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# MEI P5 - January 2005 watch

1. (Original post by chubby)
thanks a lot fermat!!!! i learnt something new today. thanks
could u also do 2 (iii). i know it is a show that question but i got the end answer as (11/18) instead of (26/9). pls help if u can.
int 7sinhx-sinh2x dx
=1/2int 7 {e^x-e^(-x)}-e^(2x)+e^(-2x)
=1/2 {[7(e^x+e^(-x))]-1/2(e^(2x)+e^(-2x))}
x=0 gives 1/2{7+7-1/2(2)]=13/2
x=ln 3
e^x=e^(ln3)=3
e^(-x)=e^-ln3=e^ln(1/3)=1/3
e^(2x)=e^(2ln3)=e^ln9=9
e^(-2x)=e^-2ln3=e^ln(1/9)=1/9
so have 1/2{7(3+1/3)-1/2(9+1/9)}=1/2(70/3-41/9)=169/18
hence int=
169/18-(13/2)=52/18
=26/9
2. thanks. did it. made a very silly mistake. forgot to mulitiply by 0.5 in the second term. thanks anyway
3. Oooh, I actually did that paper (questions 1, 2, 3). It wasn't nice.
4. (Original post by chubby)
Also fermat in 4(i) the curve is between 0 and pi as mentioned in the question so i am guessing all the four loops lie above the xaxis (two between 0 and pi/2 and 2 between pi/2 and pi). the second and fourth loop starting from the 0 angled line are dotted lines. am i right?
Hi chubby,
these things can sometimes be a bit confusing.
The curve doesn't lie between 0 and pi.
That is the range of the argument , theta.

Between 0 and pi/4, sin4θ is +ve so r is +ve and the 1st loop is gotten.
Between pi/4 and pi/2, sin4θ is -ve, so r is -ve and the curve now appears as a 2nd loop on the other side of the origin - denoted as a broken line.
Between θ/2 and 3pi/4, sin4θ is +ve again, r is +ve, and the 3rd loop is gotten.
Between 3pi/4 and pi, sin4θ is -ve again, r is -ve and the curve again appears as a 4th loop on the other side of the origin - denoted as a broken line.

So, the curve appears in all quadrants with the 1st and 3rd loops above the x-axis and the 2nd and 4th loops below the x-axis.

The Fig attached shows how the loops would look if you were taking r as +ve all the time. eg. r = k|sinθ|.
Attached Images

5. YAY. thanks fermat!!!! get it now.

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