H2 Math : Algebra

last SG · 2013-03-02T08:24:54+00:00

Hi, For (b): 1. Rewrite α/(2i) by multiplying top and bottom by i. 2. Convert the complex number that is in exponential form to the trigonometric form. 3...

Theplayfulgirl brainy SG

Someone help me? I need it now

Mad Hat SG

Q1. Since f(2006) = 0, the min. value is at least 0. So the answer is either A or E. I suspect it's A but I don't know how to prove it in any simple way. Does anyone know? Is this an IB question? It's not A-level material.

Q5. Since f^-1(³/₂) = 7, this means that f(7) = ³/₂. After that, it's easy to find w.

Q8. This depends on the min value of (2004 cos2005x − 2006)². Can reason like this:

−1 ≤ cos2005x ≤ 1
⇒ −2004 ≤ 2004 cos2005x ≤ 2004
⇒ −4010 ≤ 2004 cos2005x − 2006 ≤ −2

So the min. value of (2004 cos2005x − 2006)² should be 4. Then just add the 2007.

Theplayfulgirl brainy SG

10 min before the exam...

I have an idea for the question 8.

(2004 cos2005x − 2006)^2 >=0

=> the min value when 2004 cos2005 =2006.

=> normal solution

Theplayfulgirl brainy SG

How to solve this:

We have parabola : y =4ax^2. The region is limited by parabola, line y =b, x-axis is rotated 4 right angle about:

a. x -axis

b. y axis.

Find them.

If the second one is 10 times the first one, find b followed a

wee_ws SG

Hi,

Suppose a > 0 and b > 0 in the context of the problem.

To find (a):

1. Draw a diagram to identify the region.

2. Find the x-coordinate of a suitable point of intersection between the parabola and the line y = b.

3. Required volume = volume of cylinder (rectangular region)

- integral of (pi)(y^2) dx (region under the parabola).

To find (b):

1. Required volume = volume of cylinder (rectangular region)

- integral of (pi)(x^2) dy (region to the left of the parabola).

Thanks.

Cheers,
Wen Shih
wenshih.wordpress.com

Mad Hat SG

I'm still curious about the minimum value of the function:

f(x) = (x-2006)(x-2007)(x-2008 )(x-2009).

Wen Shih, what is the best way to handle this? Probably there is some quick and clever way to see the minimum value of the function but I can't find it?

The actual question is from playfulgirl here: http://www.flickr.com/photos/mabuthaodang/8515758304/in/photostream

Some things are obvious, like the min. value is at least 0. Also the function clearly can take negative values, e.g. choose x such that the first term is positive and the other three negative. E.g. x = 2006.5.

Also, the 2006, 2007, 2008, 2009 are not important. The question is equivalent to finding the minimum value of: f(x) = x(x-1)(x-2)(x-3). because the difference is just a horizontal translation of the graph along the x-axis.

We can do the question the long way, e.g. expand the function, differentiate, and set to 0, but I'm curious what is the short way.

wee_ws SG

Hi,

The minimum value of

f(x) = (x-2006)(x-2007)(x-2008)(x-2009)

is the same as the minimum value of f(x) = (x-1)(x-2)(x-3)(x-4).

I considered the graphing approach (due to its ease of visualising any pattern) and observed that the minimum value of the curve

y = (x-1)(x-2)(x-3)(x-4)

occurred at one point where the line

y = x-1

intersects the curve y = (x-1)(x-2)(x-3).

Solving, we obtain x = (5/2) +/- (sqrt(5)/2).

Take x = (5/2) - (sqrt(5)/2), since the curve

y = (x-1)(x-2)(x-3)(x-4)

is symmetrical about the line x = 5/2 and we arrive (after some algebra) at the minimum value of -1.

P.S. I think that the olympiad question setter typically develops interesting questions (like tha one above) that rely on some mathematical pattern or special mathematical behaviour.

Thanks.

Cheers,
Wen Shih
wenshih.wordpress.com

Mad Hat SG

Wen Shih thanks. So it's an Olympiad question? I was trying to find a "short way" with the question but maybe there's none! If I use the online GC, it also does show two minimum points, both of them -1.

wee_ws SG

Hi,

Looks like one and tedious to solve if we were to use traditional methods :P

Cheers,
Wen Shih
wenshih.wordpress.com

« Previous Page