Qualification > Math

C3 and C4 DOUBTS HERE!!!

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The Golden Girl =D:
Give me a second ang3l :)

EDIT :

Using Quotient Rule , We differentiate :   { =>  Which is :  dy/dx = [ u'V - V'u ] / V2 }

u = 5x2 - 10x + 9  =>  u' = 10 x - 10

v = (x-1)2 =>  v' = 2 (x-1)

dy/dx =>  [ { (10x-10) [ (x-1)2 } - { 2 (x-1) [ 5x2 - 10 x + 9] } ] /  [( x-1)4 ]

I took (x-1) as a Common factor so I got ;

dy/dx =>  (x-1) { [ (10x-10)(x-1) ] - [2(5x2 - 10x + 9)] } / [ (x-1)4 ]

=> I cancelled out the (x-1) from the numerator and then expanded what is inside the brackets (numerator) and got the following :


dy/dx =>  [ 10x2 - 20x +10 - 18 - 10x2 + 20x] / [ ( x-1 )3 ]

since the 10x and -10x  , the 20x and -20x cancel out we have 10-18 = -8 ;)

So we get ;

dy/dx =>  -8 / [ (x-1 )3 ]

I hope I helped (:


Don't forget to include me in your prayers =D

Malak:
Thanks GG :D

I have another doubt :(

Same paper c3 2005 June Q 5 part d (attached)

I get everything, except that in part d i am getting a negative answer for theta

for theta + 0.588 i get --> 0.429 and 0.5709..

and then for theta i get a negative answer :S

The Golden Girl =D:
Since you knew how to solve the first part then I'll just go into the theta thing (:

the Equation you'll use is =>  sin theta [ squareroot52 sin (theta + 0.588)] =0 from part (c)

so sin theta = 0

we know that sin theta is 0 , at  pie and 2pie .... only 0 is accepted => check  the range ;)

squareroot52 sin (theta + 0.588) = 0

theta + 0.588 = 0 , pie , 2 pie  (cuz I told you ,we know that sin theta is ZERO at these three angles )

so theta = pie - 0.588 = 2.553

theta =2pie -0.588 = 5.70  { rejected TOO big not within the range }


therefore Theta = 0 rad , 2.553 rad


This is one of the Tricks my teacher uses , I hope you got it (:

[I might have to go any minute cuz my bro's  got a project -.- ]

Malak:
but isnt the range greater han 0 and less than pie then why do we consider 2 pie :-\

The Golden Girl =D:
I considered it but I  didn't put it in the final answer you see :)

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