Mathematics: Post your doubts here!

[ffaadyy]

It's pi/4. though I don't really know it's right or wrong.

Are both of these arguments 'arg(z-1)= pi/6' and 'arg(z-1)= pi/3' correct?

 

[rowinekenny]

Are both of these arguments 'arg(z-1)= pi/6' and 'arg(z-1)= pi/3' correct?

Yes, it's correct.

 

[anonymous123]

thxx 1 more question.....i just went through a ms and deduced dis property from a questn..
f(x) = asdsad for the domain a > x > b
now the inverse will hav the the range a > f(x) > b
is dis correct?

help plzz

 

[ffaadyy]

Yes, it's correct.

I presume there's something wrong with this question. If you try doing this question by sketching an argand diagram, there's no way you can do it because the lines won't ever meet as both the angles will be made from (1,0). I am attaching a similar question which I found in my book, you might want to read it.

 

[rowinekenny]

I presume there's something wrong with this question. If you try doing this question by sketching an argand diagram, there's no way you can do it because the lines won't ever meet as both the angles will be made from (1,0). I am attaching a similar question which I found in my book, you might want to read it.

View attachment 6685

sorry, there's a mistake here: it's arg(z-i)=pi/6 and arg(z-1)= pi/3. I am really sorry, I made a mistake while typing the question.

 

[ffaadyy]

sorry, there's a mistake here: it's arg(z-i)=pi/6 and arg(z-1)= pi/3. I am really sorry, I made a mistake while typing the question.

Okay, so this's how you'll do this question. We have been told that arg(z-i)=pi/6 and arg(z-1)= pi/3. Let's assume that 'z = x + i y'.

z - i

z = x + i y

z - i
x + i y - i
x + i (y - 1)

The argument of 'z - i' is pi/6, therefore:

tan^-1 [(y-1)/x] = pi/6
(y-1)/x = 1/(√3)
(√3)y - √3 = x

We now have an equation for 'x' in terms of 'y'. Next, we'll find another equation using 'arg(z-1)= pi/3'.

z - 1
x + i y - 1
(x - 1) + i y

The argument of 'z - i' is pi/3, therefore:

tan^-1 [y/(x-1)] = pi/3
y/(x-1) = √3
(√3)x - √3 = y

We now have two equation's, one for 'x' and one for 'y' and we'll be solving them simultaneously.

x = (√3)y - √3 ------ i
y = (√3)x - √3 ------ ii

Substitute 'x = (√3)y - √3 ' in the place of 'x' in the second equation.

y = (√3)x - √3
y = (√3) (√3)y - √3] - √3
y = 3 y - 3 - √3
3 + √3 = 2y
(3 + √3)/2 = y

Substitute '(3 + √3)/2 = y' in the first equation to find the value of 'x'.

x = (√3)y - √3
x = (√3)[(3 + √3)/2] - √3
x = (3 + √3)/2

Once we've found out the values of 'x' and 'y', we'll put them back in the equation 'z = x + i y' and find the 'arg z'.

z = x + i y
z = [(3 + √3)/2] + i [(3 + √3)/2]

tan (arg z) = b/a
tan (arg z) = [(3 + √3)/2] / [(3 + √3)/2]
tan (arg z) = 1
arg z = (tan^-1) 1
arg z = pi/4

Therefore, arg z = pi/4.

 

[anonymous123]

9709_s11_qp_11
Q8 part ii
in ms the answer seems wrong sm1 check it plz

 

[smzimran]

9709_s11_qp_11
Q8 part ii
in ms the answer seems wrong sm1 check it plz

No, its not wrong!
Here goes:
a = 1000
n = 40
r = 110 / 100 = 1.1

Sn = a (r^n - 1) / (r - 1)
Sn = 1000 (1.1^40) / (1.1 - 1)
Sn = 443,000 [3 s.f]
Now,
Total charity = 5% of Sn = 5/100 * 443,000 = 22,100

 

[anonymous123]

9709_s11_qp_41
q7 part iii
plz explain it wid solution...the one in ms is too obscure

 

[ffaadyy]

9709_s11_qp_41
q7 part iii
plz explain it wid solution...the one in ms is too obscure

The speed with which 'B' hits the floor is equal to the initial velocity of 'A'. Therefore, we'll find the final velocity of 'B'.

v = u + at
v = 0 + (2.5)(1.6)
v = 4 m/s

The string will become slack when it reaches its maximum height. At the maximum height, A's final velocity will be 0 m/s and its acceleration will be equal to -10. Using the formula 'v = u + gt', we can find out the time taken for 'A' to reach its maximum height.

v = u + gt
o = 4 - 10t
4 = 10t
0.4s = t

Now, this time we have found out is the time taken until 'B' becomes slack. As the question has asked us to include the time until the string becomes taut again, we'll be multiplying this time by '2' .

t = 0.4 x 2
t = 0.8s

 

[hassam]

Okay, so this's how you'll do this question. We have been told that arg(z-i)=pi/6 and arg(z-1)= pi/3. Let's assume that 'z = x + i y'.

z - i

z = x + i y

z - i
x + i y - i
x + i (y - 1)

The argument of 'z - i' is pi/6, therefore:

tan^-1 [(y-1)/x] = pi/6
(y-1)/x = 1/(√3)
(√3)y - √3 = x

We now have an equation for 'x' in terms of 'y'. Next, we'll find another equation using 'arg(z-1)= pi/3'.

z - 1
x + i y - 1
(x - 1) + i y

The argument of 'z - i' is pi/3, therefore:

tan^-1 [y/(x-1)] = pi/3
y/(x-1) = √3
(√3)x - √3 = y

We now have two equation's, one for 'x' and one for 'y' and we'll be solving them simultaneously.

x = (√3)y - √3 ------ i
y = (√3)x - √3 ------ ii

Substitute 'x = (√3)y - √3 ' in the place of 'x' in the second equation.

y = (√3)x - √3
y = (√3) (√3)y - √3] - √3
y = 3 y - 3 - √3
3 + √3 = 2y
(3 + √3)/2 = y

Substitute '(3 + √3)/2 = y' in the first equation to find the value of 'x'.

x = (√3)y - √3
x = (√3)[(3 + √3)/2] - √3
x = (3 + √3)/2

Once we've found out the values of 'x' and 'y', we'll put them back in the equation 'z = x + i y' and find the 'arg z'.

z = x + i y
z = [(3 + √3)/2] + i [(3 + √3)/2]

tan (arg z) = b/a
tan (arg z) = [(3 + √3)/2] / [(3 + √3)/2]
tan (arg z) = 1
arg z = (tan^-1) 1
arg z = pi/4

Therefore, arg z = pi/4.

brother the picture u posted.....cn u solve that question in this way...i guess there's a problem in this way on that question

 

[ffaadyy]

brother the picture u posted.....cn u solve that question in this way...i guess there's a problem in this way on that question

I guess the method explained in that example is pretty easy as compared to the one which I've used to solve the other question and it won't take much time too. Secondly, you are correct; you can't calculate 'tan pi/2' and that's where we have to stop. If you try solving the question which was asked by sketching an argand diagram, the answer will be same i.e 'pi/4'.

 

[anonymous123]

9709_w10_qp_11
q7 part i
how to find the range?? in ms its f(x) =< 3

 

[hassam]

I guess the method explained in that example is pretty easy as compared to the one which I've used to solve the other question and it won't take much time too. Secondly, you are correct; you can't calculate 'tan pi/2' and that's where we have to stop. If you try solving the question which was asked by sketching an argand diagram, the answer will be same i.e 'pi/4'.

yea brother ....i understood that method of diagram...yea its hell easier....well i have one question from p3 book....
cn u help me on that.....(btw i hav done the past papers ;they are quite easy so thought of doing some p3 questions)
if arg(1/3 - z)=pi/6 then what is arg(3z-1)

 

[anonymous123]

9709_w10_qp_41
Q3
i need a detailed solution to this plz..

 

[Arjun Dhanak]

Hi I need help with differentiation and integration..
Pls tell me how we find dy/dx and d^2y/dx^2 of a curve y =(2x-3)^3-6x. This is Qs8 from 9709/01/O/N/07
AND how do we use integration to find the area of a region... like y=(2x^3+5x)/3x. limits x=2 x=5. This isn't from a past paper... its made up...
PLS HELP ME!!

 

[hassam]

is like (2x-3)^3 - 6x

 

[atomique]

Can anyone help me?
Question 3, May/June 2011 Paper 41.

 

[ffaadyy]

yea brother ....i understood that method of diagram...yea its hell easier....well i have one question from p3 book....
cn u help me on that.....(btw i hav done the past papers ;they are quite easy so thought of doing some p3 questions)
if arg(1/3 - z)=pi/6 then what is arg(3z-1)

This question is about scaling of complex numbers and I don't think that its included in our syllabus. Still, this's how you'll do it:

arg (1/3 - z) = pi/6

Assume that z' = 1/3 - z.

z' = 1/3 - z
3z' = 1 - 3z
3z - 1 = -3z'

Scaling a complex number (by a positive real number) doesn't change the argument, just the modulus. Scaling by a negative real number just rotates the vector by 'pi' so you just need to account for this. In the case of '3z - 1 = -3z'', we can either add 'pi' to 'pi/6' or subtract 'pi' from 'pi/6'; the answer will be correct either way.

'pi/6' - 'pi'
-5pi/6

'pi/6' + 'pi''
7pi/6

Both the answers, '-5pi/6' and '7pi/6', are correct.

Check this diagram for a better understanding of this question.

 

[ffaadyy]

Hi I need help with differentiation and integration..
Pls tell me how we find dy/dx and d^2y/dx^2 of a curve y =(2x-3)^3-6x. This is Qs8 from 9709/01/O/N/07
AND how do we use integration to find the area of a region... like y=(2x^3+5x)/3x. limits x=2 x=5. This isn't from a past paper... its made up...
PLS HELP ME!!

y =(2x-3)^3-6x
dy/dx = 3(2x-3)^(2) (2) - 6
dy/dx= 6(2x-3)^(2) - 6

-

dy/dx= 6(2x-3)^(2) - 6
d^2y/dx^2 = 6(2)(2x-3)
d^2y/dx^2 = 48x - 72

-

y=(2x^3+5x)/3x
y = (2x^3)/3x + 5x/3x
y = (2x^2)/3 + 5/3

Integrate with respect to 'x'.

y = (2x^2)/3 + 5/3
y = (2x^3)/9 + 5x/3

Put the limits.

(2x^3)/9 + 5x/3
{[2(5)^3]/9 + 5(5)/3} - {[2(2)^3]/9 + 5(2)/3}
36.1 - 5.1
31

The area of the shaded region iis 31 units.