Mathematics: Post your doubts here!

[ffaadyy]

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

If it's J10, P32; then Q4(ii) will be done this way.

x = tan^-1 (x) + π

Take the initial value as 'π' and put it in the above formula to obtain a few iterations.

1. 4.4042
2. 4.4891
3. 4.4932
4. 4.4933
5. 4.4934
6. 4.4934

The answer is '4.49' correct to two-decimal places.

If it's J10, P31; then Q4(ii) will be done this way.

Instead of integrating 'sin 3x sin x', we can integrate '(1/2)(cos 2x − cos 4x)' too as they both are equal (which was told to us in the first part).

(1/2)(cos 2x − cos 4x)

Integrate '(cos 2x − cos 4x)'

(cos 2x − cos 4x)
[(sin 2x)/2] - [(sin 4x)/4]

Once we've done the integration, we'll plug in the upper limit (pie/3) and the lower limit (pie/6) and obtain a final answer.

(1/2)[(sin 2x)/2] - [(sin 4x)/4]
(1/2)[(sin 2π/3)/2] - [(sin 4π/3)/4] - [(sin 2π/6)/2] - [(sin 4π/6)/4]
(1/2)[(2√3)/8]
(√3)/8

Therefore, the final answer is '(√3)/8'.

 

[Sajjad Hossain]

Guys please help Oct/Nov 2008 Paper 1 Q2 prove the identity sum1 please help fast its freaking me out

 

[ffaadyy]

Guys please help Oct/Nov 2008 Paper 1 Q2 prove the identity sum1 please help fast its freaking me out

[(1+sin x)/cos x] + [cos x/(1+sin x)]
[(1+sin x)^2 + cos^2 x]/[(cos x)(1+sin x)]
[1 + 2 sin x + sin^2 x + cos^2 x]/[(cos x)(1+sin x)]

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

[1 + 2 sin x + 1]/[(cos x)(1+sin x)]
[2 + 2 sin x]/[(cos x)(1+sin x)]
[2 (1 + sin x)]/[(cos x)(1+sin x)]

As '1 + sin x' is present in both the numerator and the denominator, it cancel's out and you are left with '2/(cos x)'.

2/(cos x)

Therefore, LHS=RHS.

 

[zain786]

ans9) integrate the derivative given.
dy = (-k/x^3) dx
dy = -k(x^-3) dx
integrate the whole equation
y = [-k(x^-2)] / -2 +c
y = k/(2x^2) +c
18 = k/ (2) +c
3 = k/(2*4^2) +c
solve both simultaneously
and keep the values of k and c in the integrated equation

thnx alott

 

[zain786]

In the first part of Q8 you found K = 5 or K = -7 so wat u have to do in 8)ii) is substitue the k values in the equation (36/2-x)-2k=x and get the answers. Ans is x=-4 x=8
In Q9)i) integrate -k/x^3 u will get y=k/x^2+c then they gave u the ponts (1,18) and (4,3) substitue this points in place of x and y u will get two equtaions then solve them simultaneously Hope it helped

thanxx broooo

 

[unique840]

thnx alott

ur welcum

 

[aksameerkhan27]

Oct/Nov 2011, Paper 11, Question number 3(ii).
Write down the number of roots of the equation 2 cos 2theta − 1 = 0 in the interval 0 ≤ theta ≤ 2pi.

This is only a 1 mark question.
Please suggest easy method to solve this question.

Urgent.

 

[smzimran]

Q7 (ii) help needed
qp: http://www.xtremepapers.com/CIE/International A And AS Level/9709 - Mathematics/9709_w08_qp_3.pdf
ms: http://www.xtremepapers.com/CIE/International A And AS Level/9709 - Mathematics/9709_w08_ms_3.pdf

 

[smzimran]

Oct/Nov 2011, Paper 11, Question number 3(ii).
Write down the number of roots of the equation 2 cos 2theta − 1 = 0 in the interval 0 ≤ theta ≤ 2pi.

This is only a 1 mark question.
Please suggest easy method to solve this question.

Urgent.

2cos2@ = 1
cos2@ = 0.5
new range is
0 ≤ 2@ ≤ 4pi

in one cycle (0 to 2pi), there are 2 positive values of 2@ (1st and 4th quadrant)
so, for 2 cycles(0 to 4pi), there will be 4 roots...

 

[smzimran]

Q7 (ii) help needed
qp: http://www.xtremepapers.com/CIE/International A And AS Level/9709 - Mathematics/9709_w08_qp_3.pdf
ms: http://www.xtremepapers.com/CIE/International A And AS Level/9709 - Mathematics/9709_w08_ms_3.pdf

^ Anyone???

 

[ffaadyy]

Q7 (ii) help needed
qp: http://www.xtremepapers.com/CIE/International A And AS Level/9709 - Mathematics/9709_w08_qp_3.pdf
ms: http://www.xtremepapers.com/CIE/International A And AS Level/9709 - Mathematics/9709_w08_ms_3.pdf

First of all, we'll find the common perpendicular of the two planes in question.

i j k
2 -1 -3
1 2 2

i ( -2 + 6 ) + j ( -3 -4 ) + k ( 4 + 1 )
4i - 7j + 5k

'4i - 7j + 5k'; This is the common perpendicular of the two planes or more precisely, the direction vector of their line of intersection.

Next, we'll be needing to find a point on this line.

(x).(2)
( y ).(-1) = 7
(z).(-3)

2x - y -3z = 7

(x).(1)
( y ).(2) = 0
(z).(2)

x + 2y + 2z = 0

As it's quiet difficult/impossible to find 3 variables by solving 2 simultaneous equations, we'll take any one of the 3 values ( x, y or z) to be equal to '0'. I have assumed that 'y=0', so we are left with:

2x -3z = 7 and x + 2z = 0

Solve simultaneously to find the values of 'x' and 'z'.

x + 2z = 0
x = -2z

2x -3z = 7
2(-2z) - 3z = 7
-7z = 7
z = -1

x = -2z
x= -2(-1)
z=2

Therefore, the coordinates of a point on this line are:

(2)
(0)
(-1)

Once we've found the coordinates, we can write the line's vector equation.

r = 2i – 1k + t(4i – 7j + 5k)

 

[ninjas4life]

can someone please explain how to do part (ii)?? the answer is 61.2 degrees

 

[smzimran]

First of all, we'll find the common perpendicular of the two planes in question.

i j k
2 -1 -3
1 2 2

i ( -2 + 6 ) + j ( -3 -4 ) + k ( 4 + 1 )
4i - 7j + 5k

'4i - 7j + 5k'; This is the common perpendicular of the two planes or more precisely, the direction vector of their line of intersection.

Next, we'll be needing to find a set of points on this line.

(x).(2)
( y ).(-1) = 7
(z).(-3)

2x - y -3z = 7

(x).(1)
( y ).(2) = 0
(z).(2)

x + 2y + 2z = 0

As it's quiet difficult/impossible to find 3 variables by solving 2 simultaneous equations, we'll take any one of the 3 values ( x, y or z) to be equal to '0'. I have assumed that 'y=0', so we are left with:

2x -3z = 7 and x + 2z = 0

Solve simultaneously to find the values of 'x' and 'z'.

x + 2z = 0
x = -2z

2x -3z = 7
2(-2z) - 3z = 7
-7z = 7
z = -1

x = -2z
x= -2(-1)
z=2

Therefore, the coordinates of a point on this line are:

(2)
(0)
(-1)

Once we've found the coordinates, we can write the line's vector equation.

r = 2i – 1k + t(4i – 7j + 5k)

if we assume x =0 or z = 0,
will the final answer be same?

 

[ffaadyy]

if we assume x =0 or z = 0,
will the final answer be same?

The points will be different but the overall answer will be correct. (2, 0 ,-1) is just one point, it is a line so there'll be many points on it. For e.g; if you take z=0, the point you'll get is (14/5, -7/5, 0) and this is correct too.

 

[smzimran]

The points will be different but the overall answer will be correct. (2, 0 ,-1) is just one point, it is a line so there'll be many points on the line. For e.g; if you take z=0, the point you'll get is (14/5, -7/5, 0) and this is correct too.

Thanx a lot...

 

[ffaadyy]

View attachment 5686can someone please explain how to do part (ii)?? the answer is 61.2 degrees

R=√10
a= 71.56

(√10) cos (2x - 71.57) = 2

The range in which we've to find the angle is 0<Θ<90 but we'll be needing to modify it.

0<Θ<90

multiply it by '2'.

0<2Θ<180

Subtract 71.56 from both the limits.

-71.56<2Θ<108.44

So according to this range, we'll be needing to find the angle in three quadrants; -1, 1 and 2.

(√10) cos (2Θ - 71.57) = 2
cos (2Θ - 71.57) = 0.6324
2Θ - 71.57 = 50.78

As the angle we've found is positive which in turn means that 'cos' is positive, we'll find the angle in the quadrant in which 'cos' is positive. 'Cos' is positive in the '-1' quadrant and the '1' quadrant; thus, it is quiet easy to find the value of Θ now.

'-1' quadrant:

2Θ - 71.57 = -50.78
2Θ = 20.79°
Θ = 10.4°

'1' quadrant:

2Θ - 71.57 = 50.78
2Θ = 122.35
Θ = 61.1°

Therefore, the 2 values of Θ are 10.4° and 61.1°.

 

[ffaadyy]

Thanx a lot...

You're welcome.

 

[ninjas4life]

thank you!!

 

[sea_princess]

can someone please solve:
novemeber 2006 paper 3 : quetion 9 (iv)
june 2007: paper 3:question 8 (i): (only the argument of u)

 

[ffaadyy]

can someone please solve:
novemeber 2006 paper 3 : quetion 9 (iv)

This is the answer to a question asked by smzimran (November 2006, P3, Q9iv).



First of all, we would've already constructed a circle of radius '1 unit' with centre at (1,1) as we were told to do so in the third part of this question. Coming back to the fourth part, we have to find the least value of |z| for points on this locus. If you take a look at the diagram above, the red line which is starting from the origin and ending as soon as it is touching the circle, its length equals to the 'least value of |z|'. We'll also drop a line from the centre of the circle to the x-axis and sort of create a right angled triangle as depicted in the diagram above. Now to calculate the least value of |z|, we'll do the following calculations.

Using the pythagora's theorem, we can easily find the length of the hypotenuse (side AB).

AB^2 = AC^2 + BC^2
AB= 1.41 units.

Once the length of the hypotenuse has been found out, the least value of |z| can be found out by subtracting the length of 'AB' by the radius of the circle (1 unit). By subtracting the radius of the circle (1 unit) from the length of 'AB', we've found out the length of the red line (shown in the diagram) which was the least value of |z|.

1.41 - 1
0.414

Therefore, the least value of |z| equals to '0.414'.