Ok now let me recall all the various questions...
Qn 11)
I got the graph wrong by including the bottom part too. 1 mark
gone. Anyway for the 2nd part, it is possible to show the
expression although the step is tedious (I did and cancelled my
original answer twice. Did in pencil on the qn paper before copying
the solution onto my paper later on). Integration one was rather
easy. Just need to eliminate the cos t term by multiplying by dt/du
which is 1/cos t.
Qn 10)
I believe the majority were asking about the values for n which S
> 4a. Ok at first, u have to find ur d in terms of a. Use ur ar
- a = 3d to get the result that d = -1/9 a. Then sub this
expression back into the inequality. You will get a quadratic
inequality of something like an^2 - 19an + 72a < 0. You can
divide by a as a > 0 as defined by the question. Therefore u get
n^2 - 19n + 72
Qn 9)
2nd part: When X1=0, X will increase towards alpha. X1 = 1, X will
decrease towards alpha. X1 = 2, X will increase to infinity.
3rd part: Proving that X(n+1)>Xn for Xn between values of alpha
and beta. To prove it, first express X(n+1)-Xn = 1/3 * e^(Xn) - Xn.
Let Xn be x, therefore expression will give you 1/3 of the equation
of the graph (e^x - 3x). When Xn is between alpha and beta, e^x -
3x is always less than zero. Since 1/3 * anything negative is less
than zero, therefore X(n+1) - Xn < 0, X(n+1) < Xn. You can
prove the other case using this method too.
That's all for qn 9, 10 and 11.
*EDIT* Range of values for n should be 6 to 13, not 5 to 13. Sorry
if I made anyone panic.
