P1 hep needed.urgently!

[nisurju]

can anyone help me in question no 4 of may june 2009 of mathematics p1?it's abt a graph!

 

[arlery]

y = asinbx +c

If you substitute 0 in place of x then answer will be y = 0 +3

so c = 3

y= asinbx + 3

b is the frequency, i.e. the no of waves or cycles in the interval given (0 ≤ x ≤ 2p.)

since there are 2 waves so b = 2

a is the amplitude, i.e., the difference between maximum value to the starting point

so you could either obtain value of a by 9-3 = 6
or 3+3 = 6

y = 6sin2x + 3

 

[nisurju]

awww..thank u so much..!!

 

[arlery]

You're welcome. =)

 

[nisurju]

can u also solve qsn no 10(ii) and 11(i) of may june 2006~(p1)???
in qsn 11,i cud find the value of k but then i cudnt do further

 

[arlery]

All right I'm on it.

 

[arlery]

June 2006 Q 10
(i) k = 27
(ii) dy/dx = 3x^2 -6x-9

for stationary points dy/dx=0

3x^2 - 6x -9 = 0
solve it and you will get x=-1, x = 3
since the minimum point was at x=3,
the maximum point will be at x= -1

therefore, y= (-1)^3 -3(-1)^2-9(-1)+27
y=32

co-ordinates of the maximum point = (-1,32)

(iii) If you look at the graph you can see that the graph is decreasing from the max. pt till the min. pt.
So the set of values for which graph is a decreasing function is -1<k<3

 

[arlery]

11(i) if you've found the values of k then you've solved part one.
(ii) If k=6 ---> f(x) = 6-x
fg(x)=5
6-(9/(x+2))=5
-9/(x+2) = -1
-9=-x-2
x=7
(iii) g(x) = 9/(x+2)
let y = 9/(x+2)
x+2 = 9/y
x=9/y -2
g^-1(x) = 9/x -2

 

[nisurju]

i found the value of k but then what about the root??
and thank u so so so so so sooooo much!

 

[nisurju]

arley,i am already ur fan!!! ur blog is awesome!!! do reply my PM!!