The Student Room Group
Competition - post a photo you've taken of nature >>
Competition - post a photo you've taken of nature >>
Competition - post a photo you've taken of nature >>

C3 Range Question

Hey guys, stuck on q.9(ii) on this paper:
http://pdf.ocr.org.uk/download/pp_11_jan/ocr_61427_pp_11_jan_gce_472301.pdf?

So here's what I've done:

Differentiate g(x) to get 2e^2x - 2ke^-2x and equate to 0
I solved the equation and got x = 1/4 ln k (this is correct), but I dunno what to do from here; I try subbing this in but to no avail :frown: Any ideas guys?

Cheers! :smile:
Reply 1
Avatar for olipal
olipal
1
Original post by Next Level
Hey guys, stuck on q.9(ii) on this paper:
http://pdf.ocr.org.uk/download/pp_11_jan/ocr_61427_pp_11_jan_gce_472301.pdf?

So here's what I've done:

Differentiate g(x) to get 2e^2x - 2ke^-2x and equate to 0
I solved the equation and got x = 1/4 ln k (this is correct), but I dunno what to do from here; I try subbing this in but to no avail :frown: Any ideas guys?

Cheers! :smile:

What's the problem with subbing it in? I haven't tried it, but remember eln⁡ke^{\ln k} is just kk (that might help in simplifying your answer!).
Reply 2
Avatar for Next Level
Next Level
OP
1
Original post by olipal
What's the problem with subbing it in? I haven't tried it, but remember eln⁡ke^{\ln k} is just kk (that might help in simplifying your answer!).


I get e^0.5lnk + ke^-0.5lnk

I have no idea how you simplify that :s-smilie:
Remember rule of logs from C2

alogb = logb^a

Original post by Next Level
I get e^0.5lnk + ke^-0.5lnk

I have no idea how you simplify that :s-smilie:
Reply 4
Avatar for Next Level
Next Level
OP
1
Original post by Brand New Eyes
Remember rule of logs from C2

alogb = logb^a


then I get e^lnk^0.5 + ke^lnk^-0.5
I can simplify the first part to get ln root k, but dunno how to simplify the second part...help? :smile:
Original post by Next Level
then I get e^lnk^0.5 + ke^lnk^-0.5
I can simplify the first part to get ln root k, but dunno how to simplify the second part...help? :smile:


Treat k and e^lnk^-0.5 as separate parts

then remember rule of powers when you times them you add the power :wink: if that makes sense.
Reply 6
Avatar for Next Level
Next Level
OP
1
Original post by Brand New Eyes
Treat k and e^lnk^-0.5 as separate parts

then remember rule of powers when you times them you add the power :wink: if that makes sense.


so you mean:

e^lnk. e^lnk^-0.5
= e^lnk + ln(1/root k)
= lnk + ln (1/root k)
?
Original post by Next Level
so you mean:

e^lnk. e^lnk^-0.5
= e^lnk + ln(1/root k)
= lnk + ln (1/root k)
?


e^lnk^0.5 + ke^lnk^-0.5

you've already got the answer for the bold bit so

k(e^lnk^-0.5) using the same rule as before is k(k^-0.5)

then add root k to that which you have already found.
Original post by Next Level
so you mean:

e^lnk. e^lnk^-0.5
= e^lnk + ln(1/root k)
= lnk + ln (1/root k)
?


x = 1/4lnk

f(x) = e^2x + ke^-2x

so you know the x co-ordinate of the minimum point is 1/4lnk

y co-ordinate is given by f(x) = e^(2)(1/4 lnk) + k(e^(-2)*1/4lnk)

f(x) = e^1/2lnk + k(e^-1/2lnk)

1/2 lnk = ln k^1/2

so f(x) = e^lnk^1/2 + (k)(e^lnk^-1/2)

e^lna = a

so we get k^1/2 + (k)(k^-1/2)

==> k^1/2 + k^1/2 = 2k^1/2

==> g(x) >= 2k^1/2
Reply 9
Avatar for Next Level
Next Level
OP
1
Original post by Extricated
x = 1/4lnk

f(x) = e^2x + ke^-2x

so you know the x co-ordinate of the minimum point is 1/4lnk

y co-ordinate is given by f(x) = e^(2)(1/4 lnk) + k(e^(-2)*1/4lnk)

f(x) = e^1/2lnk + k(e^-1/2lnk)

1/2 lnk = ln k^1/2

so f(x) = e^lnk^1/2 + (k)(e^lnk^-1/2)

e^lna = a

so we get k^1/2 + (k)(k^-1/2)

==> k^1/2 + k^1/2 = 2k^1/2

==> g(x) >= 2k^1/2


Thanks bro!

Quick Reply

Related discussions

Latest

Trending

Trending

Articles for you