Edexcel FP1 Thread - 20th May, 2016

Reply 840

iMacJack

16

Original post by thelegend99

Oh ok, have you got it now? I might give that one a try.

Got it in the end! Cheers

Reply 841

crashMATHS

16

Original post by iMacJack

Like this Zacken/Aaran??
Apologies for the mess but I rushed it because you know I'm trying to get it resolved quickly :P
@Zacken @kingaaran
ImageUploadedByStudent Room1463679237.941819.jpg

ImageUploadedByStudent Room1463679237.941819.jpg

Posted from TSR Mobile

That's what I got :biggrin:

Reply 842

Zacken

22

Original post by iMacJack

Like this Zacken/Aaran??
Apologies for the mess but I rushed it because you know I'm trying to get it resolved quickly :P
@Zacken @kingaaran
ImageUploadedByStudent Room1463679237.941819.jpg

Posted from TSR Mobile

Yes.

Original post by Hot&SpicyChicken

Can someone please check my working her:

2x2x7 = 4x7 = 28

Rep me if im right

Thanks!!!! :biggrin:

Bit worried about FP1

It's fine but no rep.

Reply 843

username1445684

11

Really sorry if this has been asked before, but can anyone help explain this? I've looked at the mark scheme but can't see where they're getting the answer from, and can't find any youtube tutorials either :frown:

Thanks so much if someone can help!

Screen Shot 2016-05-19 at 18.34.51.png21.8KB

Reply 844

iMacJack

16

Original post by kingaaran

That's what I got :biggrin:

Awesome - cheers guys! @kingaaran @Zacken

Reply 845

Zacken

22

Original post by Strom

Really sorry if this has been asked before, but can anyone help explain this? I've looked at the mark scheme but can't see where they're getting the answer from, and can't find any youtube tutorials either :frown:

Thanks so much if someone can help!

If the argument of a complex number is

-\frac{\pi}{2}

then it must mean that it lies on the imaginary axis (specifically the negative part of the axis). i.e: it has a real part equal to 0.

Find the real part of

w

in terms of

\lambda

and set it equal to 0 and solve.

Original post by iMacJack

Awesome - cheers guys! @kingaaran @Zacken

No problem.

Reply 846

Hot&SpicyChicken

3

Original post by Strom

Really sorry if this has been asked before, but can anyone help explain this? I've looked at the mark scheme but can't see where they're getting the answer from, and can't find any youtube tutorials either :frown:

Thanks so much if someone can help!

I had the exact same problem - no idea how to do it.

I though about using the inverse of tan but no good?

Reply 847

Zacken

22

Original post by Hot&SpicyChicken

I had the exact same problem - no idea how to do it.

I though about using the inverse of tan but no good?

See above.

Reply 848

crashMATHS

16

Original post by Strom

Really sorry if this has been asked before, but can anyone help explain this? I've looked at the mark scheme but can't see where they're getting the answer from, and can't find any youtube tutorials either :frown:

Thanks so much if someone can help!

I like to draw a sketch of the argument. If you try to imagine (or sketch) it, you will notice that if the arg is -pi/2, the complex number lies on the Im axis, in other words, the real part must equal 0.

Now Re(4-5i+3(lamda-3i)) = 4+3lamda.

Hence, since we know the Re = 0 --> 4=-3lamda --> lama = -4/3 :smile:

Reply 849

indecisiveness

7

Original post by Windowswind123

If you work from the unit vectors.
The i vector if you rotate it 135 degrees for example anticlockwise will end up going around until it reaches the (+x, -y) quadrant. The x coordinate of that will be -cos45 and the y coordinate sin45.

The j vector will rotate until it reaches the (-x, -y) quadrant. The x coordinate will be -cos45 and the y coordinate will be -sin45.

So (1,0) will change to (-sin45, cos45) and (0,1) will change to (-sin45, -cos45).
Alternatively you can use the formula sheet. :biggrin:

wait where does the sin and cos come from? I dont understand how you know what to use?