Maths P3 s10_qp_31.

[deathvalley]

Can anyone explain the question 7, part ii and iii briefly for me . I looked on the marksheme and still don't get it if you don't mind, can you show it on the diagram for the part ii ?
Here is the link
http://www.xtremepapers.net/CIE/Interna ... _qp_31.pdf
Many thanks

 

[Zishi]

Although, I'm in AS Level but as I've done O level add maths, so I thought that I'd be able to help you. But the link isn't working for me. :s

 

[DragonCub]

The solution is shown in the attached image. Guess it does not differ much from the MS's answer

In the Argard-diagram complex number field, any notations like |A-B| represents the distance between point A and B.
So the question can be interpreted as: the distance from 1 to Z should not exceed that from i to Z; the distance from u to Z should not exceed 1.
For the first condition, you should quickly think of the perpendicular bisector (p.b.) - the line on which each point has an equal distance from the two terminals of the bisected line segment. So in the diagram, the points should exist on the right side of the p.b. in order to get closer to point 1.
For the next condition: FIXED point u, FIXED distance 1 - a circle is demonstrated out. So draw a circle with u as the centre and 1 as radius. The point should lie inside the circle.
Then here comes the answer: Right side of the p.b. AND inside the circle.

 

[WellWIshER]

wt abt solution to part 3?

 

[rz123]

can anyone tell me whats chain rule, did not understood from the book...

 

[OakMoon!]

In calculus, the chain rule is a formula for computing the derivative of the composition of two or more functions.
The chain rule formula states dy/dt=dy/dx*dx/dt... This formula is used when there are three variables instead of two. Different rules of calculas are derived from this rule.

 

[usman]

wt abt solution to part 3?

In this part, u need to consider a point in the shaded region such that the line joining the origin (i.e. the intersection of the imaginary and real axes) to the selected point makes the least angle with the real axis [i.e. arg(z) is min]. Applying common sense, one almost immediately realizes that such a point shud lie on the intersection of a tangent to the bottom part of the circle i.e. a line wich passes through the origin and touches (only once) the bottom part of the circle.

Now, since the tangent to a circle is normal to its radius (i.e. tan_I_rad), the line joining the origin to "u" will be a hypotenuse; the line joining "u" to intersection of -->tangent (to the circle) and radius (of the cricle)<-- will be a perpendicular while the line joining the origin to the intersection point will be a base.
Hence, (radius of the circle)^2 + [min{Arg(z)}]^2 = [sqrt(8)}^2
Since the radius of the circle = 1 unit ----> min{Arg(z)} = sqrt(7)


I m afraid this is too "long and tedious" an explaination......;p


Anyways i hop this helps..!