Author Topic: 4U Maths Question Thread (Read 826510 times) Tweet Share

phunky · « **Reply #1965 on:** August 18, 2018, 11:24:32 pm »

hey guys,
could someone please explain how to graph the last one? Nut quite sure how to do those types of questions!
lol figuring out how to paste an image in was such a struggle

RuiAce · « **Reply #1966 on:** August 19, 2018, 09:51:13 am »

Quote from: phunky on August 18, 2018, 11:24:32 pm

hey guys,
could someone please explain how to graph the last one? Nut quite sure how to do those types of questions!
lol figuring out how to paste an image in was such a struggle
(Image removed from quote.)

Firstly, similar to $y = f(|x|) $ you block out the part of the curve to the left of the $y$-axis. This is because regardless of what $x$ is, $x^2$ will always be positive, so you're always taking $f$ of a positive number.

In particular, it's somewhat similar to $ y = f(|x|) $ in that you need to reflect the right-portion of the curve onto the left. However the fact that it involves $x^2$ means that the graph becomes somewhat distorted as well, because it's not an ordinary dilation like $ y = f(3x)$.

Furthermore, note that since $ (4, 3) $ lies on $ y = f(x)$, we should expect $ (2,3) $ AND $ (-2,3) $ to lie on $ y = f(x^2) $. Since $ \left( \frac74, 0 \right) $ lies on $y = f(x)$, we should expect $ \left( \frac{\sqrt7}{2}, 0 \right) $ AND $ \left( -\frac{\sqrt7}{2}, 0 \right) $ to lie on $ y = f(x^2)$.

(Picture was way too late but edited in anyway for completeness)

phunky · « **Reply #1967 on:** August 19, 2018, 04:37:14 pm »

ahhh thanks, got it!

yammy · « **Reply #1968 on:** August 19, 2018, 06:08:11 pm »

Hey Rui, can you please help me with Q15a) i and iii? -I don't understand the solutions provided
For i) why is p a double root if the hyperbola touches the ellipse at P?
For iii) what do they mean by the parameter at the point Q? Why is it -p? And how do they know that O is the midpoint of PQ?
Thank you in advance(:

RuiAce · « **Reply #1969 on:** August 19, 2018, 07:41:37 pm »

Quote from: yammy on August 19, 2018, 06:08:11 pm

Hey Rui, can you please help me with Q15a) i and iii? -I don't understand the solutions provided
For i) why is p a double root if the hyperbola touches the ellipse at P?
For iii) what do they mean by the parameter at the point Q? Why is it -p? And how do they know that O is the midpoint of PQ?
Thank you in advance(:

Because $P$ is the point $ \left(cp, \frac{c}{p} \right) $, the parameter at $P$ is $p$. That's literally it - the parameter at a point is just the parameter that represents the coordinates of that point. The reason why $Q$ must therefore be $-p$ is because clearly the ellipse and the hyperbola intersect at only two distinct points. So because $ \left( cp, \frac{c}{p} \right) $ corresponds to $P$, this leaves us with $ \left( -cp, -\frac{c}{p} \right) $ corresponding to $Q$. So clearly $-p$ must be the parameter representing the point $Q$.

(It's the same as the parabola $x^2 = 4ay$. If a point is marked $ P(2ap, ap^2)$, then the parameter at $P$ is just $p$.)

It is then easy to show by the midpoint formula that the midpoint of $P$ and $Q$ is the origin.
$\text{Now, note that because the curves touch at }P\left(cp, \frac{c}{p} \right),\\ \text{the curves must have a }\textbf{common tangent}\text{ at }P.$
$\text{In general, for any conic section, if }y=mx+b\text{ is a tangent to that conic,}\\ \text{then intersecting that line with the conic produces a quadratic equation.}\\ \text{In particular, this quadratic has one unique solution, because a tangent to a conic section}\\ \text{never intersects that conic again.}$
$\text{Since that quadratic equation has one unique solution,}\\ \text{we can immediately infer that its unique solution must be a double root.}$
For example, the quadratic equation $ x^2 - 2x + 1 = 0$ has only one unique solution. (In particular, for that one it will be $x = 1$.) That solution is also a double root of the equation.
$\text{But here, the two conics have a }\textbf{common}\text{ tangent.}\\ \text{It turns out, that the investigation made with the line is transitive.}$
$\text{For }\textbf{each}\text{ point of intersection between the two conics,}\\ \text{where in particular, the conics touch each other,}\\ \text{if we attempt to intersect either of the conics with its common tangent,}\\ \text{there will be a double root at that intersection.}$
$\text{As it turns out, thanks to higher-level algebra,}\\ \text{this means that when we obtain the equation from intersecting the conics themselves,}\\ \text{they will }\textbf{also have a double root at that point of contact.}$
It's quite abstract to argue formally, but essentially the idea is to generalise the intersection between a conic and a tangent line at the point of contact, to the intersection between two "touching" conics at their point of contact instead.

yammy · « **Reply #1970 on:** August 19, 2018, 08:48:54 pm »

Quote from: RuiAce on August 19, 2018, 07:41:37 pm

Because $P$ is the point $ \left(cp, \frac{c}{p} \right) $, the parameter at $P$ is $p$. That's literally it - the parameter at a point is just the parameter that represents the coordinates of that point. The reason why $Q$ must therefore be $-p$ is because clearly the ellipse and the hyperbola intersect at only two distinct points. So because $ \left( cp, \frac{c}{p} \right) $ corresponds to $P$, this leaves us with $ \left( -cp, -\frac{c}{p} \right) $ corresponding to $Q$. So clearly $-p$ must be the parameter representing the point $Q$.

(It's the same as the parabola $x^2 = 4ay$. If a point is marked $ P(2ap, ap^2)$, then the parameter at $P$ is just $p$.)

It is then easy to show by the midpoint formula that the midpoint of $P$ and $Q$ is the origin.
$\text{Now, note that because the curves touch at }P\left(cp, \frac{c}{p} \right),\\ \text{the curves must have a }\textbf{common tangent}\text{ at }P.$
$\text{In general, for any conic section, if }y=mx+b\text{ is a tangent to that conic,}\\ \text{then intersecting that line with the conic produces a quadratic equation.}\\ \text{In particular, this quadratic has one unique solution, because a tangent to a conic section}\\ \text{never intersects that conic again.}$
$\text{Since that quadratic equation has one unique solution,}\\ \text{we can immediately infer that its unique solution must be a double root.}$
For example, the quadratic equation $ x^2 - 2x + 1 = 0$ has only one unique solution. (In particular, for that one it will be $x = 1$.) That solution is also a double root of the equation.
$\text{But here, the two conics have a }\textbf{common}\text{ tangent.}\\ \text{It turns out, that the investigation made with the line is transitive.}$
$\text{For }\textbf{each}\text{ point of intersection between the two conics,}\\ \text{where in particular, the conics touch each other,}\\ \text{if we attempt to intersect either of the conics with its common tangent,}\\ \text{there will be a double root at that intersection.}$
$\text{As it turns out, thanks to higher-level algebra,}\\ \text{this means that when we obtain the equation from intersecting the conics themselves,}\\ \text{they will }\textbf{also have a double root at that point of contact.}$
It's quite abstract to argue formally, but essentially the idea is to generalise the intersection between a conic and a tangent line at the point of contact, to the intersection between two "touching" conics at their point of contact instead.

ahh i understand, thank youu!!
While you're at it, can you please help me with this question as well?
bi, ii and iv)

For i) i found f'(x) and let f'(x)=0 to find the stat point, but I cant really get an answer

RuiAce · « **Reply #1971 on:** August 19, 2018, 09:12:28 pm »

Quote from: yammy on August 19, 2018, 08:48:54 pm

ahh i understand, thank youu!!
While you're at it, can you please help me with this question as well?
bi, ii and iv)

For i) i found f'(x) and let f'(x)=0 to find the stat point, but I cant really get an answer

Hint for iv): You want $n!$ to appear under the power. But you know that $ n! = 1\times 2\times \dots \times n$....................

If you've correctly found the stationary point $ x = \frac{c}{n} $, you should then proceed to prove that it is a local minimum and argue that it is a global minimum. You're essentially doing the same things I did in my trial survival lecture regarding the global minimum.
$\text{Sub }c = \sum_{i=1}^n x_i\text{ and }x = x_{n+1}\\ \text{into the formula in part i). Then, }\\ \begin{align*}LHS &= \frac{1}{x} \left(\frac{x+c}{n+1} \right)^{n+1}\\ &= \frac{1}{x_{n+1}} \left( \frac{x_{n+1} + \sum_{i=1}^n x_i}{n+1} \right)^{n+1}\\ &= \frac{1}{x_{n+1}} \left( \frac{\sum_{i=1}^{n+1}x_i}{n} \right)^{n+1} \end{align*}$
$\text{and }RHS = \left( \frac{1}{n} \sum_{i=1}^n x_i \right)^n\text{ immediately. So therefore moving }x_{n+1}\text{ over},\\ \left( \frac{1}{n+1} \sum_{i=1}^{n+1} x_i \right)^{n+1} \geq x_{n+1} \frac{1}{n} \left( \sum_{i=1}^n x_i \right)^n$

yammy · « **Reply #1972 on:** August 20, 2018, 12:09:56 am »

Quote from: RuiAce on August 19, 2018, 09:12:28 pm

Hint for iv): You want $n!$ to appear under the power. But you know that $ n! = 1\times 2\times \dots \times n$....................

If you've correctly found the stationary point $ x = \frac{c}{n} $, you should then proceed to prove that it is a local minimum and argue that it is a global minimum. You're essentially doing the same things I did in my trial survival lecture regarding the global minimum.
$\text{Sub }c = \sum_{i=1}^n x_i\text{ and }x = x_{n+1}\\ \text{into the formula in part i). Then, }\\ \begin{align*}LHS &= \frac{1}{x} \left(\frac{x+c}{n+1} \right)^{n+1}\\ &= \frac{1}{x_{n+1}} \left( \frac{x_{n+1} + \sum_{i=1}^n x_i}{n+1} \right)^{n+1}\\ &= \frac{1}{x_{n+1}} \left( \frac{\sum_{i=1}^{n+1}x_i}{n} \right)^{n+1} \end{align*}$
$\text{and }RHS = \left( \frac{1}{n} \sum_{i=1}^n x_i \right)^n\text{ immediately. So therefore moving }x_{n+1}\text{ over},\\ \left( \frac{1}{n+1} \sum_{i=1}^{n+1} x_i \right)^{n+1} \geq x_{n+1} \frac{1}{n} \left( \sum_{i=1}^n x_i \right)^n$

ohhh thanks rui i get it

Can you also help me with the 2nd part of di)?

RuiAce · « **Reply #1973 on:** August 20, 2018, 08:51:10 am »

Quote from: yammy on August 20, 2018, 12:09:56 am

ohhh thanks rui i get it
Can you also help me with the 2nd part of di)?

$\text{Note that for each }k= 1, \dots, n-1\\ \begin{align*}PP_k^2 &= |1 - \omega^k|^2\\ &= (1- \omega^k) \overline{(1 - \omega^k)}\\ &= (1 - \omega^k)(1 - \overline{\omega^k})\\ &= 1 - 2\text{Re}(\omega^k) + |\omega^k|^2\\ &= 1 - 2\text{Re}(\omega^k) + 1 \end{align*}$
$\text{Therefore}\\ PP_1^2 + PP_2^2 + \dots + PP_{n-1}^2 = (2 - 2 \text{Re}(\omega)) + (2- 2\text{Re}(\omega^2)) + \dots + (2 - 2\text{Re}(\omega^{n-1})\\ \text{and it's clear that the constants sum to }2(n-1).$
$\text{As for all the real parts,}\\ \begin{align*}&\quad \text{Re}(\omega) + \text{Re}(\omega^2) + \dots + \text{Re}(\omega^{n-1})\\ &+ \text{Re}(\omega+ \omega^2 + \dots + \omega^{n-1})\\ &=\text{Re}(1+\omega+\omega^2+\dots + \omega^{n-1}) - 1\\ &= \text{Re}\left( \frac{\omega^n - 1}{\omega - 1} \right) - 1\\ &= 0 - 1\\ &= -1 \end{align*}$
$\text{Therefore the required expression equals to }2(n-1) - 2(-1)\\ \text{which equals }2n.$

3.14159265359 · « **Reply #1974 on:** August 23, 2018, 12:19:32 pm »

hey can someone please help me with q10
thank you

RuiAce · « **Reply #1975 on:** August 23, 2018, 05:44:18 pm »

Quote from: 3.14159265359 on August 23, 2018, 12:19:32 pm

hey can someone please help me with q10
thank you

I’ve attempted the question but because of its very vague wording there was very little that I could salvage from it. Please provide the final answer for me to confirm with before I post any solution.

3.14159265359 · « **Reply #1976 on:** August 23, 2018, 09:09:04 pm »

Quote from: RuiAce on August 23, 2018, 05:44:18 pm

I’ve attempted the question but because of its very vague wording there was very little that I could salvage from it. Please provide the final answer for me to confirm with before I post any solution.

there is no solutions

RuiAce · « **Reply #1977 on:** August 23, 2018, 11:43:07 pm »

Quote from: 3.14159265359 on August 23, 2018, 09:09:04 pm

there is no solutions

That's cool, but like was looking for just the final **answer**. Anyway here is a sketch solution.

yammy · « **Reply #1978 on:** August 23, 2018, 11:50:51 pm »

hello, can someone please help me with d) - mathematical induction question and 16a)? i cant seem to get a solution and there are no solutions provided

RuiAce · « **Reply #1979 on:** August 24, 2018, 08:47:10 am »

Quote from: yammy on August 23, 2018, 11:50:51 pm

hello, can someone please help me with d) - mathematical induction question and 16a)? i cant seem to get a solution and there are no solutions provided

The induction question is already addressed in the compilation as it is a part of the 2016 paper.

That is a very long question. What have you attempted thus far?

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