Author Topic: 3U Maths Question Thread (Read 1518427 times) Tweet Share

itssona · « **Reply #3105 on:** December 02, 2017, 11:38:32 pm »

Quote from: Opengangs on December 02, 2017, 10:28:27 pm

$\text{We consider the nature of coefficients in a typical binomial expansion, and we can observe different properties inside binomial expansions.}$ $\text{In particular, there reaches a maximum coefficient before the coefficients begin to decrease.}$ $\text{If this ratio follows: } \frac{T_{k+1}}{T_{k}} \leq 1 \text{, then we can begin to find some property for the greatest coefficient.}$
$\text{We begin by considering a typical binomial expansion:} (a + b)^n = \sum_{r=0}^n a^{n-r} \cdot b^r$ $\text{Using this expansion, consider some value for k such that: } T_k = \binom{n}{k-1} a^{n-(k-1)} \cdot b^{k-1}$ $\text{This occurs when we consider that a binomial expansion begins at 0 and not 1.}$ $\text{Using this idea, we can generalise an expression for the next term: } T_{k+1} = \binom{n}{k} a^{n-k} \cdot b^{k}$ $\text{Using the ratio, we can begin to find the greatest coefficient: } \frac{\binom{20}{k} 3^{20-k} \cdot 5^{k}}{\binom{20}{k-1} 3^{20-(k-1)} \cdot 5^{k-1}} \leq 1$

See if you can find a value for k, and find the greatest coefficient. Remember that k is an integer.

so to get greatest term, I would add 1 to my value of k?

Opengangs · « **Reply #3106 on:** December 02, 2017, 11:52:14 pm »

Quote from: sssona09 on December 02, 2017, 11:38:32 pm

so to get greatest term, I would add 1 to my value of k?

Greatest term =/= greatest coefficient.

You will notice that the end result will be something of the form: k >= a/b.
Since k has to be an integer, it is the smallest integer that satisfies the condition for k.

itssona · « **Reply #3107 on:** December 03, 2017, 12:14:05 am »

Quote from: Opengangs on December 02, 2017, 11:52:14 pm

Greatest term =/= greatest coefficient.

You will notice that the end result will be something of the form: k >= a/b.
Since k has to be an integer, it is the smallest integer that satisfies the condition for k.

oh so then I sub in the value of k into the general form yeah? thank youuuu

katgrace1 · « **Reply #3108 on:** December 03, 2017, 05:36:45 pm »

Hey. Quick question. I have the question, : ‘ Find the equations of the tangents to the curve x^2 =8y at the pts P: (4p,2p^2) , Q (4q,aq^2). Find the equation of the locus of their pt of intersection if PQ is a focal chord.’

I have found the top part which is
Y=qx -2q^2
Y=px - 2p^2

Then I went on to find the intersection to be (2(p+q), 2pq)
Also wouldn’t pq=-1 as it is a focal chord?
How would I find the equation of the locus from here. Thx.

RuiAce · « **Reply #3109 on:** December 03, 2017, 05:42:00 pm »

Quote from: katgrace1 on December 03, 2017, 05:36:45 pm

Hey. Quick question. I have the question, : ‘ Find the equations of the tangents to the curve x^2 =8y at the pts P: (4p,2p^2) , Q (4q,aq^2). Find the equation of the locus of their pt of intersection if PQ is a focal chord.’

I have found the top part which is
Y=qx -2q^2
Y=px - 2p^2

Then I went on to find the intersection to be (2(p+q), 2pq)
Also wouldn’t pq=-1 as it is a focal chord?
How would I find the equation of the locus from here. Thx.

$\text{As always, for locus problems we need to }\textbf{eliminate the parameters}.\\ \text{We keep in mind that }\boxed{pq=-1}\text{ as correctly stated.}$
$\text{Our point of intersection satisfies}\\ \begin{align*}x&=2(p+q)\\ y&=2pq\end{align*}$
$\text{Given this, for this particular question the locus is easy to compute}\\ \text{because subbing directly into the second equation gives}\\ \boxed{y=-2}\text{, and we are done.}$
Remark: This is not a coincidence. We know, for a fact, that the tangents at the ends of a focal chord intersect (at right angles) on the directrix.

katgrace1 · « **Reply #3110 on:** December 03, 2017, 05:49:59 pm »

😂I looked way too hard into this. Thank you.

itssona · « **Reply #3111 on:** December 04, 2017, 07:28:48 pm »

show that the line x-y-2=0 is a tangent to the parabolas x^2=8y

thank you!!

RuiAce · « **Reply #3112 on:** December 04, 2017, 07:40:17 pm »

Quote from: sssona09 on December 04, 2017, 07:28:48 pm

show that the line x-y-2=0 is a tangent to the parabolas x^2=8y

thank you!!

$\text{This is a classic 2U band 6 trick that you should memorise.}\\ \text{Whereas for cubics this isn't necessarily true,}\\\text{for circles and parabolas, if any line intersects it only once}\\ \text{it }\textbf{must}\text{ be tangential to the curve.}$
(This is also true for ellipses and hyperbolas, actually.)
$\text{Suppose we wish to find the point of intersection.}\\ \text{We would do simultaneous equations to obtain}\\ x^2 = 8(x-2) \implies x^2-8x+16=0$
$\text{The discriminant of this quadratic equation is }\Delta = 8^2-4(16) = 0\\ \text{Since the discriminant is 0, this quadratic only has one solution}\\ \text{and therefore, there is only one point of intersection between the line and the parabola}\\ \text{which is what we wished to prove.}$
Alternatively, for this particular question we could've just explicitly computed the unique point of intersection, at $x=4$.

martinarena_ · « **Reply #3113 on:** December 04, 2017, 07:50:35 pm »

Hi!
I was wondering if someone could help me out with this question:
'Use the sum of roots to find the other point where the graph cuts the x-axis'
The function is f(x)=(x cubed) - 9x + (6 root 3)

Thank you!!!!

RuiAce · « **Reply #3114 on:** December 04, 2017, 08:05:33 pm »

Quote from: martinarena_ on December 04, 2017, 07:50:35 pm

Hi!
I was wondering if someone could help me out with this question:
'Use the sum of roots to find the other point where the graph cuts the x-axis'
The function is f(x)=(x cubed) - 9x + (6 root 3)

Thank you!!!!

There is not enough information provided to answer this question; your question says other point but we don't know anything about the original point.

martinarena_ · « **Reply #3115 on:** December 04, 2017, 08:26:29 pm »

Here's the complete question:
For the function f(x) = (x^3) - 9x + (6 root 3)
(a) show that the function has a turning point at (root 3, 0) and (- root 3, 12 root 3)
(b) Use the 'sum of roots' to find the other point where the graph cuts the x-axis.
I just need some help for part b

Thanks!

RuiAce · « **Reply #3116 on:** December 04, 2017, 08:44:25 pm »

Quote from: martinarena_ on December 04, 2017, 08:26:29 pm

Here's the complete question:
For the function f(x) = (x^3) - 9x + (6 root 3)
(a) show that the function has a turning point at (root 3, 0) and (- root 3, 12 root 3)
(b) Use the 'sum of roots' to find the other point where the graph cuts the x-axis.
I just need some help for part b
Thanks!

$\text{The key aspect of this question is that we know}\\ \text{we have a }\textbf{turning point}\text{ at }(\sqrt3, 0).$
$\text{The fact that }y=f(x)\text{ passes through }(\sqrt3,0)\text{ first tells us that}\\ \sqrt3\text{ is an }x\text{-intercept.}\\ \text{This means that our first root is }\alpha = \sqrt{3}.$
$\text{But the fact that it's also a turning point tells us that as a matter of fact,}\\ \text{we have a }\textbf{double root.}$
(You explored double roots in the preliminary Polynomials topic.)
$\text{Since }\sqrt3\text{ is a double root, i.e. a root AGAIN,}\\ \text{we now know that }\beta = \sqrt{3}.$
$\text{The sum of roots now gives}\\ \sqrt{3}+\sqrt{3}+\gamma = 0 \implies \boxed{\gamma = -2\sqrt{3}}.$
$\text{So the other intercept must be at }x=-2\sqrt{3}.$

itssona · « **Reply #3117 on:** December 06, 2017, 07:13:20 pm »

heey, any direction on the steps to do this please?
the parabola x^2=4ay has focus S and vertex O. At a point P(2ap,ap^2) which moves on the parabola, tangents are drawn and from S, a perpenficular S is drawn to meet the tangent at M.
Find the locus of M.
thank you

RuiAce · « **Reply #3118 on:** December 06, 2017, 07:23:11 pm »

Quote from: sssona09 on December 06, 2017, 07:13:20 pm

heey, any direction on the steps to do this please?
the parabola x^2=4ay has focus S and vertex O. At a point P(2ap,ap^2) which moves on the parabola, tangents are drawn and from S, a perpenficular S is drawn to meet the tangent at M.
Find the locus of M.
thank you

$\text{From the reference sheet,}\\ \text{the tangent has equation }\boxed{y=px-ap^2}\text{ and gradient }m_1 = p.$
$\text{Hence, any line perpendicular to this tangent must have gradient }-\frac1p.\\ \text{We want this line to pass through }S(0,a).\text{ Hence the point-gradient form gives}\\ y-a = -\frac{1}{p}x\implies \boxed{y=-\frac{x}{p}+a}$
$\text{Since }M\text{ is the point of intersection of these lines, we commence }\\ \text{simultaneous equations. Upon equating,}\\ \begin{align*}px-ap^2 &= -\frac{x}p + a\\ px+\frac{x}{p}&=a+ap^2\\ x\left(p+\frac{1}{p}\right)&= a(p^2+1)\\ &= ap\left(p+\frac{1}{p}\right)\\ x&=ap\end{align*}\\ \text{and subsequently }y=0.$
$\text{So the coordinates of }M\text{ satisfy }x=ap\text{ and }y=0.\\ \text{Since }\boxed{y=0}\text{ does not involve the parameter }p,\\ \text{it must, by default, be the equation of the locus.}$

itssona · « **Reply #3119 on:** December 06, 2017, 07:33:52 pm »

Quote from: RuiAce on December 06, 2017, 07:23:11 pm

$\text{From the reference sheet,}\\ \text{the tangent has equation }\boxed{y=px-ap^2}\text{ and gradient }m_1 = p.$
$\text{Hence, any line perpendicular to this tangent must have gradient }-\frac1p.\\ \text{We want this line to pass through }S(0,a).\text{ Hence the point-gradient form gives}\\ y-a = -\frac{1}{p}x\implies \boxed{y=-\frac{x}{p}+a}$
$\text{Since }M\text{ is the point of intersection of these lines, we commence }\\ \text{simultaneous equations. Upon equating,}\\ \begin{align*}px-ap^2 &= -\frac{x}p + a\\ px+\frac{x}{p}&=a+ap^2\\ x\left(p+\frac{1}{p}\right)&= a(p^2+1)\\ &= ap\left(p+\frac{1}{p}\right)\\ x&=ap\end{align*}\\ \text{and subsequently }y=0.$
$\text{So the coordinates of }M\text{ satisfy }x=ap\text{ and }y=0.\\ \text{Since }\boxed{y=0}\text{ does not involve the parameter }p,\\ \text{it must, by default, be the equation of the locus.}$

omg thank you rui

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