Mathematics: Post your doubts here!

[aaakhtar19]

please I hope u will solve question 9)iii) nov qp 33 2011 mathematics A2 9709
but please as soon as possible because my exam is on 16 May
thanks

 

[2pac]

Whoaa that was a hectic task to write all that at MS Word :\
By the way here u go !!

You sir,deserve a Trophy and a million likes.
thanks a lot mate.

 

[aaakhtar19]

You sir,deserve a Trophy and a million likes.
thanks a lot mate.

Ahhh No problem !!
Prayers Needed

 

[riry]

i)
Taking theta = x
a=x rads
a6= 4x rads

S6 = 6/2 (2(x) + 5d)
We know that S6 = 2pi
2pi = 6x + 15d ----1

a6 = a + 5d
4x = x + 5d
d=3x/5 ----2

Put 2 in 1 and you should get the value of x.

The perimeter of the smallest sector = rx + r + r

thank u for help easy

 

[PhoenixAsh12]

Does anyone know how to differentiate this? Thanks

 

[Most_UniQue]

Find g-1(x) and replace x by 3.

Yah thats my question , How can I do that?

 

[Most_UniQue]

Does anyone know how to differentiate this? Thanks

Is the answer (2x(pi +4) -160)/pi?

 

[usmiunique]

O/N/2011 P33 q 10(i)
http://www.xtremepapers.com/papers/...S Level/Mathematics (9709)/9709_w11_qp_33.pdf
how do we integrate the tan equation?
i cant go any farther than ((u^n+2+u^n)/(u^2+1))du. the limits are 1 and 0

 

[Dug]

Yah thats my question , How can I do that?

y=4-3sinx
3sinx=4-y
sinx=4-y /3
x=sin-1(4-y/3)

g-1(3) = sin-1(4-3/3)
g-1(3) = 0.340

 

[PhoenixAsh12]

Is the answer (2x(pi +4) -160)/pi?

Here's the answer

 

[Most_UniQue]

Here's the answer

isnt this the question?

 

[Most_UniQue]

y=4-3sinx
3sinx=4-y
sinx=4-y /3
x=sin-1(4-y/3)

g-1(3) = sin-1(4-3/3)
g-1(3) = 0.340

Ah man thats so simple! Lol tnx a lot!!

 

[Dug]

Does anyone know how to differentiate this? Thanks

To find derivatives of equations like this one, its best to break it up.
A= [(pi x^2 +4x^2) / pi] - (160x/pi) + (1600/pi)
Simplifying further,
A = [ x^2 + {(4x^2)/pi}] - 160x/pi + 1600/pi
dA/dx = 2x + 8x/pi -160/pi

 

[2pac]

Ahhh No problem !!
Prayers Needed

okay correct me if I am wrong but AB=OB-OA right?

 

[Most_UniQue]

To find derivatives of equations like this one, its best to break it up.
A= [(pi x^2 +4x^2) / pi] - (160x/pi) + (1600/pi)
Simplifying further,
A = [ x^2 + {(4x^2)/pi}] - 160x/pi + 1600/pi
dA/dx = 2x + 8x/pi -160/pi

I gave the same answer But I didn't simplify it yours is long , just simply seperate 1/pi frn the equation and then find the derivative and put back 1/pi again

 

[PhoenixAsh12]

isnt this the question?

Sorry uploaded the wrong one

 

[PhoenixAsh12]

To find derivatives of equations like this one, its best to break it up.
A= [(pi x^2 +4x^2) / pi] - (160x/pi) + (1600/pi)
Simplifying further,
A = [ x^2 + {(4x^2)/pi}] - 160x/pi + 1600/pi
dA/dx = 2x + 8x/pi -160/pi

How do we get it in this form though?

Thanks

 

[mominzahid]

People please help i cant figure out how to solve the second part. the answer in marking scheme is 7pi/12 but some how i end up getting pi/12.. please help asap i have amy maths exam tomorrow. :'(
this question is from summer 09 qp1.

 

[Dug]

I gave the same answer But I didn't simplify it yours is long , just simply seperate 1/pi frn the equation and then find the derivative and put back 1/pi again

I think the 'long' method reduces the chances of errors because everything is kinda 'opened'. Plus, I am sure you've heard the saying "Old habits die hard".

 

[Dug]

How do we get it in this form though?

Thanks

If the question asked you to get it in this form, then my method is not suitable. You should use the formal method just like Most_UniQue did.