# A level Maths

• May 22nd 2007, 09:18 AM
A level Maths
i have some problems with these attached file , need help as soon as possible need to make sure i did them correctly:p
• May 22nd 2007, 10:14 AM
janvdl
$\cos 2x = 0.5$

$2x = 60^{\circ}$

$2x = 60^{\circ} \pm 360 k$

$x = 30^{\circ} \pm 180 k$

$\mbox{Or } x = 150^{\circ} \pm 180k
$
• May 22nd 2007, 10:21 AM
janvdl
1. cos 2x = 0.5
2x = 60 degrees
2x = 60 degrees \pm 360 k
x = 30 degrees \pm 180 k
Or x = 150 degrees \pm 180k

------------------------------------

2. Look at the RHS, where you have $cos^2 \theta$
Remember that $cos^2 = 1 - sin^2$

Therefore:

$4 sin^2 \theta - 2 sin \theta = 4 - 4 sin^2 \theta - 1$
$8 sin^2 \theta - 2 sin \theta - 3 = 0$

By solving for theta we find:
$sin \theta = 0.75 \ OR \ sin \theta = -0.5$
• May 22nd 2007, 10:26 AM
janvdl
3. $A = \pi R^2$

But Perimeter $P = 2 \pi R$

$20 = 2 \pi R$

$\frac{10}{\pi} = R$

Set R into A

$A = \pi (\frac{10}{\pi})^2$
$A = \frac{100}{\pi}$

Now my area formula does not seem at all correct. See if you can figure out where i made a mistake.

-----------

Now lets ASSUME we did get the right formula.
To maximise, we must find the derivative and equal it to 0

So dA/dr = 0

$25 - (r - 5)^2 = 0$

$25 - r^2 + 10r - 25 = 0$

$r^2 + 10r = 0$

$r(r + 10) = 0$

Therefore $r = 0 \ or \ r = -10$

Set both 0 and -10 into the ORIGINAL equation

If r = 0
Area = 0

If r = -10
Area = -200
• May 22nd 2007, 10:53 AM
janvdl
For the Architect Sum:

a) $4x + 4y + x \pi$

b) $4xy + \frac{1}{2} x^2 \pi$

c) $100 = 4x + 4y + 2x \pi$
$y = 25 - (x + \frac{x \pi}{2})$

Set y into the Area formula

Then we will eventually get: $A = 50x - 2x^2 + \frac{1}{2} x^2 \pi$

---------------

Why cant i get these sums right??
Right lets assume the given values again

$\frac{dA}{dx} = 100 - 4x + x \pi = 0$
$100 = 4x - x \pi$
$100 = x(4 - \pi)$
$x = \frac{100}{(4 - \pi)}$

Now just set $x$ into the ORIGINAL equation to find $y$
• May 22nd 2007, 11:11 AM
janvdl
For the Graph sum:

a) To find where they intersect we must set y = y
$x^2 - 2x + 2 = -x + 4$
$x^2 - x - 2 = 0$
$x = 2 \ or \ x = -1$

Set these values into an original equation.

Then the points are: y = -(2) + 4
And y = -(-1) + 4

So point P = (-1 ; 5)
Point Q = (2 ; 2)

b) For MPQ to be right-angled, the product of the two formulas must equal -1

I do not get the product = -1, so i am assuming that they are not right angled.
• May 22nd 2007, 11:13 AM
janvdl
I will see if i can answer the rest tomorrow. I hope that the things that i could get right myself helped you.

:)
• May 22nd 2007, 01:28 PM
Jhevon
Quote:

Originally Posted by janvdl
For the Graph sum:

a) To find where they intersect we must set y = y
$x^2 - 2x + 2 = -x + 4$
$x^2 - x - 2 = 0$
$x = 2 \ or \ x = -1$

Set these values into an original equation.

Then the points are: y = -(2) + 4
And y = -(-1) + 4

So point P = (-1 ; 5)
Point Q = (2 ; 2)

b) For MPQ to be right-angled, the product of the two formulas must equal -1

I do not get the product = -1, so i am assuming that they are not right angled.

we have M(1 , 1), P(-1 , 5) and Q(2 , 2)

The slope of the line connecting P and Q is given by:
$m_{1} = \frac {2 - 5}{2 + 1} = \frac {-3}{3} = -1$

The slope of the line connecting M and Q is given by:
$m_{2} = \frac {2 - 1}{2 - 1} = 1$

since 1 is the negative reciprocal of -1, the line connecting P and Q is perpendicular to the line connecting M and Q. Thus this forms a right angle at Q. If we draw a line connecting M and P therefore, we form a right-triangle
• May 22nd 2007, 02:06 PM
Jhevon
Quote:

Originally Posted by janvdl
3. $A = \pi R^2$

But Perimeter $P = 2 \pi R$

$20 = 2 \pi R$

$\frac{10}{\pi} = R$

Set R into A

$A = \pi (\frac{10}{\pi})^2$
$A = \frac{100}{\pi}$

Now my area formula does not seem at all correct. See if you can figure out where i made a mistake.

Let's find the arc length first of all.

Recall:

$s = r \theta$
where s is the arc length and $\theta$ is the angle subtending the arc in radians

We are told that the perimeter is 20 cm this means that:
$P = r + r + s = 2r + r \theta = r(2 + \theta)$, where $P$ is the perimeter

$\Rightarrow r(2 + \theta) = 20 \Rightarrow 2 + \theta = \frac {20}{r} \Rightarrow \theta = \frac {20}{r} - 2$

Now, the area A of the sector will be given by:

$A = \frac {1}{2} r^2 \theta$
$= \frac {1}{2} r^2 ( \frac {20}{r} - 2)$
$= r^2 ( \frac {10}{r} - 1)$
$= 10r - r^2$

By completing the square, you will get the desired result

Quote:

Now lets ASSUME we did get the right formula.
To maximise, we must find the derivative and equal it to 0

So dA/dr = 0

$25 - (r - 5)^2 = 0$

$25 - r^2 + 10r - 25 = 0$

$r^2 + 10r = 0$

$r(r + 10) = 0$

Therefore $r = 0 \ or \ r = -10$

Set both 0 and -10 into the ORIGINAL equation

If r = 0
Area = 0

If r = -10
Area = -200
Umm, you forgot to find dA/dr janvdl, you did the manipulations on A, which is why you had weird values. (The negative valus should have hinted to you that you did something wrong)

We wish to maximize A, so we must find A' and set it equal to zero, a value for r that causes this to happen is the value we are looking for.

From above, $A = 25 - (r - 5)^2$
$\Rightarrow A = 10r - r^2$
$\Rightarrow A' = 10 - 2r$
Set $A' = 0$ we get:
$10 - 2r = 0$
$\Rightarrow r = 5$

Note, the formula for a is a downward opening parabola, all we needed to do was find the value of r that gives the vertex. we didn't even need calculus here, we could have found this value by using:
$r = \frac {-b}{2a}$, where $a$ is the coefficient of $r^2$ and $b$ is the coefficient of $r$.
• May 22nd 2007, 02:19 PM
Jhevon
Quote:

Originally Posted by janvdl
$\cos 2x = 0.5$

$2x = 60^{\circ}$

$2x = 60^{\circ} \pm 360 k$

$x = 30^{\circ} \pm 180 k$

$\mbox{Or } x = 150^{\circ} \pm 180k
$

Remember to only choose the values such that $- \pi < x < \pi$, you should list them, it'll only be like four or so, also, use radians instead of degrees, since the limits were given in radians
• May 22nd 2007, 02:52 PM
Jhevon
I assume you actually know how to draw the graphs, draw them both on the same pair of axis. see the graphs below.

a)

Note that we have the roots of $x^2 + e^x - 4 = 0$ where the graphs intersect (do you see why?). from the graphs we see that the intersections occur such that one is for a positive x-value and the other is for a negative x-value, thus we will have a positive and negative root.

b)
I can't really see the power of e very well, i think it is $x_{n}$ so let's run with that.

$x_{n+1} = - \sqrt {4 - e^{x_{n}}}$
$\Rightarrow x_{1} = - \sqrt {4 - e^{x_{0}}} = - \sqrt {4 - e^{-2}} = -1.96588$
$\Rightarrow x_{2} = - \sqrt {4 - e^{x_{1}}} = - \sqrt {4 - e^{-1.96588}} = -1.96468$
$\Rightarrow x_{3} = - \sqrt {4 - e^{x_{2}}} = - \sqrt {4 - e^{-1.96468}} = -1.96463$
so the approx for the negative root to three decimal places is -1.965

c)
the approximation will go towards the negative root, since within any reasonable range $\sqrt {4 - e^{x_{n}}$ will give a positive result. so with the minus on the outside, we will get a negative result

EDIT: So i believe i have taken care of all the problems janvdl didn't do or the ones he expressed doubt about, for the others, i think you're in good hands with him on the job, so i won't even check them
• May 23rd 2007, 11:06 AM
janvdl
Quote:

Originally Posted by Jhevon
Remember to only choose the values such that $- \pi < x < \pi$, you should list them, it'll only be like four or so, also, use radians instead of degrees, since the limits were given in radians

• May 23rd 2007, 11:22 AM
topsquark
Quote:

Originally Posted by janvdl
$\theta = \frac{s}{r}$
For conversion purposes, you can use this formula to determine that there are $\pi$ radians in 180 degrees.