Edexcel A2 C4 Mathematics June 2016 - Official Thread

hi, can someone help me drawing this graph

y=ln(2-4x)

how would you know that the graph shifts two to the left?

Reply 1683

kiasne

3

Original post by imran_

hi, can someone help me drawing this graph

y=ln(2-4x)

how would you know that the graph shifts two to the left?

Because the 2 in the brackets is positive so it moves the graph two to the left.

Reply 1684

AeroCarrot

6

Original post by imran_

hi, can someone help me drawing this graph

y=ln(2-4x)

how would you know that the graph shifts two to the left?

Here you go

Algebra and Functions-page-001.jpg

Reply 1685

imran_

Original post by kiasne

Because the 2 in the brackets is positive so it moves the graph two to the left.

y=ln(x-2), would be shifting the graph to the right.
Can you explain how y=ln(2-x) shifts the graph to the right?

Reply 1686

edothero

15

Original post by imran_

y=ln(x-2), would be shifting the graph to the right.
Can you explain how y=ln(2-x) shifts the graph to the right?

What do we get for

x

when we let

y=0

And what do we get for

y

when we let

x=0

?

Then you know that its a

\mathrm{ln(-x)}

graph so you know the general shape.

Draw a line between the two points you found with the general shape of the

\mathrm{ln(-x)}

graph

Voilà

Also note

\mathrm{ln(2-x)=ln[-(x-2)]}

(edited 7 years ago)

Reply 1687

abbey1

1

Please could someone help me with this question?? I'm stuck doing the first differentiation part. Having the t as the power is throwing me a bit and I can't seem to get the answer on the mark scheme?? Thank you!! (It's from C4 Jan 2011)

Reply 1688

Original post by abbey1

Please could someone help me with this question?? I'm stuck doing the first differentiation part. Having the t as the power is throwing me a bit and I can't seem to get the answer on the mark scheme?? Thank you!! (It's from C4 Jan 2011)

0.5^t = e^{t \ln 0.5}

, straight from the formula booklet. Can you differentiate this now?

Reply 1689

Kevin De Bruyne

21

Original post by abbey1

Please could someone help me with this question?? I'm stuck doing the first differentiation part. Having the t as the power is throwing me a bit and I can't seem to get the answer on the mark scheme?? Thank you!! (It's from C4 Jan 2011)

Can you remember a result for differentiating a^x with respect to x? (It is in your textbook :h:

)

Admittedly it is better to understand it as @Zacken does :tongue:

Reply 1690

abbey1

1

Original post by Zacken

0.5^t = e^{t \ln 0.5}

, straight from the formula booklet. Can you differentiate this now?

Ohhhh yeah, didn't even think to use that😂 thank you!!

Reply 1691

Original post by abbey1

Ohhhh yeah, didn't even think to use that😂 thank you!!

Hmm, fitting in with the theme of 'understanding' as per Sean, I figure I might as well explain why that particular result is true.

e^{x}

and

\ln x

are inverse functions so that when you take the composition of them, they return just

x

. So for example:

\ln e^x = x \ln e = x

because they are inverses. It's kind of like why taking the cube root of the cube function gives you just the number, i.e:

\sqrt[3]{x^3} = x

because the cube root and the cube function are inverses.

So, we also have that

e^{\ln x} = x

, and because of this:

e^{t \ln 0.5} = e^{\ln 0.5^t} = 0.5^t

.

Reply 1692

Glavien

11

For part b of the question attached, would the integral from -4 to 41 give the area bounded between the positive side of the curve and the x-axis or the negative side of the curve and the x- axis? Or could it give the total area bounded, but this would result in the integral being 0 due to a positive area and equal negative area cancelling out, right?
ImageUploadedByStudent Room1462208844.228888.jpg

ImageUploadedByStudent Room1462208844.228888.jpg

Posted from TSR Mobile

(edited 7 years ago)

Reply 1693

Danllo

Are there IYGB papers for S1 and S2 ?

Reply 1694

Original post by Glavien

Or could it give the total area bounded, but this would result in the integral being 0 due to a positive area and equal negative area cancelling out, right?

Precisely this. Which is why you should find the area of the "top shaded" region and then double it. Or the "bottom shaded" region and double it.

Reply 1695

Glavien

11

Original post by Zacken

Precisely this. Which is why you should find the area of the "top shaded" region and then double it. Or the "bottom shaded" region and double it.

But, the limits for the top shaded region are -4 to 41 which are the same as the limits for the bottom shaded region. So, how could you integrate w.r.t x without getting 0?

Reply 1696

Imperion

17

Original post by Danllo

Are there IYGB papers for S1 and S2 ?

Doubt... I know there's C1-C4 and not so sure about M1+

Reply 1697

Original post by Glavien

But, the limits for the top shaded region are -4 to 41 which are the same as the limits for the bottom shaded region. So, how could you integrate w.r.t x without getting 0?

Ah, sorry. When you integrate y dx between -4 to 41, you are finding only one half of the area (the top shaded region). Hence you need to double it to get the full area.

Reply 1698

imran_