Mathematics: Post your doubts here!

[saimaiftikhar92]

here you go

THANKYOU SOOOOOOOOOOOOO MUCH !!!

 

[saimaiftikhar92]

I NEED HELP IN THIS ONE:

P3 MAY HUNE 2008 QUESTION NUMBER 2, 3 , 4, 5

 

[ffaadyy]

I NEED HELP IN THIS ONE:

P3 MAY HUNE 2008 QUESTION NUMBER 2, 3 , 4, 5

Q2:

e^x + e^2x = e^3x

Take 'y=e^x'.

e^x + e^2x = e^3x
y + y^2 - y^3 = 0
y ( 1 + y - y^2) = 0
- y^2 + y + 1 = 0
y^2 - y - 1 = 0

a=1, b=-1, c=-1

y = [-b+-√(b^2 - 4ac)]/2a
y = [1+-√(5)]/2

y = 1.62

Put this value of 'y' back into the equation 'y = e^x' to calculate the value of 'x'.

y = e^x
1.62 = e^x
ln 1.62 = ln e^x
0.481 = x

Therefore, 'x=0.481'.

Q3(i):

Arc length of the sector:

s=rθ
s=rx

Perimeter of the sector:

r + r + s
2r + rx

Perimeter of the rectangle:

3a + 3a + a + a
8a

We need to eliminate 'a' & 'r' and introduce 'sin x' in this equation. We can do so by considering the triangle ANX and solving for sin x = opp/hyp.




sin x = NX/AN
sin x = a / r
r = a / sin x

We've been told that the perimeter of the sector is half the perimeter of the recttangle.

2r + rx = (1/2)(8a)
2 ( a / sin x ) - ( a / sin x ) x = 4a
( 2a / sin x ) - ( xa / sin x ) = 4a
a [ (2 + x)/sin x ] = 4a
(2 + x)/sin x ] = 4
(1/4)(2+x) = sin x

Hence, shown.

Q3(ii):

Keep 'x=0.8' in the equation '(sin^−1)[(2 + x)/4]' as the initial value and show the result of each iteration.

1. 0.8
2. 0.7754
3. 0.7668
4. 0.7638
5. 0.7628
6. 0.7625
7. 0.7623

Therefore, the value of the root is '0.76'.

Q4(i):

tan(30◦ + θ) = 2 tan(60◦ − θ)
(tan 30 + tan θ)/(1 - tan30 tanθ) = 2 [(tan 60 - tan θ)/(1 + tan 60 tan θ)]

[(1/√3) + tan θ)]/[1 - (1/√3) tanθ] = 2 [(√3 - tan θ)/(1 + √3 tan θ)]
[(1/√3) + tan θ)]/[1 - (1/√3) tanθ] = 2 [(√3 - tan θ)/(1 + √3 tan θ)]
[1 + √3 tan θ]/[√3 - tanθ] = 2 [(√3 - tan θ)/(1 + √3 tan θ)]
[1 + √3 tan θ]^2 = 2 [√3 - tanθ]^2
1 + 2√3 tan θ + 3 tan^2 θ = 6 - 4√3 tanθ + 2 tan^2 θ
1 + 2√3 tan θ + 3 tan^2 θ - 6 + 4√3 tanθ - 2 tan^2 θ = 0
tan^2 θ + 6√3 tan θ - 5 = 0

Q4(ii):

tan^2 θ + 6√3 tan θ - 5 = 0

a=1, b=6√3, c=-5

tan θ = [-b +- √(b^2 - 4ac)]/2a
tan θ = [-6√3 +- √(6√3)^2 + 20)]/2a
tan θ = [ -6√3 +- √128 ]/2

When 'tan θ = 0.46', θ = 24.7◦.
When 'tan θ = -10.6', θ = 5.3◦.

Q5(i):

z = 2 cos θ + i(1 − 2 sin θ)

|z − i| = 2

z - i
2 cos θ + i(1 − 2 sin θ) - 1i
2 cos θ + 1i − i 2 sin θ - 1i
2 cos θ - i 2 sin θ

Find the modulus of '2 cos θ - i 2 sin θ ' and show that it equals to '2'.

√[(2 cos θ)^2 + (- 2 sin θ)^2]
√[4 cos^2 θ + 4 sin^2 θ]
√[4 (cos^2 θ + sin^2 θ)]

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

√[4 (cos^2 θ + sin^2 θ)]
√[4 (1)]
2

Therefore, the modulus is '2'.

For the sketching part, draw a circle of radius '2' with its' center at (0,1).

Q5(ii):

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 )]

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.

 

[1357913579]

You are not wrong you just need to further simplify your answer...

thanks man.

 

[saimaiftikhar92]

THANKYOU.....
I HAVE SOME MORE:

MAY JUNE 2007 QUESTION 6 AND 8
NOVEMBER 2007 QUESTION 7 AND 10

 

[ffaadyy]

june o7 question no. 9 part two! p3 I know that haw to apply formula and to get the angle .
but unfortunately m unable to get the other direction vector of the plane OAB


we are with the eq. of the plane ABC

how to get the other one?? to get both direction ratios


A detailed exlaination will be good!

If you've done the first part of this question, you must have found out the direction vector of the plane ABC which comes out to be (4,2,1). We'll use the same method to find the direction vector of the plane OAB and then find the acute angle between these two planes.

Plane OAB.

AO = (-2, 0, 0)
AB = (-1, 2, 0)

Using these two directions, we'll find their common perpendicular.

i j k
-2 0 0
-1 2 0

i(0) + j(0) + k(-4)

Therefore, the direction vector of the plane OAB=(0, 0, -4).

Now we'll simply use the angle formula to find the acute angle between them.

(direction vector of OAB) x (direction vector of ABC) = [modulus of the direction vector of OAB) x [modulus of the direction vector of ABC] x cos θ

(0) x (4)
(0) x (2) = [4]x[21]^(1/2) cos θ
(-4) x (1)

cos θ = -0.23
θ = 180 - 102.6
θ = 77.4°

 

[ffaadyy]

THANKYOU.....
I HAVE SOME MORE:

MAY JUNE 2007 QUESTION 6 AND 8
NOVEMBER 2007 QUESTION 7 AND 10

Q6(i), June 2007

Area of triangle AOB:

A = (1/2) a b sin x
A = (1/2) r^2 sin x

Area of the sector:

s = r x

A = (1/2) r x s
A = (1/2) r x rx
A = (1/2) r^2 x

Area of the Triangle = (1/2) Area of the sector

(1/2) r^2 sin x = (1/4) r^2 x
sin x = x/2
2 sin x = x

Q6(ii), June 2007

2 sin x = x
2 sin x - x

Substitute 'π/2' in the place of 'x'.

2 sin x - x
2 sin π/4 - π/4
0.628

Substitute '2π/3' in the place of 'x'.

2 sin x - x
2 sin 2π/3 - 2π/3
-0.36

Q6(iii), June 2007

x = (1/3)(x + 4 sin x)
3x = x + 4 sin x
3x - x = 4 sin x
2 x = 4 sin x
x = 2 sin x

Q6(iv), June 2007

x = (1/3)(x + 4 sin x)

1. 1.8
2. 1.8985
3. 1.8952
4. 1.8955
5. 1.8955
6. 1.8955
7. 1.8955

Therefore, the value of 'x' is '1.90'.

Q8(i), June 2007

u = 2/(-1+i)

2/(-1+i) x (-1 - i)/(-1 - i)
( - 2 - 2i )/2
-1 - i

Therefore 'u = -1 - 1i'.

Modulus of 'u' = 1.41
argument of 'u' = tan^-1 (1/1) + π = 3.93 rad.

u^2:

(-1 - 1i)(-1 - 1i)
1 + 1i + 1i -1
2i

Modulus of u^2 = 2
Argument of u^2 = tan^-1 (2/0) = (infinite). 'Tan' is infinite at 90° therefore the argument of u^2 is '90°'.

Q8(ii), June 2007.

This's the diagram with the correct shading:



First of all, you'll construct a circle of radius '2' with centre at the origin (0,0). According to the inequality '|z|<2', area inside the circle will be shaded. Next, you'll construct a perpendicular bisector for the inequality '|z-u^2|<|z-u|'. Mark the complex numbers '-1,-1' (u) and '0+2i' (u^2) on the argand diagram. To construct a perpendicular bisector, place the compass on the point '-1,-1' , open it more than half and mark 2 arcs (one above and one below). Repeat the same procedure for the point '0+2i' and join both the arcs by a straight line (this line is the perpendicular bisector). Now comes the shading part. For the inequality '|z-u^2|<|z-u| , area above the perpendicular bisector will be shaded and according to the inequality '|z|<2', area inside the circle will be shaded. Therefore, the area which we'll be shading should be above the perpendicular bisector and inside the circle.

Also note that the circle and the perpendicular bisector are dotted lines, not solid lines. We can recognize whether it'll be a dotted line or a solid line by reading the inequality.

 

[1357913579]

iam not able to prove in mayjune 2010 paper-31 number-9 please have a look at what i have done and tell me where iam wrong and solve it to get the answer only for first part.
iam getting the answer as (1-x)^1/2 times by (1+x)^3/2 not able to get the same aswer required to be proved here is the attached files please reply as soon as possible.

I need help in same question which is mayjune 2010 paper-31 number-9 but this time part-2
though i know the concept like they have said the p is where the normal has maximum gradient wich i think is like 1/0 (infinite) and it will occur when the gradient of tangent=0 iam not sure whether this is how to solve so please solve this problem asap.

 

[ffaadyy]

I need help in same question which is mayjune 2010 paper-31 number-9 but this time part-2
though i know the concept like they have said the p is where the normal has maximum gradient wich i think is like 1/0 (infinite) and it will occur when the gradient of tangent=0 iam not sure whether this is how to solve so please solve this problem asap.

All what you'll need to do is differentiate the gradient of the normal to the curve and to find its maximum point, equate it to '0' and obtain two values of x.

y = (1 + x)√(1 − x^2)

(dy/dx) = (1+x)[(-2x)/(2√(1 − x^2)] + 1[√(1 − x^2]
0 = (-x)(1+x) + 1 - x^2
0 = -x - x^2 + 1 -x^2
0 -2x^2 - x + 1
2x^2 + x - 1 = 0
2x^2 + 2x - 1x - 1 = 0
2x(x + 1) - 1(x + 1) = 0
x = 1/2 & x=-1

As 'P' lies in the positive x-axis, we'll disregard the value 'x=-1' and the x-coordinate of 'P' is '1/2'.

 

[1357913579]

thanks i got what to do
but didnt get why we have to do that.
because we already get the gradient of normal then why again differentiation? concept?

 

[1357913579]

though i saw the marking sceme and what you said was written there but for me it doesnt make logic like why do to that?
(never studied that)

 

[saimaiftikhar92]

THANKYOUU


PLEASE GIUDE ME IN THIS ONE;

MAY JUNE 2008 QUESTION NUMBER 6 AND 8

AND OCT/NOV 2008 7 AND 10

 

[leadingguy]

though i saw the marking sceme and what you said was written there but for me it doesnt make logic like why do to that?
(never studied that)

well to prove that the curve is having a maximum or minimum value at the point p we have to double differentiate!

i.e

d^2y/dx^2.... thats why he has differentiated this again

if simple gradient was asked than we can simply take the differentiation (no need for doing it again)




d^2y/dx^2 > 0 then itx a maximum value
d^2y/dx^2 < 0 then itx a minimum value ( it could be maximum for < or minimum fr > I m not sure but this so.)

now it has been told in the question that p is having a maximum value here so we have to show it by differentiating it again

I hope U understand it.

refer book as well, ITX in the book of ADDMATHS in olevels andin Alevels core maths too.

itx a simple rule

 

[1357913579]

well to prove that the curve is having a maximum or minimum value at the point p we have to double differentiate!

i.e

d^2y/dx^2.... thats why he has differentiated this again

if simple gradient was asked than we can simply take the differentiation (no need for doing it again)




d^2y/dx^2 > 0 then itx a maximum value
d^2y/dx^2 < 0 then itx a minimum value ( it could be maximum for < or minimum fr > I m not sure but this so.)

now it has been told in the question that p is having a maximum value here so we have to show it by differentiating it again

I hope U understand it.

refer book as well, ITX in the book of ADDMATHS in olevels andin Alevels core maths too.

itx a simple rule

the thing is i know all that youre saying ofcourse i have studied that
but
the questions doesnt say maximum of the curve because we d d^2/dx^2 so see the nature of the curve whether its maxima or minima but the question says gradient of the NORMAL to the curve HAS ITS maximum value at p
bu

 

[leadingguy]

the thing is i know all that youre saying ofcourse i have studied that
but
the questions doesnt say maximum of the curve because we d d^2/dx^2 so see the nature of the curve whether its maxima or minima but the question says gradient of the NORMAL to the curve HAS ITS maximum value at p
bu

notice that the gradient of the normal to the curve is always the same to the gradient of the curve at that point!!!

MEAN to say that whatever is the gradient of the curve at a particular point will surely be the gradient of the tangent at that point on the curve.

this clears that gradient of tangent at a point on curve will be the gradient fo curve too; at that point

look at the question it says that the gradient of the normal is maximum at point "P" MEans that at point P gradient of the curve will also be maximum.!!!!!


now how to find a maximum point????

jxt double differentiate the eq. to get the x coordinate of P.

I hope to be write but m not sure??? as m too giving the same paper this year, (IT should be true )

 

[saimaiftikhar92]

PLEASE HELP ME IN THE QUESTIONS I HAVE POSTED................................

 

[aksameerkhan27]

june o7 p3 question 6 part 2 (6ii)
Pls help how to verify.

 

[saimaiftikhar92]

PLEASE GIUDE ME IN THESE:

MAY JUNE 2008 QUESTION 6 AND 8

OCT NOV 2008 QUESTION 7 AND 10

 

[leadingguy]

PLEASE GIUDE ME IN THESE:

MAY JUNE 2008 QUESTION 6 AND 8

OCT NOV 2008 QUESTION 7 AND 10

may june o8 question 6:
its says that there is a point on which the tangent is paralel. the tangent is paralel to x axis means the gradient will be zero.!!!!!!!!!!!!!
therefore frst differentiate the eq. xy(x+y) =2a^3


while differetiating remember that its an implicit function so will be differentiated at itx pattern.

aftr U get UR differentiation put it equal to zero as tangent paralel to x axis is always with zero gradient.

U will be having UR eq. as ......... y= -2x

substitute this value of y in the original eq. U will have the value of x = a

hope U are asking for the same?? or U need full working??

 

[ffaadyy]

june o7 p3 question 6 part 2 (6ii)
Pls help how to verify.

Q6(ii), June 2007

2 sin x = x
2 sin x - x

Substitute 'π/2' in the place of 'x'.

2 sin x - x
2 sin π/4 - π/4
0.628

Substitute '2π/3' in the place of 'x'.

2 sin x - x
2 sin 2π/3 - 2π/3
-0.36

There's a simple rule of verifying. In iteration, you'll be given two values and a formula which in this case was 'π/2' & '2π/3' and 'x = 2 sin x' respectively. Arrange the formula as this 'x - 2 sin x', insert the two values one by one and calculate a final answer. If you are getting a negative answer and a positive answer, then it's verified.