Mathematics: Post your doubts here!

[unique840]

Please Help !
November 2003 Maths P3 Question 7 (iii)

the circle is the locus (z-u) = 2
{z-(1+2i)}
means the coordinate will be 1,2 from where we will originate our locus
the line is the locus arg z
arg {z - (0+0i)}
means the coordinate will be 0,0 from where we will originate our locus
the greatest value of angle will be the angle between x-axis and the locus of arg z
it will be twice of theta...
find theta by the triangle and multiply it by 2 to find the greatest angle

 

[unique840]

Q10 Part (b) http://www.mediafire.com/download.php?krh8zc8hr8xm25h
Jazak Allah

locus is {z-(0+3i)} is less than equal to 2.
a circle is made with the centre (0,3) with radius 2. the inside of the circle is shaded
the other locus is arg z
arg {z-(0+0i)}
the line is drawn from the origin making tangents with the circle at two points. the line towards right is the one with least argument and the one on the left is the one with the greatest argument. greatest argument will be the least argument + 2(theta).
theta will be (sin inverse of (2/3)) = 41.8
2 theta will be 83.6 degrees
least argument will be 90 - theta = 48.1 degrees
greatest argument will be 48.1 + 83.6 = 131.8 degrees

 

[ffaadyy]

Please Help !
November 2003 Maths P3 Question 7 (iii)

Somebody had asked the same question few pages back, this's how it'll be done.

We'll construct a circle with center 1,2 and radius 2. Using that same diagram for the 'iii' part to find the greatest arg z, we'll draw a tangent to the circumference of the circle and make the x-axis our base line, like the one illustrated in this diagram (tangent and the base line indicated by the red colour):

View attachment 3713

Now to find the arg z, we'll use this diagram:

View attachment 3714

To find the angle, we'll simply do the following calculation:

tan θ = (2/1)
θ = 1.11 rad.

This angle is of one half only so we'll multiply it by '2' to obtain a final answer of '2.21 rad'.

 

[atomique]

Please help me. How do you solve this question?

 

[Doctor Nemo]

Please help me. How do you solve this question?

I will give it a try.

Separate variables:

dx/(x+1) = cos(2O)/sin(2O)dO I do not seem to have the symbols on my computer so O = theta

integrating left side is just ln(x+1)

apply double angle formula to two terms on right side.

cos2(O) -sin2(O) d(O)
2sin(O)cos(O)

simplify

cos(O) - sin(O) d(O)
2sin(O) 2cos(O)

Each of these terms can now be integrated because the top term is the derivative of the bottom, giving

ln(x+1) = 1/2 ln(sin(O)) + 1/2ln(cos(O)) +C = 1/2ln(sin(O)cos(O)) +C = 1/2ln(1/2sin(2O))+ C

Now take a deep breath before continuing.

Substitute

x = 0 when O = pi/12

0 = 1/2 ln (1/2 sin(pi)/6) + C

0 = 1/2 ln(1/4) +C

0 = ln(1/2) + C

C = -ln (1/2) = ln(2)

ln(x+1) = 1/2ln1/2sin(2O) + ln(2)

ln(x+1) = 1/2ln1/2sin(2O) + 1/2ln(4)

ln(x+1) = 1/2ln2sin(2O)

x+ 1 = sqrt(2sin(2O)

(x+1)2 = 2sin(2O)

 

[atomique]

I will give it a try.

Separate variables:

dx/(x+1) = cos(2O)/sin(2O)dO I do not seem to have the symbols on my computer so O = theta

integrating left side is just ln(x+1)

apply double angle formula to two terms on right side.

cos2(O) -sin2(O) d(O)
2sin(O)cos(O)

simplify

cos(O) - sin(O) d(O)
2sin(O) 2cos(O)

Each of these terms can now be integrated because the top term is the derivative of the bottom, giving

ln(x+1) = 1/2 ln(sin(O)) + 1/2ln(cos(O)) +C = 1/2ln(sin(O)cos(O)) +C = 1/2ln(1/2sin(2O))+ C

Now take a deep breath before continuing.

Substitute

x = 0 when O = pi/12

0 = 1/2 ln (1/2 sin(pi)/6) + C

0 = 1/2 ln(1/4) +C

0 = ln(1/2) + C

C = -ln (1/2) = ln(2)

ln(x+1) = 1/2ln1/2sin(2O) + ln(2)

ln(x+1) = 1/2ln1/2sin(2O) + 1/2ln(4)

ln(x+1) = 1/2ln2sin(2O)

x+ 1 = sqrt(2sin(2O)

(x+1)2 = 2sin(2O)

Thank you so much. I like the way you explained it.

 

[zain942010]

the circle is the locus (z-u) = 2
{z-(1+2i)}
means the coordinate will be 1,2 from where we will originate our locus
the line is the locus arg z
arg {z - (0+0i)}
means the coordinate will be 0,0 from where we will originate our locus
the greatest value of angle will be the angle between x-axis and the locus of arg z
it will be twice of theta...
find theta by the triangle and multiply it by 2 to find the greatest angle

Thanks !

 

[zain942010]

Somebody had asked the same question few pages back, this's how it'll be done.

Thanks !

 

[jason234]

0709/01/M/J07 May june 2007 p1 , Q1, 3, 4, and 11

 

[unique840]

Thanks !

ur welcum

 

[unique840]

0709/01/M/J07 May june 2007 p1 , Q1, 3, 4, and 11

ans 3
{1+[(sin^2)x / (cos^2)x]} / {1 - [(sin^2)x / (cos^2)x]}
take the lcm in numerator and denominator respectively
{[(cos^2)x + (sin^2)x] / (cos^2)x} / {[(cos^2)x - (sin^2)x] / (cos^2)x}
(cos^2)x will be cancelled
(cos^2 x + sin^2 x ) / (cos^2 x - sin^2 x)
1 / [(1-sin^2 x) - sin^2 x)]
1 / [1 - 2sin^2 x]

 

[mrpaudel]

0709/01/M/J07 May june 2007 p1 , Q1, 3, 4, and 11

Q1) y=sqrt(4x) and find dy/dx. dy/dx gives the slope of the curve which is also the same for the tangent. So, the slope of tangent is 2(comparing y=2x+c with y=mx+c) so, dy/dx=2 ...and solve it to get the value of x. After u get the value of x, put it in the curve's equation to get the value of y. Putting both values to the equation of tangent gives u a value of c.

Q4) Suppose x^2=a, which makes the equation as 18/a^2 + 1/a = 4.Take LCM and u will get a quadratic equation, factorise and solve it to get the value of a. Put the value of a as x^2 and get the value of x..

Q11)
i) Find f '(x) using chain rule.. Put the value greater than zero in the f '(x) which gives u negative value.thus decreasing.
ii) I hope u can find Inverse of f(x) by interchanging the place of x and y..and solve for y...this is inverse. and domain of inverse of f(x)= range of f(x)....Since we are given the domain of f(x) i.e x greater or equal to zero, put that value carefully to find out the least value..or greater value...Anyways, put the value and find range of f(x). and then u can straightly write domain of Inverse of f..dont need to show it..
iii) Copy the graph of f(x)..and then draw a line y=x. The reflection of f(x) on y=x is the graph of f inverse.
iv)fg(x)=2
or, f(0.5x)=2
or, 6/(2* 0.5x + 3) =2..solve for x.

Well, i assume u just need help..so didnot do all process.. If u need, reply me, i will do the whole process..Also, if i started doin it all, it takes a lot of time to me...hope u understand them...

 

[ousamah112]

maths p4 may/june 2011...p42 q6 part ii and iii...
and maths p3 o/n 06 p3 q9 part iv

 

[mrpaudel]

maths p4 may/june 2011...p42 q6 part ii and iii...
and maths p3 o/n 06 p3 q9 part iv

I assume you did iii)...U should have made a circle with radius 1 and centre at u...what u may be, i hope u have calculated earlier. The point u need to recall is, z is a point anywhere at the circumference of that circle. and calculating |Z| means...calculate the distance from origin to the circumference of circle such that it gives less result..Which means draw a line which joins origin and diamater of circle....the line cuts the circle at two points. now, the one which looks short btwn them is the shortest one...

 

[mrpaudel]

For calculation, u can use distance formula...from origin to U....u may have some pt..it its 2+3i, the co ordinate is (2,3) so...u can calculate the distance i hope...and that distance- radius of a circle gives u a least distance...hope that helped..here's a figure for little more help..

 

[ffaadyy]

Please help me. How do you solve this question?

I will give it a try.

Separate variables:

dx/(x+1) = cos(2O)/sin(2O)dO I do not seem to have the symbols on my computer so O = theta

integrating left side is just ln(x+1)

apply double angle formula to two terms on right side.

cos2(O) -sin2(O) d(O)
2sin(O)cos(O)

simplify

cos(O) - sin(O) d(O)
2sin(O) 2cos(O)

Each of these terms can now be integrated because the top term is the derivative of the bottom, giving

ln(x+1) = 1/2 ln(sin(O)) + 1/2ln(cos(O)) +C = 1/2ln(sin(O)cos(O)) +C = 1/2ln(1/2sin(2O))+ C

Now take a deep breath before continuing.

Substitute

x = 0 when O = pi/12

0 = 1/2 ln (1/2 sin(pi)/6) + C

0 = 1/2 ln(1/4) +C

0 = ln(1/2) + C

C = -ln (1/2) = ln(2)

ln(x+1) = 1/2ln1/2sin(2O) + ln(2)

ln(x+1) = 1/2ln1/2sin(2O) + 1/2ln(4)

ln(x+1) = 1/2ln2sin(2O)

x+ 1 = sqrt(2sin(2O)

(x+1)2 = 2sin(2O)

The final answer isn't correct. This's how this question will be done:

sin 2Ө dx = (x + 1) cos 2Ө dӨ
(dx)/(x+1) = [(cos 2Ө) / (sin 2Ө)] dӨ
ln (x + 1) = (1/2) ln (sin 2Ө) + k

Now to find the value of the constant 'k', we'll substitute 'Ө' with 'pie/12' and 'x' with '0'.

0 = (1/2) ln (1/2) + k
- (1/2) ln (1/2) = k

Put the value of 'k' back in the equation 'ln (x + 1) = (1/2) ln (sin 2Ө) + k'.

ln (x + 1) = (1/2) ln (sin 2Ө) - (1/2) ln (1/2)
ln (x + 1) = (1/2) [ ln (sin 2Ө) - ln (1/2)]
ln (x + 1) = (1/2) ln (2 sin 2Ө)
ln (x + 1) = ln (2 sin 2Ө)^(1/2)

Both the "ln's" cancel out and we are left with:

x+1 = (2 sin 2Ө)^(1/2)
x = (2 sin 2Ө)^(1/2) - 1

Therefore, the final answer is x = (2 sin 2Ө)^(1/2) - 1.

 

[Doctor Nemo]

The final answer isn't correct. This's how this question will be done:

sin 2Ө dx = (x + 1) cos 2Ө dӨ
(dx)/(x+1) = [(cos 2Ө) / (sin 2Ө)] dӨ
ln (x + 1) = (1/2) ln (sin 2Ө) + k

Now to find the value of the constant 'k', we'll substitute 'Ө' with 'pie/12' and 'x' with '0'.

0 = (1/2) ln (1/2) + k
- (1/2) ln (1/2) = k

Put the value of 'k' back in the equation 'ln (x + 1) = (1/2) ln (sin 2Ө) + k'.

ln (x + 1) = (1/2) ln (sin 2Ө) - (1/2) ln (1/2)
ln (x + 1) = (1/2) [ ln (sin 2Ө) - ln (1/2)]
ln (x + 1) = (1/2) ln (2 sin 2Ө)
ln (x + 1) = ln (2 sin 2Ө)^(1/2)

Both the "ln's" cancel out and we are left with:

x+1 = (2 sin 2Ө)^(1/2)
x = (2 sin 2Ө)^(1/2) - 1

Therefore, the final answer is x = (2 sin 2Ө)^(1/2) - 1.

I agree that yours is a better solution since I applied the double angle formula when I didn´t need to and I did not solve for x like the question said. In math questions it is always good to go back and reread the question after you have done the calculations to make sure that when you think you are finished that you really have arrived at the answer that you were asked to obtain.

 

[ninjas4life]

how do you do part (ii)?

 

[elaine]

Help!!! I don't understand the working for paper 7 may june 2003 Q7 part ii....
why they need to divide surd n???

 

[ffaadyy]

how do you do part (ii)?

If we compare the equation 'f(x) = 12x^3 + 25x^2 − 4x − 12' with the equation
'12 × 27^y + 25 × 9^y − 4 × 3^y − 12', it is seen that 'x' has been replaced with '3^y' in the second equation; therefore, we assume that 'x=3^y'. Now to find the value of 'y', we'll use the equation 'x=3^y' with '2/3' as the value of 'x'. This value of 'x' is the factor which has been obtained in the first part of the question after factorising the cubic equation, we'll not use the other two values (factors) of 'x' because they are negative. Coming back to the question, after substituting 'x=2/3' in the equation 'x=3^y', this's how we'll solve it:

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

Put 'ln' on both the sides:

ln (2/3) = y ln 3
[ ln (2/3) ] / ( ln 3 ) = y
- 0.369 = y