OCR (not MEI) Core 4 January 2013

I think it was 8 + ln27/4

The co-ordinates i'm sure were: 1,root2 and 1,-root2
.

Areas can be negative, think about if the curve was below the x-axis for the limits of x.

You can have negative definite integrals but you cant have negative areas.
You can work out area using a negative integral but its not a negative area.

Think about it for a second, "a negative area"
Its like saying the length of the sides of a cube is -4cm.

And i think you are right to do so, as y²=2x, where x = - 1 or +1

therefore, y = root of -2
therefore, no solution

Surely it should be 1,root 2 and -1, root 2 ?


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As far as a I can remember:
1) x/3 *sinx +1/9 *Cosx +C
2) 27 -72x +32X^2 With |x|9/16

4) Vector Question... Point of intersection was (5,4,6) but I can't remember the constants. Angle was~68

The co-ordinates were (1, -Root(2)) (1, Root(2)) (stat point question)

Parametrics: 2/3Tan(Theta) was dy/dx, theta was .6435

AB=AD=Root(24)
Vector question can't remember. But when the constant (s,t, lambda, etc) was 5 c lay on the line.
It was a kite.

Substitution was 1/64 (I got this wrong, does anybody know how many marks it was?)

ln(theta +20)=-Kt +c

C=ln 60

t=-20ln(1/3) = 22

Last question:
X-1 +(x+7)/(X^2 -x -6)

X-1 +2/(X-3) -1/(x+2)

After integration and limits, 8+ln(27/4)

i read the question very carefully, it specifically said stationary points and in past papers when it has said stationary point, there is one answer, and when it says stationary points its more than one answer.

however i only got (1,root 2)
y^2=2x , if x=-1 y=root-2 which is impossible

Nope then its 2i .


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Nope then its 2i

i read the question very carefully, it specifically said stationary points and in past papers when it has said stationary point, there is one answer, and when it says stationary points its more than one answer.

however i only got (1,root 2)
y^2=2x , if x=-1 y=root-2 which is impossible

For the parametrics question, wasn't finding theta unnecessary?

if you got to 2/3tan(theta) =1/2
then tan(theta) = 3/4

By knowing your 3,4,5 Triangle, exact values of sin(theta) and cos(theta) Just pops out nicely from there, which you could put back into x and y.

It was 1,root2 AND 1, (-1*root2)

How did you get the 2nd one?

X=1

2x= y^2

2=y^2

y=+or-root2

D'oh! I forgot about that! Kept thinking people derived -root two from the x value -1 . there goes my a *
What mark is 90 ums usually?

D'oh! I forgot about that! Kept thinking people derived -root two from the x value -1 . there goes my a *
What mark is 90 ums usually?