Additional Math Help HERE ONLY...!

[J.Darren]

I got it as <=-6 and >9. I don't know why the mark scheme says its <=-6 but <=10

1-(x^2)+6x

-x^2 + 6x + 1

x = -0.16 or 6.16

dy/dx = -2x + 6

-2x + 6 = 0

x = 3

(-3)^2 + 6 (3) + 1

y = 28

maximum point (3, 28)

<= 28

[adrian1993]

I don't really get what you wrote.

Anyways I managed to figure it out. The maximum point was (-3, 10)

So obviously the domain would be less than 10.

[J.Darren]

I don't really get what you wrote.

Anyways I managed to figure it out. The maximum point was (-3, 10)

So obviously the domain would be less than 10.

Uh I was trying to rewrite it in the quadratic form, as the parabola is going to be an inverted bowl shape, we can conclude that the range has to be smaller than the y coordinates of the maximum point ... Obviously I have made a computation error in working out the ... maximum point.

[J.Darren]

12 hours to go ... argh !

[adrian1993]

I am very excited and tense. XD

[xlane]

could sum1 help me wid oct/nov 09 paper 2..or part (ii)...hw to find da E coordinates..

[theaguia]

could sum1 help me wid oct/nov 09 paper 2..or part (ii)...hw to find da E coordinates..

can u post the question, i dont have the paper

[xlane]

m really sorry.m usin a cellfone..so its not possible

[syedz123]

I did variant 2. What topics came up in your variant 1 for paper 1?

ermm...sets,binomial theorems,indices(surds),functions,diferentiation,polynomial,findin a coordinate at a point of contact of the tangent,rate of change n small of change,i dnt rmmbr even d0in any integrating question..i think da onli one wuz da second one of da last q..i chose da first one which involved difrentiation
like i sed our variant 1 wasnt as tought altough i still made silly mistakes...bt i think our p2 will b harder bt da p2 of variant 2 will b as easy as our p1 to make it fair then...i advice doin da topics i stated above dat came in p1 v1 for us

[jellybeans]

heyyyyyy, i've got a doubt

to find when the curve crosses the x axis, we sub in y = 0 right?

so 6 sin(3x + pi/4) = 0
       sin(3x + pi/4) = 0
           (3x + pi/4) = sin^-1 0
           (3x + pi/4) = 0
                        3x = -pi/4
                         x = -1/12pi

but why does the MS say sin^-1 0 = PI? isnt it supposed to be 0?? help please! thanks in advanceeeee.

[BlackBunny103]

Guys, I need help with this question.
Thanxx

[J.Darren]

heyyyyyy, i've got a doubt

to find when the curve crosses the x axis, we sub in y = 0 right?

so 6 sin(3x + pi/4) = 0
       sin(3x + pi/4) = 0
           (3x + pi/4) = sin^-1 0
           (3x + pi/4) = 0
                        3x = -pi/4
                         x = -1/12pi

but why does the MS say sin^-1 0 = PI? isnt it supposed to be 0?? help please! thanks in advanceeeee.

You are absolutely correct.

[J.Darren]

Guys, I need help with this question.
Thanxx

Construct a right-angle triangle with the longest side pointing north-west (this is the compensate for the tide flow from west to east). The horizontal side with have a length of 2, whereas the vertical line with have a length of 1.5 (90/60), the longest side is the speed the boat travels when it is in still water. The angle would be the one between the vertical side and the longest side.

[J.Darren]

heyyyyyy, i've got a doubt

to find when the curve crosses the x axis, we sub in y = 0 right?

so 6 sin(3x + pi/4) = 0
       sin(3x + pi/4) = 0
           (3x + pi/4) = sin^-1 0
           (3x + pi/4) = 0
                        3x = -pi/4
                         x = -1/12pi

but why does the MS say sin^-1 0 = PI? isnt it supposed to be 0?? help please! thanks in advanceeeee.

Ah, I think I have figured out what has gone wrong, in this case you must use the second solution which is 360 degrees. When x = 0, 6 sin (45), for sure when y = 0, x cannot be smaller than 45 degrees

[jellybeans]

*dum dum dummmmm*
ADD MATH'S OVERRRRRRRR, WOOP