Okay here are a few things i need help with please...
1. The roots of the equation $\displaystyle 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. $\displaystyle g(x) = 2x + 1$ and $\displaystyle p(x) = g(x + k)$ where $\displaystyle k $ is a constant and $\displaystyle x - 1$ is a factor of $\displaystyle p(x)$. Calculate $\displaystyle k.$
I got $\displaystyle k = -3$
3. We have a graph $\displaystyle y = -2x + 6$ and we are asked to give the equation of the graph symmetrical to the above one, with respect to the line $\displaystyle y = x$.
I was thinking it could be: $\displaystyle x = -2y + 6$ ?
But i'm really not sure.
4. The function $\displaystyle f$ is defined by $\displaystyle f(x) = x^{log 100 x} $
Determine $\displaystyle x$ if $\displaystyle f(x) = 1000$
I said that:
$\displaystyle 10^3 = x^{2x} $
$\displaystyle 3 = 2x log x $
But i dont know what to do next.
Thanks for any help guys.
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.
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
k = -3/2 ------------------------answer.
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.
We are given the following function:
$\displaystyle f(x) = 2x^3 + px^2 + qx + 3$
And we are told that point F with co-ordinates of (2 ; -9) is a local minimum.
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1. Show that $\displaystyle p = -5$ and $\displaystyle q = -4$
I set point F into $\displaystyle f(x)$ and got: $\displaystyle 2q + 4p = -28$
When they say show then it means something completely different than prove doesnt it? Does it mean i can just go ahead and set $\displaystyle p = -5$ and $\displaystyle q = -4$ into $\displaystyle 2q + 4p = -28$ and see if the equation is true?
Sorry this is so late lol
Let L - x and x be the lengths for the circle and the square respectively.
Then,
$\displaystyle 2\pi r=L-x$
$\displaystyle r=\frac{L-x}{2\pi}$
$\displaystyle r^2=\frac{(L-x)^2}{4\pi^2}$
$\displaystyle Area_{Circle}=\pi r^2=\frac{(L-x)^2}{4\pi}$
Also,
$\displaystyle Area_{Square}=\left( \frac{x}{4} \right)^2=\frac{x^2}{16}$
So
$\displaystyle Area_{Circle}+Area_{Square}=\frac{(\pi+4)x^2-8Lx+4L^2}{16\pi}$
$\displaystyle \frac{dy}{dx}=\frac{(4+\pi)x-4L}{8\pi}$
Equating to zero and solving: $\displaystyle x=\frac{4L}{4+\pi}$
$\displaystyle \frac{dy}{dx}=6x^2+2px+q$
Equating to zero and using the quadratic formula we get
$\displaystyle x=\frac{-p \pm \sqrt{p^2-6q}}{6}$
We have two solutions, but since we know that in a cubic graph when $\displaystyle x^3$ has a coefficient $\displaystyle >0$ 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:
$\displaystyle \frac{-p+\sqrt{p^2-6q}}{6}$
So you have two simultaneous equations:
$\displaystyle \frac{-p+\sqrt{p^2-6q}}{6}=2$
$\displaystyle 2q+4p=-28$
Show assumes you don't know that p=-5 and q=-4 and you have to work them out from scratch.
I 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...
Well tomorrow is D-Day for me...
Anyway, I have one more question.
For which values of $\displaystyle k$ will $\displaystyle k = 2x^3 + px^2 + qx + 3$ have only one root? (There has been proven that $\displaystyle p = -5$ and $\displaystyle q = -4$)
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.