Mathematics: Post your doubts here!

[MEGUSTA_xD]

This might be a senseless question but how did you take square of (x^2-4x+4)?

(x^2-4x+4)(x^2-4x+4), i think the question should be why you took the square of (x^2-4x+4)

 

[CaptainDanger]

This might be a senseless question but how did you take square of (x^2-4x+4)?

Wrote it twice and multiplied... There must be something that we are not applying here...

 

[CaptainDanger]

(x^2-4x+4)(x^2-4x+4), i think the question should be why you took the square of (x^2-4x+4)

To get volume we have to square the given equation... If I take it, shouldn't we end up with the same answer?

 

[Gémeaux]

Wrote it twice and multiplied... There must be something that we are not applying here...

the answer comes the same this way too, i.e 32π/5
writing it twice n multiplying gives
x^4 - 8x^3 + 24x^2 - 32x + 16
and integrating it we get, [(1x^5)/5 - 2x^4 +8x^3 - 16x^2 + 16x] and the limits are 2 and 0, putting these in the integrated form,
we get (32/5 -32 +64 -64 +32 )π
32π/5

 

[MEGUSTA_xD]

To get volume we have to square the given equation... If I take it, shouldn't we end up with the same answer?

Oh ye. Well, intregating (x-2)^4 gives me the same result as mubarka got.

 

[XPFMember]

9709_s11_qp_11.pdf

9709_s11_ms_11.pdf

Q3 ii) I do get the answer by just squaring the equation (x-2)^4 and integrating it...
But what if I open the equation y= (x-2)^2, take its square (x^2-4x+4)^2 and now integrate it? I don't end up getting the same answer. WHY?
If I further open it (x^2-4x+4)^2 and then integrate it I do get the answer... But why not without opening it?

Assalamoalaikum wr wb!

u can integrate the equation directly if it's in the form (ax + b)^n and NOT otherwise unlike differentiation!
so u don't need to open the brackets!
integrate (x-2)^4 directly
=[(x-2)^5 ]/(5 x 1 )
and simplify...

 

[student92]

http://www.xtremepapers.com/CIE/International A And AS Level/9709 - Mathematics/9709_s10_qp_11.pdf
q1,
asalam alikum, help me please
and what are asending powers of x?
thanks alot
marking scheme
http://www.xtremepapers.com/CIE/International A And AS Level/9709 - Mathematics/9709_s10_ms_11.pdf

 

[CaptainDanger]

the answer comes the same this way too, i.e 32π/5
writing it twice n multiplying gives
x^4 - 8x^3 + 24x^2 - 32x + 16
and integrating it we get, [(1x^5)/5 - 2x^4 +8x^3 - 16x^2 + 16x] and the limits are 2 and 0, putting these in the integrated form,
we get (32/5 -32 +64 -64 +32 )π
32π/5

I said I do get it this way... You opened it there... I mean if we don't open it and integrate it directly... Integrate this (x^2-4x+4)^2

Oh ye. Well, intregating (x-2)^4 gives me the same result as mubarka got.

Yes I know that... But if you don't open it (x^2-4x+4)^2 and integrate it... Then its not the same...

Assalamoalaikum wr wb!

u can integrate the equation directly if it's in the form (ax + b)^n and NOT otherwise unlike differentiation!
so u don't need to open the brackets!
integrate (x-2)^4 directly
=[(x-2)^5 ]/(5 x 1 )
and simplify...

Yes. But when you open the first equation... Square it and then integrate this (x^2-4x+4)^2 without opening... The answer isn't same... Where as it should come that way too?

 

[smzimran]

http://www.xtremepapers.com/CIE/International A And AS Level/9709 - Mathematics/9709_s10_qp_11.pdf
q1,
asalam alikum, help me please
and what are asending powers of x?
thanks alot
marking scheme
http://www.xtremepapers.com/CIE/International A And AS Level/9709 - Mathematics/9709_s10_ms_11.pdf

Use the double angle formula in (i) and (ii)
Use a right angled triangle in (iii)
Its very simple.
And ascending powers of x mean that moving from left to right in an equation, the terms should have x with its power increasing
For example:
1 + 2x + 4x^2 - 6x^3 + 8x^4

 

[smzimran]

9709_s11_qp_11.pdf

9709_s11_ms_11.pdf

Q3 ii) I do get the answer by just squaring the equation (x-2)^4 and integrating it...
But what if I open the equation y= (x-2)^2, take its square (x^2-4x+4)^2 and now integrate it? I don't end up getting the same answer. WHY?
If I further open it (x^2-4x+4)^2 and then integrate it I do get the answer... But why not without opening the second equation?

That is because there is no formula applicable for integration of (ax^2 + bx +c)^n

 

[CaptainDanger]

That is because there is no formula applicable for integration of (ax^2 + bx +c)^n

Thank you!

 

[smzimran]

Thank you!

U r welcome!

 

[sea_princess]

I need help with part (ii), I don't understand how he got the differentaition to
"0.1 ln(a)-0.1 ln(10 -a) " where did the negative sign come from?
I attached the question and the markscheme below

 

[XPFMember]

aoa wr wb!

Yes. But when you open the first equation... Square it and then integrate this (x^2-4x+4)^2 without opening... The answer isn't same... Where as it should come that way too?

yup...that's what i said UNLIKE differentiation
in integration that thing is ONLY applicable, when it is in the form (ax+b)^n..!
though in differentiation, u can follow it for any form...(i hope u get what i mean by this )

 

[Mustehssun Iqbal]

Assalamu alaikum,
November 2003, Paper 3, Q.4, 6 (2) , 7 (3) , 8 and 10.

 

[ffaadyy]

I need help with part (ii), I don't understand how he got the differentaition to
"0.1 ln(a)-0.1 ln(10 -a) " where did the negative sign come from?
I attached the question and the markscheme below

When you'll integrate '1/(10-a)', the integrand will come out to be '-ln (10-a)'.

 

[smzimran]

I need help with part (ii), I don't understand how he got the differentaition to
"0.1 ln(a)-0.1 ln(10 -a) " where did the negative sign come from?
I attached the question and the markscheme below

First list the partial fractions:
1/a(10-a) = A/a + B/(10-a)
1/a(10-a) = [A(10-a) + B(a)]/a(10-a)
Eliminate denominator
1 = A(10-a) + B(a)---eq1
let a=10 in eq1
1 = A(10-10) + B(10)
B = 1/10
let a=0 in eq1
1 = A(10-0) + B(0)
A = 1/10

So,
1/a(10-a) = 1/10a + 1/10(10-a)

Now, back to the differential equation
da/dt = 0.004a(10-a)
Separating variables
da/a(10-a) = 0.004 dt
Substitute the partial fraction results
1/10(1/a )da + 1/10(1/10-a) da = 0.004 dt
Integrating:
1/10(lna) - 1/10(ln(10-a)) = 0.004t + ln c
[sign change because of formula as it is ln(ax+b) and a=-1 and while integreting a is divided by the integral]
We know that t=0, a=5 So,
ln5/10 - ln5/10 = 0.004(0) + lnc
0 = 0 + ln c
So ln c = 0

1/10(lna) - 1/10(ln(10-a)) = 0.004t
1/10(lna - ln(10-a)) = t/250
lna - ln(10-a) = t/25
ln(a/10-a) = t/25
t = 25 ln(a/10-a)

Solved...

 

[ffaadyy]

Assalamu alaikum,
November 2003, Paper 3, Q.4, 6 (2) , 7 (3) , 8 and 10.

Q4.

We'll be needing to separately differentiate both 'x' and 'y'.

√x +√y = √a

Differentiating 'x' and 'y':

(-1/2√x) + (-1/2√y)(dy/dx) = 0
-1/2√x = (1/2√y)(dy/dx)

We'll now take the term in 'y' on the other side so as the equation will be dy/dx in terms of 'x' and 'y'.

-(2√y)/(2√x) = dy/dx
-√(y/x) = dy/dx

Therefore, the answer is ' dy/dx = -√(y/x)'.

Q6(ii)

We'll first find the x-coordinate of 'A' because that'll be our upper limit when we'll be intergrating y = (3 − x)e^−2x.

y = (3 − x)e^−2x
0 = (3 − x)e^−2x
0 = 3 - x
x = 3

Therefore, we'll be integrating 'y = (3 − x)e^−2x' from '0' (lower limit) to '3' (upper limit).

Integrating the curve equation:

(3 − x)e^−2x
3e^-2x - (x)(e^-2x)

'3e^-2x' can be integrated easily and the answer will come out to be '-(3e^-2x)/2' but we'll be needing to use the integration by parts technique to integrate '(x)(e^-2x)'

Integrating '(x)(e^-2x)' by parts:

(x)(e^-2x)
-[(x)(e^-2x)]/2 - [ 1 x -(e^-2x)/2 ]
-[(x)(e^-2x)]/2 + (1/2) [(e^-2x)]
-[(x)(e^-2x)]/2 + (1/2) (-(e^-2x)/2)
-[(x)(e^-2x)]/2 - (e^-2x)/4

Now we'll put back both the integrands in our original equation '3e^-2x - (x)(e^-2x)'

3e^-2x - (x)(e^-2x)
-(3e^-2x)/2 - {-[(x)(e^-2x)]/2 - (e^-2x)/4}
-(3e^-2x)/2 + [(x)(e^-2x)]/2 + (e^-2x)/4

Next, plug in the upper (3) and the lower limit (0) in this equation '-(3e^-2x)/2 + [(x)(e^-2x)]/2 + (e^-2x)/4'

-(3e^-2x)/2 + [(x)(e^-2x)]/2 + (e^-2x)/4
-(3e^-6)/2 + 3(e^-6)/2 + (e^-6)/4} - [ (-3/2) + (1/4) ]
(e^-6)/4 + (5/4)
( 5 + e^-6)/4

Therefore, the answer is '(5 + e^-6)/4'

Q7(iii)



Sketch an Argand diagram showing the point representing the complex number u (1 + 2i). Next draw a circle of radius '2' with center at (1 + 2i). Now to find the greatest value of arg z for points on this locus, please refer to the above diagram. We'll draw two tangents from (0,0) to the circle of which one will touch the circle at x=1 and the other will touch it at some value of y. Draw two lines from the center of the circle to the tangents, this will sort of create a triangle. Further drawing a line from (0,0) to the center of the circle will create two right angle triangles. Now if you see one of the two right angled triangles, the base is 1 unit and the opposite side is 2 units. From this, we can easily find the angle between the hypotenuse and the base using the 'tan Θ' formula.

tan Θ = 2 / 1
Θ = 1.11 rad.

The angle we have found out is of one half only therefore we'll multiply it by two in order to find the greatest value of arg z.

1.11 x 2 = 2.21 rad.

Therefore, the greatest value of arg z is 2.21 rad.

Q8(i)

Expand '(x −1)(x^2 +1)' to 'x^3 - x^2 + x -1'. Divide 'x^3 + 0x^2 −x −2' by 'x^3 - x^2 + x -1' and the value of quotient will be the value of 'A'. Next, use the value of the remainder 'x^2 - 2x - 1' as the numerator and '(x −1)(x^2 +1)' as the denominator and then solve the partial fractions way.

(x^2 - 2x - 1)/(x −1)(x^2 +1) = B/(x-1) + (Cx + D)/(x^2 + 1)

keep the value of x=1

(1 - 2 - 1) / 2 = B
-1 = B

(x^2 - 2x - 1)/(x −1)(x^2 +1) = -1/(x-1) + (Cx + D)/(x^2 + 1)
(x^2 - 2x - 1)/(x −1)(x^2 +1) + -1/(x-1) = (Cx + D)/(x^2 + 1)
(2x^2 - 2x)/(x-1) = Cx + D
2x(x - 1)/(x - 1) = Cx + D
2x + 0 = Cx + D

Therefore, C = 2 and D = 0.

Q8(ii)

Integrate '1 - 1/(x-1) + 2x/(x^2+1)'.

1 - 1/(x-1) + 2x/(x^2+1)
x - ln x + ln (x^2+1)

Plug in the upper limit (3) and the lower limit (2).

[3 - ln 2 + ln 10] - [2 - ln 1 + ln 5]
3 + ln 5 - 2 - ln5
1

Therefore, '1' is the final answer.

 

[smzimran]

^^ ???

 

[ffaadyy]

Assalamu alaikum,
November 2003, Paper 3, Q.4, 6 (2) , 7 (3) , 8 and 10.

Q10(i)

r = i −2k +s(2i +j +3k)

x(L) = 1 + 2s
y(L) = s
z(L) = -2 + 3s

r = 6i −5j +4k +t(i −2j +k)

x(M) = 6 + t
y(M) = -5 - 2t
z(M) = 4 + t

Equate x(L) with x(M).

1 + 2s = 6 + t
-5 + 2s = t

Equate y(L) with y(M).

s = -5 - 2t
s = -5 - 2(-5 + 2s)
5s = 5
s=1

Substitute s=1 in '-5 + 2s = t'.

-5 + 2s = t
t = -3

Substitute the value 't=-3' in 'x(L)', 'y(L)' & 'z(L)' and the value of 's=1' in 'x(M)', 'y(M)' & 'z(M)' to show that they intersect.

x(L) = 3
y(L) = 1
z(L) = 1

x(M) = 3
y(M) = 1
z(M) = 1

Therefore, the point of intersection is ( 3, 1, 1).

Q10(ii)

First we'll find the direction of the plane using the common perpendicular rule.

i j k
2 1 3
1 -2 1

i (1+6) + j (3-2) + k (-4-1)
7x + y - 5z = k

To find the value of 'k', substitute the point of intersection of the lines in '7x + y - 5z = k'

7(3) + 1 - 5 = k
17 = k

Therefore, the equation of the plane is 7x + y - 5z = 17.