# Math Help - Writing Math Paper 1 on Friday - Help Please

1. Originally Posted by DivideBy0
Are you sure $Area_{Circle+Square}$ is right?
To be honest, no i don't. But i don't know what else to do.

2. Originally Posted by DivideBy0
Are you sure $Area_{Circle+Square}$ is right?
No it is not right. See if you can do it for him.

Hint: You make a cut. That length is the total perimeter (circumfurence) of the shape you are about to bend into. Since you know its perimeter you can express its area.

3. Of course! Thats where i went wrong with the square. I made the perimeter the area. I will try it again later.

4. Okay here are a few things i need help with please...

1. The roots of the equation $x^2 - kx + k + 3 = 0$ is equal. Calculate the roots of the equation.

I keep getting that the roots of this equation doesnt exist...

2. $g(x) = 2x + 1$ and $p(x) = g(x + k)$ where $k$ is a constant and $x - 1$ is a factor of $p(x)$. Calculate $k.$

I got $k = -3$

3. We have a graph $y = -2x + 6$ and we are asked to give the equation of the graph symmetrical to the above one, with respect to the line $y = x$.

I was thinking it could be: $x = -2y + 6$ ?
But i'm really not sure.

4. The function $f$ is defined by $f(x) = x^{log 100 x}$
Determine $x$ if $f(x) = 1000$

I said that:
$10^3 = x^{2x}$
$3 = 2x log x$

But i dont know what to do next.

Thanks for any help guys.

5. 1. The roots of the equation x^2 -kx +k + 3 = 0 is equal. Calculate the roots of the equation.

You mean the roots are equal. Then the equation is a perfect square, in the form (x-a)^2.

So it can be done by "completing the square", or by the Qudratic Formula.
In the x = {-b +,-sqrt[b^2 -4ac]} / 2a,
if the [b^2 -4ac] is zero, then x = -b/(2a) only. Two of them (equal), because a quadratic always has two roots.

x^2 -kx +k + 3 = 0 --------------(i)
x^2 -kx +(k+3) = 0
The [b^2 -4ac] is k^2 -4(1)(k+3).
equate that to zero,
k^2 -4k -12 = 0
Factor that,
(k-6)(k+2) = 0
k = 6 or -2.

When k = 6, in (i),
x^2 -6x +6 +3 = 0
x^2 -6x +9 = 0
(x-3)^2 = 0
x = 3

When k = -2, in (i),
x^2 +2x -2 +3 = 0
x^2 +2x +1 = 0
(x+1)^2 = 0
x = -1

Therefore, for the two roots to be equal, the equation (i) has 2 roots of 3, or 2 roots of -1. ----answer.

6. Thanks ticbol.
I realised my mistake. I dont know why, but i accidentally wrote that Delta = $b^2 + 4ac$

7. 2.
g(x) = 2x +1 and p(x) = g(x+k) where k is a constant and (x-1) is a factor of p(x). Calculate k.

p(x) = 2(x+k) +1 ------(i)

if (x-1) is a factor of p(x), then the other factor must have a degree of zero because p(x) is linear only (degree of 1) and (x-1) is of degree 1 already. So the other factor is a constant, say "a".

a(x-1) = 2(x+k) +1
ax -a = 2x +2k +1
ax -a = 2x +(2k +1)
For them to be equal, it must be that
ax = 2x ---------------(1)
-a = 2k +1 -----------(2)

ax = 2x
So, a = 2
Substitute that into (2),
-2 = 2k +1
-2 -1 = 2k

8. 3. We have a graph y = -2x +6 and we are asked to give the equation of the graph symmetrical to the above one, with respect to the line y = x.

I was thinking it could be: x = -2y +6?
But i'm really not sure.

y = -2x +6
y = -2(x-3)
Swap x and y,
x = -2(y -3)
-x/2 = y -3
-x/2 +3 = y
or,
y = -(1/2)x +3 ----------------the equation of that line.

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4. The function f is defined by f(x) = x^(log[100x])
Determine x if f(x) = 1000.

I said that: 10^3 = x^(2x)

But log[100x] is not equal to 2x.
It is equal to log[100] +log[x] = 2 + log[x].

x^(log[100x]) = 1000
Take the logs of both sides,
(log[100x])*log[x] = log[10^3]
(log[100] +log[x])*log[x] = 3
(2 +log[x])*log[x] = 3
2log[x] +(log[x])^2 = 3
For simplicity, let y = log[x],
2y +y^2 = 3
y^2 +2y -3 = 0
(y+3)(y-1) = 0
y = -3 or 1
Or,
log[x] = -3 or 1

log[x] = -3
x = 10^(-3) = 1/1000 --------------answer.

log[x] = 1
x = 10^1 = 10 ----------------------answer also.

9. We are given the following function:
$f(x) = 2x^3 + px^2 + qx + 3$

And we are told that point F with co-ordinates of (2 ; -9) is a local minimum.

---

1. Show that $p = -5$ and $q = -4$

I set point F into $f(x)$ and got: $2q + 4p = -28$

When they say show then it means something completely different than prove doesnt it? Does it mean i can just go ahead and set $p = -5$ and $q = -4$ into $2q + 4p = -28$ and see if the equation is true?

10. Sorry this is so late lol

Let L - x and x be the lengths for the circle and the square respectively.

Then,
$2\pi r=L-x$

$r=\frac{L-x}{2\pi}$

$r^2=\frac{(L-x)^2}{4\pi^2}$
$Area_{Circle}=\pi r^2=\frac{(L-x)^2}{4\pi}$

Also,
$Area_{Square}=\left( \frac{x}{4} \right)^2=\frac{x^2}{16}$

So
$Area_{Circle}+Area_{Square}=\frac{(\pi+4)x^2-8Lx+4L^2}{16\pi}$

$\frac{dy}{dx}=\frac{(4+\pi)x-4L}{8\pi}$

Equating to zero and solving: $x=\frac{4L}{4+\pi}$

11. Originally Posted by janvdl
We are given the following function:
$f(x) = 2x^3 + px^2 + qx + 3$

And we are told that point F with co-ordinates of (2 ; -9) is a local minimum.

---

1. Show that $p = -5$ and $q = -4$

I set point F into $f(x)$ and got: $2q + 4p = -28$

When they say show then it means something completely different than prove doesnt it? Does it mean i can just go ahead and set $p = -5$ and $q = -4$ into $2q + 4p = -28$ and see if the equation is true?
$\frac{dy}{dx}=6x^2+2px+q$

Equating to zero and using the quadratic formula we get

$x=\frac{-p \pm \sqrt{p^2-6q}}{6}$
We have two solutions, but since we know that in a cubic graph when $x^3$ has a coefficient $>0$ the minimum is farther right than the maximum (hopefully someone with more experience can explain this better ). Hence, the x coordinate of the minimum is:
$\frac{-p+\sqrt{p^2-6q}}{6}$

So you have two simultaneous equations:

$\frac{-p+\sqrt{p^2-6q}}{6}=2$

$2q+4p=-28$

Show assumes you don't know that p=-5 and q=-4 and you have to work them out from scratch.

12. Originally Posted by DivideBy0
$\frac{dy}{dx}=6x^2+2px+q$

Equating to zero and using the quadratic formula we get

$x=\frac{-p \pm \sqrt{p^2-6q}}{6}$
We have two solutions, but since we know that in a cubic graph when $x^3$ has a coefficient $>0$ the minimum is farther right than the maximum (hopefully someone with more experience can explain this better ). Hence, the x coordinate of the minimum is:
$\frac{-p+\sqrt{p^2-6q}}{6}$

So you have two simultaneous equations:

$\frac{-p+\sqrt{p^2-6q}}{6}=2$

$2q+4p=-28$

Show assumes you don't know that p=-5 and q=-4 and you have to work them out from scratch.
I thought of doing simulataneous equations, but i was a little lazy I've been doing maths since this morning

Hey, i'll let you guys know about my math mark after the exam ok? Thanks for all the help.

13. Originally Posted by DivideBy0
Sorry this is so late lol

Let L - x and x be the lengths for the circle and the square respectively.

Then,
$2\pi r=L-x$

$r=\frac{L-x}{2\pi}$

$r^2=\frac{(L-x)^2}{4\pi^2}$
$Area_{Circle}=\pi r^2=\frac{(L-x)^2}{4\pi}$

Also,
$Area_{Square}=\left( \frac{x}{4} \right)^2=\frac{x^2}{16}$

So
$Area_{Circle}+Area_{Square}=\frac{(\pi+4)x^2-8Lx+4L^2}{16\pi}$

$\frac{dy}{dx}=\frac{(4+\pi)x-4L}{8\pi}$

Equating to zero and solving: $x=\frac{4L}{4+\pi}$
I sort of had the idea... OK, i was wrong! Good one DivideBy0!

EDIT: Hey no, wait, this morning i started with it, except that my formulas were just the other way round.
Like the square was (L - x), etc.
But i stopped halfway through it because i had work to do...
But i did it the way you did...

14. Well tomorrow is D-Day for me...

Anyway, I have one more question.

For which values of $k$ will $k = 2x^3 + px^2 + qx + 3$ have only one root? (There has been proven that $p = -5$ and $q = -4$)

I've thought about using Delta, and equating it to zero, but that only applies to quadratic equations. This question only counts 2 marks, so the solution should be quite simple.

15. Originally Posted by janvdl
Well tomorrow is D-Day for me...

Anyway, I have one more question.

For which values of $k$ will $k = 2x^3 + px^2 + qx + 3$ have only one root? (There has been proven that $p = -5$ and $q = -4$)
And what are $p\mbox{ and }q$.

I relized something interesting,
Given $x^3+ax^2+bx+c=0$ which has exactly 1 root.
It is necessary (thought maybe not sufficient) that $b^3=a^3c$.
Can you prove that?