With six points, each point is attached to five other points, so at least three lines radiating from A must be of the same colour (gold, say). The points which are joined to A must then be joined to each other by silver lines, or an entirely gold triangle would be made. However, then a triangle with entirely silver edges can be made from these three points.
Vertical component =
Using s = ut + 1/2at^2.
Quadratic in t. Ignoring the negative root.
R = horizontal component x time.
Horizontal component =
Multiply both numerator and denominator by 2.
Using
We can simplify it.
Splitting up Mr. Big Fraction.
Ignore the first fraction for the moment, we'll work on the second one.
Unparseable or potentially dangerous latex formula. Error 4: no dvi output from LaTeX. It is likely that your formula contains syntax errors or worse.
And lo and behold, there's a common factor to both fractions. Let's take it out:
And that's what we're looking for.
*I'll finish off the last small bit in a few minutes*
I think that I shall restrict myself to a maximum of three solutions per day (at least at the start), so that everyone has a chance to answer some if they want.
*bump* (just to remind people that this thread is still here ) - I have solutions that I can type up if no one else wants to, but I really don't want to take over the thread..
If we work out the area of a quarter of the rectangle, which is in the top right hand bit of the axes, and maximise this, it make calculation soooo much eaiser. (Since we can forget about modulus signs and y being negative) :
Let the area of the rectangle be .
Base of smaller rectangle = 1-x
height =
so
which we want to maximise.
Which is zero when
Finding the second derivative and subsituting x = in shows is a maximum at this point.
So .
________
We can do the second bit in the same way, since its a closed curve that is symmetrical in both axes.
Let the area of the second rectangle be .
Working in the top right quadrant again;
Base = 1-x
height = .
So .
Which is zero when x = 0 or x = .
Clearly the area is a minimum when x= 0, and so is a maximum when x =.
I've got a solution to I question 9 , but don't have a scanner available so will add graphs when I get the opportunity.
Question 9 STEP I
and Setting y'=0
f(2)=-2, f(-2)=2, also for graphing purposes it may be worth to note that f(0)=0 and f(4)=2 and f(-4)=-2. y=0 has solutions x=0 or f''(-2)=- i.e. local max f''(2)=+ i.e. local min
Graph, see attachment
(a) i.e. therefore the equation of y in this X-Y plane is setting it equal to 0 gives f(1)=-2 ans f(-1)=2
For graphing it can also be good to see that f(2)=2, f(-2)=-2 and
Graph, see attachment
(b) i.e. therefore the equation of y in this X-Y plane is Setting this equal to 0 gives . f(2)=-1 and f(-2)=1
For graphing, also see that , f(4)=1 and f(-4)=-1
Graph, see attachment.
(c) i.e. therefore the equation in the X-Y plane is Setting this equal to 0 gives f(0)=2, f(2)=-2.
For graphing, also see that and f(-1)=-2, f(3)=2
(d) i.e. therefore the equation in the X-Y plane is setting this equal to 0 gives f(2)=0 f(-2)=2
For graphing I'll check f(4) and f(-4) too, which give 2 and 0 respectively.
For the last part, we're looking for a graph in the X-Y plane with local min (0,0) and local max (1,1). To obtain this on the form X=ax+b and Y=cy+d, first observe that for a graph of a cubic, changing the sign means a reflection in the x-axis i.e. local max and min swap places, this is what we need to do. That means either a or c is negative (but not both!). a squeezes the graph along the X-axis and c squeezes it along the Y-axis, c moves the graph to the left/right and d moves the graph up/down. Knowing these we can see that and will produce the desired graph.
Okay, so I hope I've interpreted everything correct in the question. Seems pretty nice as a question.
We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.