Mathematics: Post your doubts here!

[bakhita]

Was Gave my papers this november
Wbu?

oh nice...I'll Inshallah take my exams in June of 2016 (maths, chem and Bio)
which subjects you took?

 

[nehaoscar]

can anyone show me how to do Question 4 of 9709_s14_qp_31?
It is about differential equations- I don't know why the answer has k in the answer instead of just +c.
http://maxpapers.com/syllabus-materials/mathematics-9709-a-level/attachment/9709_s14_qp_31/

 

[***amd***]

can u tell me the answer.?
im getting 42.67 for the first part. and 0.5 for the second

Accurate! very nice indeed. You got it right for the first part! but not for the second though, it's the same as first both parts have answer 42.67

the answer is same for both the parts since you are to find the area of the same region in two (almost) different ways. For the first part, i guess u both have already done it, but fr the second one, i guess u guys are getting confused by the 'horizontal components'. All u need to do is, rotate your x-y plane by 90 degrees, now your x-axis will be your y axis, and your y axis will be your x axis. I guess I made it more confusing. Just replace the y with x, and x with y. and u get the equation x = sq rt (16-y).......( as given in second part).
integrate this eq wrt y. u get (-2 (16-y)^2/3)/3. Apply the limits y = 16 (when x = 0) and y = 0 on this and u get 128/3 = 42.67
hope i made it clear enough.

 

[Anum96]

can anyone show me how to do Question 4 of 9709_s14_qp_31?
It is about differential equations- I don't know why the answer has k in the answer instead of just +c.
http://maxpapers.com/syllabus-materials/mathematics-9709-a-level/attachment/9709_s14_qp_31/

You can take c or k. Its the same thing

 

[Anum96]

the answer is same for both the parts since you are to find the area of the same region in two (almost) different ways. For the first part, i guess u both have already done it, but fr the second one, i guess u guys are getting confused by the 'horizontal components'. All u need to do is, rotate your x-y plane by 90 degrees, now your x-axis will be your y axis, and your y axis will be your x axis. I guess I made it more confusing. Just replace the y with x, and x with y. and u get the equation x = sq rt (16-y).......( as given in second part).
integrate this eq wrt y. u get (-2 (16-y)^2/3)/3. Apply the limits y = 16 (when x = 0) and y = 0 on this and u get 128/3 = 42.67
hope i made it clear enough.

Oh damn. I did exact same thing. Miscalculation of the power. I must've subtracted 1 instead of adding it. *facepalm* Moreover I was too lazy to redo it.
Thank you though!

 

[Anum96]

can anyone show me how to do Question 4 of 9709_s14_qp_31?
It is about differential equations- I don't know why the answer has k in the answer instead of just +c.
http://maxpapers.com/syllabus-materials/mathematics-9709-a-level/attachment/9709_s14_qp_31/

Oh wait. That k is a constant neha. They have just written k instead of the answer. I got 2ln(2 plus e^3x) So k is basically 2. Dont get confused. If you want me to solve it then I will.... ?

 

[nehaoscar]

Oh wait. That k is a constant neha. They have just written k instead of the answer. I got 2ln(2 plus e^3x) So k is basically 2. Dont get confused. If you want me to solve it then I will.... ?

yes could you solve it please?

 

[Anum96]

yes could you solve it please?

Sure. 5 minutes

 

[bakhita]

the answer is same for both the parts since you are to find the area of the same region in two (almost) different ways. For the first part, i guess u both have already done it, but fr the second one, i guess u guys are getting confused by the 'horizontal components'. All u need to do is, rotate your x-y plane by 90 degrees, now your x-axis will be your y axis, and your y axis will be your x axis. I guess I made it more confusing. Just replace the y with x, and x with y. and u get the equation x = sq rt (16-y).......( as given in second part).
integrate this eq wrt y. u get (-2 (16-y)^2/3)/3. Apply the limits y = 16 (when x = 0) and y = 0 on this and u get 128/3 = 42.67
hope i made it clear enough.

I'd understood that though thank you!

 

[Anum96]

yes could you solve it please?

Dy/dx = 6ye^3x/(2+e^3x)

Dy/y = 6e^3x/(2+e^3x)dx

ʃ dy/y = ʃ 6e^3x/(2+e^3x)dx

lny = 6 ʃ e^3x/(2+e^3x)dx

lny = 6 (1/3)ln(2+e^3x) + C

Now you have y = 36 and x = 0. Use them to find C.

ln36 = 2ln(2+e^3(0)) + C

ln36 = 2 ln(2 +1) + C

ln36 = 2ln3 + C

ln 36 = ln3^2 + C

ln 36 =ln9 + C

ln36 – ln9 = C

ln(36/9) = C

C = ln4

Re-write.

lny = 2ln(2 + e^3x) + ln4

lny = ln(2 + e^3x)^2 + ln4

lny = ln[4((2 + e^3x)^2]

ln cancels out.

y= [4((2 + e^3x)^2] Ans.

 

[nehaoscar]

Dy/dx = 6ye^3x/(2+e^3x)

Dy/y = 6e^3x/(2+e^3x)dx

ʃ dy/y = ʃ 6e^3x/(2+e^3x)dx

lny = 6 ʃ e^3x/(2+e^3x)dx

lny = 6 (1/3)ln(2+e^3x) + C

Now you have y = 36 and x = 0. Use them to find C.

ln36 = 2ln(2+e^3(0)) + C

ln36 = 2 ln(2 +1) + C

ln36 = 2ln3 + C

ln 36 = ln3^2 + C

ln 36 =ln9 + C

ln36 – ln9 = C

ln(36/9) = C

C = ln4

Re-write.

lny = 2ln(2 + e^3x) + ln4

lny = ln(2 + e^3x)^2 + ln4

lny = ln[4((2 + e^3x)^2]

ln cancels out.

y= [4((2 + e^3x)^2] Ans.

Thankyou soooo much! I was taking the 6y to the other side in the first step so it was becoming all complicated and i wasn't getting the answer
Thanks a lot!

 

[nehaoscar]

Part (ii) please

 

[Anum96]

View attachment 57707
Part (ii) please

ii) Maximum point. Put dy/dx = 0

dy/dx ;

y = e^(2sinx) * cosx

u = e^(2sinx)

u’ = 2cosx*e^(2sinx)

v = cosx

v’ = -sinx

dy/dx = uv’ + vu’

0 = [e^(2sinx)* -sinx ] + [cosx*2cosx*e^(2sinx)]

0 = e^2sinx [-sinx + 2cos^2(x)]

0 = -sinx + 2cos^2(x)

Sinx = 2(1 – sin^2(x))

sinx= 2 – 2sin^2(x)

2sin^2(x) + sinx -2 = 0

Apply the quadratic formulae. You will get a negative answer. Ignore that. The second answer you will get will be 0.78

Lastly,

Sinx = 0.78

x=sin^-1(0.78)

x=0.8959 Radians ̴̴ 0.896 Ans.

 

[musiclover gurl]

(x-2)^2 + 3
I am asked to sketch this graph. All that is ok. I found its minimum point and y-intercept and sketched. But what is confusing is that when I equated the equation(that is, y=0) , I got two answers for x.
But how is that possible when the minimum point of the graph is (2,3) and it doesn't touch the x-axis at all.
Help please?

 

[***amd***]

(x-2)^2 + 3
I am asked to sketch this graph. All that is ok. I found its minimum point and y-intercept and sketched. But what is confusing is that when I equated the equation(that is, y=0) , I got two answers for x.
But how is that possible when the minimum point of the graph is (2,3) and it doesn't touch the x-axis at all.
Help please?

Hello? When y=0 , (x-2)^2 = -3, and that means no real roots for that eq. 2 values for x must be in terms of i

 

[nehaoscar]

Part (ii) please

 

[Anum96]

View attachment 57728
Part (ii) please

tell me ur ans to i

 

[nehaoscar]

tell me ur ans to i

Oh yes sorry

(3/(3-2x)) + ((-x-2)/(x^2+4))
i.e.

 

[Anum96]

Oh yes sorry

(3/(3-2x)) + ((-x-2)/(x^2+4))
i.e.
View attachment 57730

Okay so here it is:

Bring the denominators up.

3(3 + 2x)^(-1) + (-x – 2)[(x^2 + 4)^(-1)]

Expand.

3(3 + 2x)^(-1) + [-x (x^2 + 4)^(-1)] – [2((x^2 + 4)^(-1)]

You need to separately expand :

(3 + 2x)^(-1) & (x^2 + 4)^(-1)

So,

(3 + 2x)^(-1) = 3^(-1)[1 +2x/3]^(-1)

= 1/3 (1 + 2x/3)^(-1)

1/3 (1 + (-1)(2x/3) + [(-1)(-1-1)]/2! (2x/3)^2

1/3 ( 1-2x/3+(-1)(-2)/2!(2x/3)^2

1/3 ( 1 – 2x/3 + 2/2! (4x^2/9)

1/3 ( 1 – 2x/3 + 4x/9)

1/3 – 2x/9 + 4x^2/27

Similarly,

(x^2 + 4)^(-1)

Is same as

(4 + x^2)^(-1)

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

¼ [1 + (-1)(x^2/4)

¼ [1 – x^2/4]

¼ - x^2/16

Plug back in.

3(1/3 – 2x/9 + 4x^2/27) – x(¼ - x^2/16) -2(¼ - x^2/16)

Simplify;

1 – 2x/3 + 4x^2/9 – x/4 + x^3/16 – ½ + x^2/8

Combine the like terms and solve.

1 – ½ - 2x/3 –x/4 + 4x^2/9 + x^2/8 [ term with power 3 is not needed]

½ - 11x/12 + 41x^2/72 Ans.

Okay, Im guessing I messed up somewhere Whats the answer?

 

[nehaoscar]

Okay so here it is:

Bring the denominators up.

3(3 + 2x)^(-1) + (-x – 2)[(x^2 + 4)^(-1)]

Expand.

3(3 + 2x)^(-1) + [-x (x^2 + 4)^(-1)] – [2((x^2 + 4)^(-1)]

You need to separately expand :

(3 + 2x)^(-1) & (x^2 + 4)^(-1)

So,

(3 + 2x)^(-1) = 3^(-1)[1 +2x/3]^(-1)

= 1/3 (1 + 2x/3)^(-1)

1/3 (1 + (-1)(2x/3) + [(-1)(-1-1)]/2! (2x/3)^2

1/3 ( 1-2x/3+(-1)(-2)/2!(2x/3)^2

1/3 ( 1 – 2x/3 + 2/2! (4x^2/9)

1/3 ( 1 – 2x/3 + 4x/9)

1/3 – 2x/9 + 4x^2/27

Similarly,

(x^2 + 4)^(-1)

Is same as

(4 + x^2)^(-1)

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

¼ [1 + (-1)(x^2/4)

¼ [1 – x^2/4]

¼ - x^2/16

Plug back in.

3(1/3 – 2x/9 + 4x^2/27) – x(¼ - x^2/16) -2(¼ - x^2/16)

Simplify;

1 – 2x/3 + 4x^2/9 – x/4 + x^3/16 – ½ + x^2/8

Combine the like terms and solve.

1 – ½ - 2x/3 –x/4 + 4x^2/9 + x^2/8 [ term with power 3 is not needed]

½ - 11x/12 + 41x^2/72 Ans.

Okay, Im guessing I messed up somewhere Whats the answer?

Thanks a lot! I'm sorry actually i seem to have typed out the question wrong in the image!
It's like this (3/(3-2x)) + ((-x-2)/(x^2+4)) and i accidentally typed the first denominator as 3+2x
I'm really sorry but i got the procedure
the answer is
1/2 + 5x/12 +41x^2/72