Mathematics: Post your doubts here!

[sea_princess]

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

View attachment 5728

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'.

thanx a lot

 

[ffaadyy]

can someone please solve:
june 2007: paper 3:question 8 (i): (only the argument of u)

Multiply the complex number '2/(-1+ i)' by the conjugate of '-1+ i' to obtain '-1 - i'. Next, find its argument using the formula 'tan x = b/a',

tan x = b/a
tan x = -1/-1
x = 0.785

'Tan' is positive in the third quadrant therefore we'll add this value of 'x' to that of 'π' to obtain a final answer '3.93 rad.'.

 

[RGBM211]

how many even numbers can be formed with the digits 3,4,5,6,7 by using some or all of the numbers(repetitions are not followed)?

 

[XPFMember]

Assalamoalaikum wr wb!

@smzimran and sea_princess: Do check this: Complex No. max/min IzI and arg(z) - P3

how many even numbers can be formed with the digits 3,4,5,6,7 by using some or all of the numbers(repetitions are not followed)?

write down diff. possibilities..1 digit, 2 digit, 3 digit, 4 digit, 5 digit.....even no. = no. ending with even digit = 4 or 6 in this case...

 

[RGBM211]

cod yu plz show the working dnt knw hw to

 

[Utsav.]

need help with this...

 

[RGBM211]

Anyone^ ?

 

[ffaadyy]

need help with this...

Solution:

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

Integrating both the sides.

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

Put x=0 and Ө=(π/12) to find the value of the constant 'c'.

ln (x+1) = (1/2) ln (sin 2Ө) + c
ln 1 = (1/2) ln (1/2) + c
- (1/2) ln (1/2) = c

Put the value of 'c' back into the integrated equation.

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

Recall the identity 'sin 2Ө=2cosӨsinӨ'.

x+1 = (2 sin 2Ө)^(1/2)
x+1 = (4 cos Ө sin Ө)^(1/2)
x+1 = 2 (cos Ө sin Ө)^(1/2)
x = 2 (cos Ө sin Ө)^(1/2) - 1

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

 

[ffaadyy]

Assalamoalaikum wr wb!
@smzimran and sea_princess: Do check this: Complex No. max/min IzI and arg(z) - P3

Talking of Argand Diagrams, the attached file might also help in the construction of Argand Diagrams and finding the least/greatest values.

 

[XPFMember]

how many even numbers can be formed with the digits 3,4,5,6,7 by using some or all of the numbers(repetitions are not followed)?

cod yu plz show the working dnt knw hw to

Assalamoalaikum wr wb!
Sorry for the late reply..
Here’s how you do it. You need to write down different possible combinations. Since they said any number of digits, you gotta consider all of them.
As you know, even number is that ending with a 0, 2, 4, 6 or 8....in this case we have 5 numbers – 3, 4, 5, 6, 7 – and that includes only 2 even digits – 4 and 6.

1 digit:

• 4
• 6

ð 2

2 digits:

• _ 4
• _ 6

ð 4P1 x 2
(cuz for each we have the permutation as 4P1 and we have one with 4 and other with 6...so we get 4P1 x 2)

3 digits:

• _ _ 4
• _ _ 6

ð 4P2 x 2

4 digits:

• _ _ _ 4
• _ _ _ 6

ð 4P3 x 2

5 digits:

• _ _ _ _ 4
• _ _ _ _ 6

ð 4P4 x 2

In each case, we chose our last digit and find the permutation for the other digits...

Hope this helps...

 

[smzimran]

@ffaadyy and @XPF Member:
Thanx a lot...

 

[Chooi1993]

Can anyone help me help me solve oct/nov 2002 Q9 (i) I can find the value of a .
May /June 2003 Q5
Please help me!
Thanks.

 

[s1294]

okay i have question- maths
i have been trying to get it but i get close n then get lost agn
the question is as follows
eqn of curve= y^2+2x=13
eqn of line= 2y+x=k
find the value of k for which the line is a tangent to the curve

now i knw that the discriminant has to be used so that the line meets the curve just once so b^2-4ac=0
now from here when i set both equation to equal to each other i get
(k-x)/2= (13-2x)^0.5 = (k-x)^2/4= 13-2x
from here i get (k-x)^2= 52-8x
what do i do from here
please explain step by step- this is a question frm oct/nov 2011 p12 9709- 9709/12/O/N/11
your help will be greatly appreciated

 

[ffaadyy]

Can anyone help me help me solve oct/nov 2002 Q9 (i) I can find the value of a .
May /June 2003 Q5
Please help me!
Thanks.

Which paper; P1 or P3? And please be a little more clear on where you are getting stuck.

 

[ffaadyy]

okay i have question- maths
i have been trying to get it but i get close n then get lost agn
the question is as follows
eqn of curve= y^2+2x=13
eqn of line= 2y+x=k
find the value of k for which the line is a tangent to the curve

now i knw that the discriminant has to be used so that the line meets the curve just once so b^2-4ac=0
now from here when i set both equation to equal to each other i get
(k-x)/2= (13-2x)^0.5 = (k-x)^2/4= 13-2x
from here i get (k-x)^2= 52-8x
what do i do from here
please explain step by step- this is a question frm oct/nov 2011 p12 9709- 9709/12/O/N/11
your help will be greatly appreciated

You don't have to set both the equations equal to each other. You'll rearrange the equation of the line so as it'll become an expression for 'x' in terms of 'y'.

2y + x = k
x = k - 2y

Replace 'x' with 'x = k - 2y' in the curve equation.

y^2 + 2x = 13
y^2 + 2 (k - 2y) = 13
y^2 + 2k - 4y - 13 =0
y^2 - 4y + (2k - 13) = 0

Now use the discriminant (b^2 - 4ac = 0) to find the value of k.

b^2 - 4ac = 0

a = 1, b = -4, c = 2k -13

b^2 - 4ac = 0
16 - 4 (1) (2k -13) = 0
16 - 8k + 52 = 0
68 = 8k
8.5 = k

Therefore, the value of 'k' is '8.5'.

 

[saimaiftikhar92]

I NEED HELP IN THIS ......PLEASE GUYS:
NOV 2006 P3 QUESTION O 3, 4 AND 5.

I HAVE A MOCK ON MONDAY ..............PLEASE BE QUICK......I MAY HAVE SOME MORE AS WELL

 

[ffaadyy]

I NEED HELP IN THIS ......PLEASE GUYS:
NOV 2006 P3 QUESTION O 3, 4 AND 5.

I HAVE A MOCK ON MONDAY ..............PLEASE BE QUICK......I MAY HAVE SOME MORE AS WELL

Q3(i):

y=6e^x - e^3x
(dy/dx) = 6e^x - 3e^2x
0 = 3e^x ( 2 - e^2x )
0 = 2 - e^2x
e^2x = 2
ln e^2x = ln 2
2x = ln 2
x= 0.35

Q3(ii):

(dy/dx) = 6e^x - 3e^2x
(d^2y/dx^2) = 6e^x - 9e^3x

Substitute the value of 'x=0/35' in the above equation which we've double differentiated. If the answer is negative, it's a maximum value. If the answer's positive, it's a minimum value.

(d^2y/dx^2) = 6e^x - 9e^3x

x=0.35

(d^2y/dx^2) = 6e^x - 9e^3x
(d^2y/dx^2) = 6e^(.35) - 9e^3(.35)
(d^2y/dx^2) = -16.97

As the answer is negative, this point is a 'maximum value'.

Q4:

( y )(dy/dx) = 1 + y^2
[( y )/(1+y^2)] dy = 1 dx

Integrate both the sides.

(1/2) ln (1+y^2) = x + c

Put 'y-2' and 'x=0' to find the value of 'c'.

(1/2) ln (1+y^2) = x + c
(1/2) ln 5 = c

Put back the value of 'c' into the integrated equation.

(1/2) ln (1+y^2) = x + (1/2) ln 5
(1/2) ln (1+y^2) - (1/2) ln 5 = x
(1/2) ln [(1+y^2)/5] = x
(1+y^2)/5 = e^2x
1+y^2 = 5e^2x
y^2 = 5e^2x -1

Therefore, the expression for y^2 in terms of 'x' is 'y^2 = 5e^2x -1'.

Q5(i):


[√(1 + x) +√(1 − x)][
√(1 + x) −√(1 − x)]
(1 + x) + (1-x)(1+x) - (1-x)(1+x) - (1 - x)
(1 + x) - (1 - x)
2x

[√(1 + x) +√(1 − x)][
√(1 + x) −√(1 − x)] = 2x
[√(1 + x) +√(1 − x)] = 2x / [
√(1 + x) −√(1 − x)]

1 / [√(1 + x) +√(1 − x)]
1 / { 2x / [
√(1 + x) −√(1 − x)] }
[
√(1 + x) −√(1 − x)] / 2x

Q5(ii):

As '1 / [√(1 + x) +√(1 − x)]' is equal to '[
√(1 + x) −√(1 − x)] / 2x', we'll instead expand
'[
√(1 + x) −√(1 − x)] / 2x'.

[
√(1 + x) −√(1 − x)] / 2x

Expanding '√(1 + x) ':

√(1 + x) = 1 + (x/2) - (x^2)/8 + (x^3)/16

Expanding '√(1 − x)]':

√(1 - x) = 1 - (x/2) - (x^2)/8 - (x^3)/16

Putting the expanded variables back into the main equation.

[
√(1 + x) −√(1 − x)] / 2x
{[ 1 + (x/2) - (x^2)/8 + (x^3)/16 ] - [ 1 - (x/2) - (x^2)/8 - (x^3)/16 ] } / 2x

Open the brackets and do addition/subtraction of the like terms and then divide them by '2x' to obtain the final answer of:

(1/2) + (x^2)/16

 

[s1294]

You don't have to set both the equations equal to each other. You'll rearrange the equation of the line so as it'll become an expression for 'x' in terms of 'y'.

2y + x = k
x = k - 2y

Replace 'x' with 'x = k - 2y' in the curve equation.

y^2 + 2x = 13
y^2 + 2 (k - 2y) = 13
y^2 + 2k - 4y - 13 =0
y^2 - 4y + (2k - 13) = 0

Now use the discriminant (b^2 - 4ac = 0) to find the value of k.

b^2 - 4ac = 0

a = 1, b = -4, c = 2k -13

b^2 - 4ac = 0
16 - 4 (1) (2k -13) = 0
16 - 8k + 52 = 0
68 = 8k
8.5 = k

Therefore, the value of 'k' is '8.5'.


wow- thanx a lot u made it look so easy. thank you sooo much.

i have one more from the same paper- f(x)=3x+a, g(x)=b-2a
given that ff(2)=10 and g^-1(2)=3 find values of a and b and an expression for fg(x)
so far i have this figured out
ff(2)=10 - 3*(2) + a= 6 +a = 3*(6+a)+a= 18 +3a+a= 18 +4a= 10 a= (10-18)/4= -2
g^-1 (2)= 3 - inverse of b-2x = (x-b)/-2= 3 (2-b)/-2 =3 -b= (3*-2)= -6 -2= -8 b=8
so then fg(x)= f(x)= 3x-2 g(x)= 8-2x so then it is 3*(8-2x) - 2= 24- 6x-2= 22-6x= answer
is it correct?

 

[ffaadyy]

You don't have to set both the equations equal to each other. You'll rearrange the equation of the line so as it'll become an expression for 'x' in terms of 'y'.

2y + x = k
x = k - 2y

Replace 'x' with 'x = k - 2y' in the curve equation.

y^2 + 2x = 13
y^2 + 2 (k - 2y) = 13
y^2 + 2k - 4y - 13 =0
y^2 - 4y + (2k - 13) = 0

Now use the discriminant (b^2 - 4ac = 0) to find the value of k.

b^2 - 4ac = 0

a = 1, b = -4, c = 2k -13

b^2 - 4ac = 0
16 - 4 (1) (2k -13) = 0
16 - 8k + 52 = 0
68 = 8k
8.5 = k

Therefore, the value of 'k' is '8.5'.


wow- thanx a lot u made it look so easy. thank you sooo much.

i have one more from the same paper- f(x)=3x+a, g(x)=b-2a
given that ff(2)=10 and g^-1(2)=3 find values of a and b and an expression for fg(x)
so far i have this figured out
ff(2)=10 - 3*(2) + a= 6 +a = 3*(6+a)+a= 18 +3a+a= 18 +4a= 10 a= (10-18)/4= -2
g^-1 (2)= 3 - inverse of b-2x = (x-b)/-2= 3 (2-b)/-2 =3 -b= (3*-2)= -6 -2= -8 b=8
so then fg(x)= f(x)= 3x-2 g(x)= 8-2x so then it is 3*(8-2x) - 2= 24- 6x-2= 22-6x= answer
is it correct?

You're welcome. As far as the other question is concerned, this's how it'll be done:

f : x → 3x + a

ff(2)=10

y = 3x + a
3 (3x + a) + a = 10
9x + 3a + a = 10
9x + 4a = 10
9(2) + 4a = 10
a = -2

g : x → b − 2x

g−1(2) = 3

y = b - 2x
2x = b - y
x = (b - y)/2

g−1 = (b - x)/2

y = (b - x)/2

g−1(2) = 3

3 = (b-2)/2
6 = b - 2
8 = b

Therefore, a=-2 and b=8.

(ii):

fg(x)

fg(x) = 3x - 2
fg(x) = 3 (8-2x) - 2
fg(x) = 24 - 6x - 2
fg(x) = 22 - 6x

 

[s1294]

You're welcome. As far as the other question is concerned, this's how it'll be done:

f : x → 3x + a

ff(2)=10

y = 3x + a
3 (3x + a) + a = 10
9x + 3a + a = 10
9x + 4a = 10
9(2) + 4a = 10
a = -2

g : x → b − 2x

g−1(2) = 3

y = b - 2x
2x = b - y
x = (b - y)/2

g−1 = (b - x)/2

y = (b - x)/2

g−1(2) = 3

3 = (b-2)/2
6 = b - 2
8 = b

Therefore, a=-2 and b=8.

(ii):

fg(x)

fg(x) = 3x - 2
fg(x) = 3 (8-2x) - 2
fg(x) = 24 - 6x - 2
fg(x) = 22 - 6x

Thanks agn- so i was right just my methods slightly different. thank you!!!!!!!!