2) / (8 + 8 cos θ)
[2 ( cos θ + 1 ) ] / [ 8 ( 1 + cos θ ) ]

'cos θ + 1' cancels out as its present in both the numerator and the denominator. We are left with '2/8' now which further simplifies to '1/4'. Therefore, '1/4' is our final answer.

 

[unique840]

THANK YOU VERY MUCH

ur welcum

 

[unique840]

HELP NEEDED.
M/J/10.
P3
Q 4 (ii)
REPLY FAST.
I HAVE MY EXAM TOMORROW

p31 or 32 or 33?

 

[mhalvi]

http://www.xtremepapers.com/communi...es-examiner-reports.11543/page-11#post-176948

thankyou soo much

 

[unique840]

thankyou soo much

ur welcum