Help on maths PAPER 1 please O/N 2008

[1992]

i cannot do the O/N 2008 Q5.i)
Its is very strange question. how could i posible find a and b.?? working out needed please

 

[1992]

Q7.i) also please. dont understand anything :S

 

[intel1993]

q5...

-1<cosx<1
b> -bcosx > -b
a+b > a- bcosx> a-b

if rearrange

a-b<a-bcox<a+b
now

a-b=-2
a+b=10

solve it similtaneously u will get the ans.................

 

[princesszahra]

what thing in Q7

 

[intel1993]

q7 i )

80= perimeter of square +n circle
80=4x+2 pi r
r=(80-4x)/2 pi

A= x^2 + pi r^2
A= x^2 + pi( 80-4x / 2pi)^2

now solve it u will get the given equation...................

 

[1992]

q5...

-1<cosx<1
b> -bcosx > -b
a+b > a- bcosx> a-b
quote]

dont undersatnd this part :S y did u put -1<cosx<1
and b> -bcosx > -b


and ty !!

 

[intel1993]

as we knoe taht the cos only exist between -1 n 1 so it is its range then......
i tery to make the equation as given.l..
and put -b with cos x due to which signs of inequality changes and further steps given........................

 

[1992]

this is th first time i see this type of question. have seen question like this in other past paper??

 

[1992]

also im dont mind, could u do the question 9.ii) from the same pastpaper O/N2008 :9

 

[intel1993]

its simple integration............

v= pi y^2
pi integral (3x+1)^2 / 2
then use limit of x frm 0 to 1.....as shown in figure....
calculate the volume under graph...
then minus the volume of cylinder frm the volume calculated..

u will get ur ans..

 

[1992]

what volume did u get for volume under the graph?? i got 3pi/2, i think its wrong because i cant get the answer :S
can u also explain the question after it please. i think this is the most dificult pastpaper ive ever done

 

[PlanetMaster]

Volume under curve = π ∫(y²); limit 0-1
Volume under curve = π ∫(3x + 1); limit 0-1
Volume under curve = 5π/2

Volume of cylinder = πr²h
Volume of cylinder = π x 2² x 1
Volume of cylinder = 4π

Volume of shaded region = 4π - (5π/2)
Volume of shaded region = 3π/2

 

[1992]

ty. got it

 

[1992]

anyone know how to do Q10.ii) ?¿?¿?
i got this 6(3x-2) - (3x-2)^2
now i dont know what to do next

 

[PlanetMaster]

Multiply the brackets which will make –9x² + 30x –16.
Now differentiate it and at turning point d/dx = 0.
Find a value of x which will be 5/3.
Show that it is maximum by showing (d²/dx²) < 0
Now substitute x back in gf(x) you found earlier (–9x² + 30x –16) to find max value of gf(x)

 

[XWTBSILT]

its simple integration............

v= pi y^2
pi integral (3x+1)^2 / 2
then use limit of x frm 0 to 1.....as shown in figure....
calculate the volume under graph...
then minus the volume of cylinder frm the volume calculated..

u will get ur ans..

Can u please elaborate more on your answer coz i can do oly upto here

pi { (3x^2 /2) + (x) }
Can u continue from here

 

[XWTBSILT]

ok i got the answer and understood
thanx

 

[zahraahmed]

ohhh sorrieee i did it bymistake......................

 

[filza94]

http://notezone.net/cambridgechem/chemi ... sis%5D.pdf can sum1 go in dix page 11 n do solve n gimme answers plzzzz i cant able to slove it n it might cum for exam help mee!!!!