Question

5i and iii pleaseee.
http://www.xtremepapers.com/papers/CIE/Cambridge%20International%20A%20and%20AS%20Level/Mathematics%20(9709)/9709_w03_qp_3.pdf

Answer 1

@ganeshie8
Maybe helpful.

Answer 2

tried Intermediate Value Theorm?

Answer 3

@divu.mkr no could you help me with this

Answer 4

@hartnn

Answer 5

if the function is continuous in one interval then it holds the condition of IVT..it simply derives that a root lies between x1 and x2 if f(x1)>0 and f(x2)<0 or the converse

Answer 6

like there cannot exist a continuous function which dont cross X-axis without changing its sign

Answer 7

you mean iii)?

Answer 8

can you pls write out the steps foh me?

Answer 9

let f(x)= sec x +x^2-3
f(0)=-ive
f(pi/2)=+ive
f(0).f(pi/2)<0
then there must exist at least one root between (0,pi/2)
secx +x^2-3=0 for some x

Answer 10

what the hell is this? :O

Answer 11

its the theorem

Answer 12

(1.025, 1.035) This is the answer. How do you work it to this?

Answer 13

we have to prove, not the exact values

Answer 14

can you write it out thoroughly for me?

Answer 15

at least, i could understand.

Answer 16

i have written all steps..there's nothing beyond that.. you better go-through the theory part

Answer 17

I don't rmb doing all this honestly. All I know it's just calculator work.

Answer 18

It's just some iterative method.

Answer 19

maybe

Answer 20

what? i'm confused?

Answer 21

@welshfella Any idea? (:

Answer 22

are you talking about number 5(iii)?

Answer 23

yes

Answer 24

OK well you need a scientific calculator to do this. There's one on your PC or on the web

You first plug 1 into the formula given and work it out

x1 = 1:-
x2 = cos-1 (1 / (3 - 1^2) = 1.04719
x3 = cos-1(1 / 3 - 1.04719^2) = 1.017638
x4 = cos-1(1 / (3 - 1.017638^2) = 1.036706

Answer 25

that's all? what about 1.025?

Answer 26

yea - i see what you mean - I'm a bit confused with this . We are looking for the root of the equation - that is the point where the curve crosses the x axis
I'll have to give this more thought....

Answer 27

x5 is 1.025 tho

Answer 28

hey, i think it's alright. would you help me with no 6 please?

Answer 29

oh ok

Answer 30

to find the x coordinate of M given its a minimum you have to find the derivative , equate this to zero and solve for x

Can you differentiate y = (3 -x)e^(-2x) ?

Answer 31

no. D:

Answer 32

You can use the product rule
dy/dx = (3-x)* e^(-2x) * -2 + -1*e^(-2x) = 0
-2e^(-2x)(3-x) - e^(-2x) = 0
e^(-2x)(-2(3-x) - 1) = 0
e^(-2x)(2x - 7) = 0

Answer 33

now solve that equation for x

Answer 34

e^(-4x^2 + 14x) = 0

Answer 35

No ignore the e^-2x
as the 2 functions are multiplied together either can equal 0
so 2x - 7 = 0
so x = ?

Answer 36

alright thanks. what about ii)?

Answer 37

you need to integrate the function between the limits x = 0 and x = value at A

find A by solving (3 - x)e^(-2x) = 0

Answer 38

how do i solve that? sorry i really have no clue about it.

Answer 39

in a similar way to last equation either e^(-2x) = 0 or 3 - x = 0

Answer 40

ignore the e part

Answer 41

3-x = 0

Answer 42

okay so A (3,0) then?

Answer 43

yes that the coordinattes of A

now you have to find the area by integrating between the limits x = 0 and x = 3

Answer 44

alright.