Question

YEAR 11 hard question(long) Help please?

Answer 1

Need help with C only but i dont think we need to do 1 or b to do c

Answer 2

This is some of my working

Answer 3

@jim_thompson5910 HELP PLS
@sami-21

Answer 4

For \(SD\), we have a rather simple difference of squares:$$SD=\frac12(2^x+2^{-x})(2^x-2^{-x})=\frac12(2^{2x}-2^{-2x})$$For \(S+D\), we combine and reduce$$S+D=\frac12(2^x+2^{-x}+2^x-2^{-x})=\frac12\times2\times2^{x}=2^x$$For \(S-D\), we take a very similar approach:$$S-D=\frac12(2^x+2^{-x}-2^x+2^{-x})=\frac12\times2\times2^{-x}=2^{-x}$$For \(S^2-D^2\), recall this difference of squares is equivalent to \((S+D)(S-D)\), which we can easily:$$S^2-D^2=2^x2^{-x}=1$$

Notice the similarity with the hyperbolic functions: http://en.wikipedia.org/wiki/Hyperbolic_function

Answer 5

Thanks for your working out but i only got stuck on number c

Answer 6

For (b), it's relatively simple as well; to get a quadratic in terms of \(2^x\), we need to get rid of our negative exponent -- by multiplying by \(2^x\):$$S(2^x)=2^{2x-1}+2^{-1}\\\frac12(2^{2x})-S(2^x)+\frac12=0\\\frac12(2^x)^2-S(2^x)+\frac12=0\\(2^x)^2-2S(2^x)+1=0$$... which is indeed quadratic in \(2^x\). We use a very similar technique to determine a quadratic form for \(D\) to yield \((2^x)^2-2D(2^x)-1=0\).

Now, to solve for \(x\) in \(S\), try to solve for \(2^x\) using the quadratic formula:$$2^x=\frac{2S\pm\sqrt{4S^2-4}}{2}=S\pm\sqrt{S^2-1}\\x=\log_2\left(S+\sqrt{S^2-1}\right)$$We now proceed to do something similar for \(D\) to yield \(x=\log_2\left(D+\sqrt{D^2+1}\right)\).

Answer 7

yes yes i did get that answer for b :)

Answer 8

i just need help for c lol

Answer 9

Now imagine what happens when we sum these last two:$$2x=\log_2\left(S+\sqrt{S^2-1}\right)+\log_2\left(D+\sqrt{D^2+1}\right)\\x=\frac12\log_2\left[(S+\sqrt{S^2-1})(D+\sqrt{D^2+1})\right]$$Now finish.

Answer 10

is this related to question C?
i only need help for part c...

Answer 11

@AonZ I don't have any solution for your problem, but when look at your attachment for problem c, i think you confuse between y and S, First, you cannot start at the result by assuming x = 1/2 log.... Second, how can you state that y= 1/2 (2^x...) . I think you should start from y = DS^-1 by find out that y from DS^-1, S^-1 = 1/2(2^y- 2^-y)* D .
try on that way. It may help.

Answer 12

Hi, I got it from that, DS^-1 =y \[y= \frac{ 1 }{ 2 }(2^x-2^-x)(\frac{ 1 }{ 2 }(2^y +2^-y)\]
--->\[4y= 2^x2^y+2^x2^-y -2^-x2^y-2^-x2^-y\]

Answer 13

@Hoa where is your ^-1? I think that refers to the inverse function of S, or it may refer to its multiplicative inverse.

Answer 14

S= 2^(x-1)+1/2^(x+1) ,

Answer 15

the inverse of exponent function is log function, is it not right?

Answer 16

since S is sum of 2 exponent functions . can we take inverse of them?

Answer 17

Ok, if bataku said so, i think i might go on the wrong way, sorry, waste your time

Answer 18

haha your actually right

Answer 19

who is right?

Answer 20

well my working is wrong lol. I though y was D :P ok ill fix that up

Answer 21

i guess ill keep on working it out and see what i get :P

Answer 22

Hi friend S^-1 is inverse of S, it is not 1/S

Answer 23

so many people solve your problem. you are so happy. no one solve mine. ok take time. I'm out. others will take care of yours.

Answer 24

i dont understand when you said inverse of S. If it is inverse of s i wouldnt know how to do it... Im pretty sure DS^-1 means D/S

Answer 25

really? if it is so. to easy to get the answer, you can do that. for sure.

Answer 26

\[x=\log_2\left(S+\sqrt{S^2-1}\right) \]
You can get the inverse of S from here - just swap x and S.