Question

can anyone help me with question 3(i) and 4???? thanks :)

http://www.xtremepapers.com/CIE/Cambridge%20IGCSE/0606%20-%20Mathematics%20%28Additional%29/0606_w05_qp_2.pdf

Answer 1

and yes, i know the mark scheme exists, but it doesn't explain HOW you get the answer.

Answer 2

xtreme and cambridge in the same url; just seems a bit odd to me :)

Answer 3

y = 1 + ln (2x – 3)
\[\frac{dy}{dx}= \frac{2}{2x+3}\]

Answer 4

they define algebra and trig; and then ask for a derivative?

Answer 5

amistre, it shouldn't sound odd to u lol, well it doesn't sound odd to me... I used that resource when I was preparing for the same exam. my teacher used to print out copies of those PDFs and give it to us :P

Answer 6

f(x)= 2 + 5 sin 3x

y=Asin(B(x+C/B))+d

where A is the amplitude, T is the period with T=2\(\pi\)/B

Answer 7

So for f(x) the amplitude is 5 and the period is \(\frac{2\pi}{3}\)

Answer 8

^I know. and wait, isn't it 2(pi) + 3??

Answer 9

C is the horizontal translation and D is the vertical translation for the equation

Answer 10

ohh, wait, you're right, it IS 2(pi)/3

Answer 11

Are you able to graph this now that you know the values? :)

Answer 12

ohh, and 3(ii), also, please :P

Answer 13

y = 1 + ln (2x – 3) and x=2+p

y = 1 + ln (2(2+p) – 3)
y = 1 + ln (4+2p – 3)
y = 1 + ln (2p + 1)

As p is small --> 2p+1=1

y=1+ln(1)
=1+0
=1

So y is approximately 1.

Answer 14

thank youu :)

Answer 15

no worries, good luck :)

Answer 16

:P