Author Topic: Mathematics Question Thread (Read 1626095 times) Tweet Share

RuiAce · « **Reply #3150 on:** November 11, 2017, 10:21:02 pm »

Quote from: owidjaja on November 11, 2017, 09:21:54 pm

Hey guys,

I need help with the following question:

Find any stationary points on the curve y=(4x^2-1)^4 and determine their nature.

Thanks!

$\frac{dy}{dx}=32x(4x^2-1)^3\\ \text{Setting this equals to 0 gives}\\ x=0\text{ and }4x^2-1=0\implies x=\pm \frac12$
$\text{One may then use the second derivative to check their nature}\\ \text{but due to the leftover power on }(4x^2-1)\text{ the test will be a bit inconclusive}\\ \text{so we simply test a bit to both sides in the first derivative.}$
$\text{When }x=-1. \frac{dy}{dx} < 0\\ \text{When }x = -\frac14, \frac{dy}{dx} > 0\\ \text{So we have a local min at }\left(-\frac12,0\right)$
$\text{When }x=-\frac14, \frac{dy}{dx} > 0\\ \text{When }x=\frac14, \frac{dy}{dx} < 0\\ \text{so we have a local max at }(0,1)$
$\text{Similarly, we have a local min at }\left(\frac12, 0\right)$

owidjaja · « **Reply #3151 on:** November 11, 2017, 10:39:33 pm »

Quote from: RuiAce on November 11, 2017, 10:21:02 pm

$\frac{dy}{dx}=32x(4x^2-1)^3\\ \text{Setting this equals to 0 gives}\\ x=0\text{ and }4x^2-1=0\implies x=\pm \frac12$
$\text{One may then use the second derivative to check their nature}\\ \text{but due to the leftover power on }(4x^2-1)\text{ the test will be a bit inconclusive}\\ \text{so we simply test a bit to both sides in the first derivative.}$
$\text{When }x=-1. \frac{dy}{dx} < 0\\ \text{When }x = -\frac14, \frac{dy}{dx} > 0\\ \text{So we have a local min at }\left(-\frac12,0\right)$
$\text{When }x=-\frac14, \frac{dy}{dx} > 0\\ \text{When }x=\frac14, \frac{dy}{dx} < 0\\ \text{so we have a local max at }(0,1)$
$\text{Similarly, we have a local min at }\left(\frac12, 0\right)$

Wait, where did the x= -1/4, 1/4 come from?

RuiAce · « **Reply #3152 on:** November 11, 2017, 10:59:38 pm »

Quote from: owidjaja on November 11, 2017, 10:39:33 pm

Wait, where did the x= -1/4, 1/4 come from?

From the fact that we needed to test both sides of 0 to determine the nature of the stationary point.

Note that choices such as -1/3 could've replaced -1/4, and 2/5 could've replaced 1/4. The only important thing was that they didn't go too far and exceeded 1/2 (or went below -1/2), which were the other stationary points.

owidjaja · « **Reply #3153 on:** November 11, 2017, 11:04:58 pm »

Hey, not entirely sure how to do this question as well:

The curve y=ax^3+bx^2-x+5 has a point of inflexion at (1,-2). Find the values of a and b.

Thanks!

jamonwindeyer · « **Reply #3154 on:** November 12, 2017, 10:48:07 am »

Quote from: owidjaja on November 11, 2017, 11:04:58 pm

Hey, not entirely sure how to do this question as well:

The curve y=ax^3+bx^2-x+5 has a point of inflexion at (1,-2). Find the values of a and b.

Thanks!

Hey! The second derivative of that function is:

$y''=6ax+2b$

Now we know there is a point of inflexion at (1,-2), so substituting $x=1$ should give a second derivative of zero:

$0=6a+2b\ \ (1)$

And we also know (1,-2) must lie on the curve, so we can substitute those into the original curve:

$-2=a+b+4\\a+b=-6\ \ (2)$

Solve (1) and (2) simultaneously

taliakaragiannis8 · « **Reply #3155 on:** November 16, 2017, 03:59:51 pm »

Jude invested $4500 five years ago at x% p.a. Evaluate x if Jude now has an amount of
(a) $6311.48

how do i do this ?

RuiAce · « **Reply #3156 on:** November 16, 2017, 05:29:49 pm »

Quote from: taliakaragiannis8 on November 16, 2017, 03:59:51 pm

Jude invested $4500 five years ago at x% p.a. Evaluate x if Jude now has an amount of
(a) $6311.48

how do i do this ?

$\text{Just plug straight into the compound interest formula}\\ 4500(1+x)^5 = 6311.48$

taliakaragiannis8 · « **Reply #3157 on:** November 19, 2017, 01:50:00 pm »

12. NSW Bank offers loans at 9% p.a. with an interest-free period of
3 months, while Sydney Bank offers loans at 7% p.a. Compare these loans on an amount of $5000 over 3 years and state which bank offers the best loan and why.

could I please get some help with this question ? thank you

RuiAce · « **Reply #3158 on:** November 19, 2017, 04:04:26 pm »

$\text{The question had intended you to assume monthly compounding}\\ \text{and monthly payments. Which is quite unacceptable.}$
But it's maths in focus.
$\text{Let }M\text{ be the monthly payment in the second case.}\\ \text{7 percent per annum is converted to }\frac{7}{12}\text{ percent per month.}$
$\text{So following the usual recursive approach, we have}\\ \begin{align*}A_0 &= 5000\\ A_1 &= A_0 \left(1+\frac{7}{1200}\right) - M\\&= 5000\left(1+\frac{7}{1200}\right) - M\\ A_2&=A_1\left(1+\frac{7}{1200}\right)-M\\ &=5000\left(1+\frac{7}{1200}\right)^2 - M\left(1+\frac{7}{1200}\right) - M\end{align*}$
$\text{Continuing, }\\\begin{align*}A_3 &= A_2\left(1+\frac{7}{1200}\right)-M \\&=5000\left(1+\frac{7}{1200}\right)^3 - M\left(1+\frac{7}{1200}\right)^2 - M \left(1+\frac{7}{1200}\right) - M\\ &\vdots \\ A_n &= 5000\left(1+\frac{7}{1200}\right)^n - M\left(1+\left(1+\frac{7}{1200}\right)+\left(1+\frac{7}{1200}\right)^2+\dots + \left(1+\frac{7}{1200}\right)^{n-1}\right)\\ &= 5000\left(1+\frac{7}{1200}\right)^n - M\left(\frac{\left(1+\frac{7}{1200}\right)^n-1}{\frac{7}{1200}}\right)\end{align*}$
$\text{Setting }A_{36}=0\text{ gives } M = 154.3854843\\ \text{and since we make 36 payments, in total we pay }\$5557.88$
Remark: I just used 7/1200 to avoid errors in rounding.
_________________________________________________________

Now, maths in focus uses 'interest free' period in the question, which is not terminology you're expected to know (and thus very unfair). An interest free-period does NOT mean that interest is not charged for the first three months, but rather that you don't PAY in the first three months. i.e. you're still being charged interest; but it sits there for three months and that's it.
$\text{Let }N\text{ be the monthly payment in the first case.}\\ \text{9 percent per annum is converted to 0.75 percent per month.}$
$\text{Since no payment is made in the first three months}\\ \begin{align*}A_0&=5000\\ A_1&=A_0(1.0075)\\ &= 5000(1.0075)\end{align*}\\ \text{Similarly }A_2 = 5000(1.0075)^2 \text{ and }A_3 = 5000(1.0075)^3$
$\text{In the fourth month, we start paying.}\\ \begin{align*}A_4 &= A_3(1.0075) - N\\ &=5000(1.0075)^4 - N\\ A_5&=A_4(1.0075)-N\\ &= 5000(1.0075)^5 - N(1.0075)-N \\ A_6&= A_5(1.0075)-N\\ &= 5000(1.0075)^6 - N(1.0075)^2-N(1.0075)-N\end{align*}$
$\text{So}\\ \begin{align*}A_n &= 5000(1.0075)^n - (1+1.0075+\dots+1.0075^{n-4})\\ &= 5000(1.0075)^n - N\left(\frac{1.0075^{n-3}-1}{0.0075}\right)\end{align*}\\ \text{Setting }A_{36}=0\text{ gives }N=175.4924818\\ \text{and since we have 33 payments, in total we pay }\$5791.25$
So upon comparison, the second choice is better.

gilliesb18 · « **Reply #3159 on:** November 22, 2017, 12:27:52 pm »

Hello!!
I just need help on a (probably fairly easy) question;
The curve y= x^5 + mx^3 - 7x +5 has a stationary point at x= -2. Find the value of m.
Thanks!

RuiAce · « **Reply #3160 on:** November 22, 2017, 12:35:09 pm »

Quote from: gilliesb18 on November 22, 2017, 12:27:52 pm

Hello!!
I just need help on a (probably fairly easy) question;
The curve y= x^5 + mx^3 - 7x +5 has a stationary point at x= -2. Find the value of m.
Thanks!

$\frac{dy}{dx} = 5x^4+3mx^2-7$
$\text{Since }\frac{dy}{dx} = 0\text{ when }x=-2\text{, upon subbing}\\ 0 = 5(-2)^4 + 3m(-2)^2 - 7\\ \text{This should be fairly straightforward to solve and gives }\boxed{m=-\frac{73}{12}}$

Dragomistress · « **Reply #3161 on:** November 23, 2017, 09:51:03 pm »

Hello!
I would like some assistance with this question.

Eric11267 · « **Reply #3162 on:** November 23, 2017, 10:05:58 pm »

Quote from: Dragomistress on November 23, 2017, 09:51:03 pm

Hello!
I would like some assistance with this question.

i) you have two equations in the form of y=
so to find the point of intersection equate the two equations so that
x²+4=x+k
then x²-x+4+k=0
ii) The equation from part 1 is given in the form of a quadratic, so to get one point of intersection the discriminant needs to be 0
In this case the discriminant is
1-4(4-k)=
4k-15=0
so k=15/4
iii) Now the equation is
x²-x+4-15/4=0
putting this into the quadratic formula gives you the x coordinate of P as 1/2
then putting 1/2 into either equation will give the y value as 17/4
so the coordinates of P are (1/2,17/4)

RuiAce · « **Reply #3163 on:** November 23, 2017, 10:08:28 pm »

Quote from: Dragomistress on November 23, 2017, 09:51:03 pm

Hello!
I would like some assistance with this question.

$\text{Part i) is just simultaneous equations}\\ \text{because that's how we always find the point(s) of intersection}.\\ \begin{align*}y&=x^2+4\\y&=x+k\end{align*}$
$\text{So }x^2+4=x+k \implies x^2-x+4-k=0$
_____________________________________________________________
$\text{This is one of the common band 6 tricks that you}\\ \text{should certainly know for the future.}$
$\text{Recall from part i) that }\boxed{x^2-x+4-k=0}\\ \text{is the equation that we must solve}\\ \text{to obtain the }x\text{-coordinate of the point of intersection.}$
$\text{But we're not interested in what that coordinate is}\\ \textbf{We're just interested in how many points there are.}$
$\text{We are told that there's only }\textbf{one}\text{ point of solution.}\\ \text{Hence, there is only }\textbf{one}\text{ solution to }\boxed{x^2-x+4-k=0}\\ \text{But this is, of course, a }\textbf{quadratic}.\\ \text{Keep in mind, that a quadratic with only one solution, }\textbf{must}\text{ satisfy }\boxed{\Delta = 0}$
$\text{Hence }(-1)^2 - 4(4-k) = 0 \implies \boxed{k=\frac{15}{4}}$
_____________________________________________________________
$\text{This is now easy. We sub }k=\frac{15}4\text{ in:}\\ x^2-x+\frac{1}{16}=0 \implies \left(x-\frac12\right)^2 = 0\implies x=\frac12\\ \text{So the point is }\left(\frac12, \frac{17}4\right)$

owidjaja · « **Reply #3164 on:** November 28, 2017, 08:16:54 pm »

Hey guys,
Not so much of me needing help with a specific question, but tomorrow's my 2U exam (topics covering quadratic function, locus and geometric application). Any last minute tips or anything I should do before my exam?

Thanks!

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