Author Topic: 3U Maths Question Thread (Read 1520856 times) Tweet Share

RuiAce · « **Reply #1020 on:** October 26, 2016, 12:38:21 pm »

Quote from: Rikahs on October 26, 2016, 12:22:49 pm

Just wondering whether i got the answer right for the following question:

"At a football club a team of 11 players is to be chosen from a pool of 30 players consisting of 18 Australian- born and 12 plays born elsewhere. What is the porbability that the team will consist of all Australian-born players?"

would the answer be : 18C11/30C11

Looks good to me

teapancakes08 · « **Reply #1021 on:** October 28, 2016, 07:36:04 pm »

I'm confused by this question:

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

I got up to 2nd derivative and got lost from there...is it possible to sub in the values of P.O.I. into the curve or first derivative to help get the answer?

Syndicate · « **Reply #1022 on:** October 28, 2016, 07:58:46 pm »

Quote from: teapancakes08 on October 28, 2016, 07:36:04 pm

I'm confused by this question:

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

I got up to 2nd derivative and got lost from there...is it possible to sub in the values of P.O.I. into the curve or first derivative to help get the answer?

Since you are required to calculate the values of two variables, you need two equations to solve for a and b.

$\text{ Substitute the coordinates of the cubic's point of inflection in to the equation of the curve} \\ -2 = a(1)^3+b(1)^2-1+5 \\ -6 = a+b \space (1)$ $\text{ Now differentiate the equation of the cubic twice, and substitute the given coordinates} \\ \frac{d^2y}{d^2x} = 6ax + 2b \\ 0 = 6 a(1) +2b \\ 0 = 6a + 2b \space (2)$ $-12=2a+ 2b \space (3)$ $(2)-(3) \\ 4a = 12 \\ \therefore a = 3 \\ \text{ sub a=3 into (1) } \\ -6 = 3+b \\ \therefore b = -9$

teapancakes08 · « **Reply #1023 on:** October 28, 2016, 08:16:48 pm »

Quote from: Syndicate on October 28, 2016, 07:58:46 pm

Since you are required to calculate the values of two variables, you need two equations to solve for a and b.

$\text{ Substitute the coordinates of the cubic's point of inflection in to the equation of the curve} \\ -2 = a(1)^3+b(1)^2-1+5 \\ -6 = a+b \space (1)$ $\text{ Now differentiate the equation of the cubic twice, and substitute the given coordinates} \\ \frac{d^2y}{d^2x} = 6ax + 2b \\ 0 = 6 a(1) +2b \\ 0 = 6a + 2b \space (2)$ $-12=2a+ 2b \space (3)$ $(2)-(3) \\ 4a = 12 \\ \therefore a = 3 \\ \text{ sub a=3 into (1) } \\ -6 = 3+b \\ \therefore b = -9$

Oh! I get it now. Thank you so much

teapancakes08 · « **Reply #1024 on:** October 29, 2016, 09:45:37 pm »

Currently working on this question:

15. Find any stationary points on the curve y= x^2/(x-2). By also considering the domain of the curve, sketch its graph.

I've found one stationary point at (0,0), but am missing the other one. Other than that, I have most of what I need to sketch it (the horizontal asymptote still confuses me, though; I divided by the highest power and got an undefined asymptote...). When doing the derivative I ended up with (-x^3+4x^2-4x)/(x-2)^2 and simplified it to x(2-x)/(x-2) (if I've done the math correctly, that is). Any advice on how to go from there?

RuiAce · « **Reply #1025 on:** October 29, 2016, 09:51:30 pm »

Quote from: teapancakes08 on October 29, 2016, 09:45:37 pm

Currently working on this question:

15. Find any stationary points on the curve y= x^2/(x-2). By also considering the domain of the curve, sketch its graph.

I've found one stationary point at (0,0), but am missing the other one. Other than that, I have most of what I need to sketch it (the horizontal asymptote still confuses me, though; I divided by the highest power and got an undefined asymptote...). When doing the derivative I ended up with (-x^3+4x^2-4x)/(x-2)^2 and simplified it to x(2-x)/(x-2) (if I've done the math correctly, that is). Any advice on how to go from there?

$\text{If you differentiated it by brute force, then via the quotient rule}\\ \begin{align*}\frac{dy}{dx}&=\frac{(x-2)(2x)-(x^2)(1)}{(x-2)^2}\\ &= \frac{2x^2-4x-x^2}{(x-2)^2}\\ &= \frac{x^2-4x}{(x-2)^2}\end{align*}$

teapancakes08 · « **Reply #1026 on:** October 29, 2016, 09:59:52 pm »

Quote from: RuiAce on October 29, 2016, 09:51:30 pm

$\text{If you differentiated it by brute force, then via the quotient rule}\\ \begin{align*}\frac{dy}{dx}&=\frac{(x-2)(2x)-(x^2)(1)}{(x-2)^2}\\ &= \frac{2x^2-4x-x^2}{(x-2)^2}\\ &= \frac{x^2-4x}{(x-2)^2}\end{align*}$

Oh! Ah...I just realised I made a differentiation error. Thanks for the help ^^

teapancakes08 · « **Reply #1027 on:** November 01, 2016, 12:34:31 pm »

Need help on another locus....

1. Find equation of the locus of midpoint M of all chords PQ where P (2ap, ap^2) and Q (2aq, aq^2) lie on the parabola x^2 = 4ay and PQ passes (0,2a).

I've found the midpoint, but I'm not sure where to go from there.

RuiAce · « **Reply #1028 on:** November 01, 2016, 12:47:12 pm »

Quote from: teapancakes08 on November 01, 2016, 12:34:31 pm

Need help on another locus....

1. Find equation of the locus of midpoint M of all chords PQ where P (2ap, ap^2) and Q (2aq, aq^2) lie on the parabola x^2 = 4ay and PQ passes (0,2a).

I've found the midpoint, but I'm not sure where to go from there.

$\text{It can be shown that the chord }PQ\text{ has equation}\\ y=\frac{p+q}{2}x-apq\\ \text{This is left as your exercise.}$
$\text{We know that the chord }PQ\text{ passes through }(0,2a)\\ \text{So upon substitution}\\ 2a=-apq \implies pq=-2$
$\text{The midpoint of }PQ\text{ is }M\left(a(p+q),\frac{a(p^2+q^2)}{2}\right)\\ \text{So the midpoint has coordinates given by}\\ \begin{align*}x&=a(p+q)\\ y&=\frac{a(p^2+q^2)}{2}\tag{1}\end{align*}$
$\text{Recall that the objective with locus is to }\textbf{eliminate the parameter}.\\ \text{The presence of the }p^2+q^2\text{ implies that we should square.}\\ x^2=a^2(p^2+2pq+q^2)$
$\text{Given that }pq=-2\text{ we have }x^2=a^2(p^2+q^2-4)\\ \text{And now, clearly, by rearranging (1) into }p^2+q^2=\frac{2y}{a}\\ \text{We can sub it in to find the equation of the locus.}$

mbdtHSC · « **Reply #1029 on:** November 09, 2016, 11:45:50 pm »

Hi, I'm having trouble with this question

a) Show that 2 sin inverse (3/5) = sin inverse (24/25)

b) Hence, prove that sin inverse (24/25) + 2 sin inverse (4/5) = pi

I've done part a by letting sin inverse 3/5 be alpha, then solving for sin 2 alpha using triangle identities of cos and sin. I'm not sure about part b though and how you'd get to pi

RuiAce · « **Reply #1030 on:** November 10, 2016, 07:38:06 am »

Quote from: mbdtHSC on November 09, 2016, 11:45:50 pm

Hi, I'm having trouble with this question

a) Show that 2 sin inverse (3/5) = sin inverse (24/25)

b) Hence, prove that sin inverse (24/25) + 2 sin inverse (4/5) = pi

I've done part a by letting sin inverse 3/5 be alpha, then solving for sin 2 alpha using triangle identities of cos and sin. I'm not sure about part b though and how you'd get to pi

Just type sin^-1(x)
$\text{By substituting the identity we proved in part (i)}\\ \text{We really are trying to prove that:}\\ \begin{align*}&&2\sin^{-1}\frac{3}{5}+2\sin^{-1}\frac{4}{5}&=\pi \\ \iff&& \sin^{-1}\frac{3}{5}+\sin^{-1}\frac{4}{5}&=\frac{\pi}{2}\end{align*}$
$\text{There are many ways of doing this.}\\ \text{One is just to draw out the triangle.}\\ \text{Another is to prove that }\sin^{-1}\frac{3}{5}=\cos^{-1}\frac{4}{5}\\ \text{and use the identity }\sin^{-1}x+\cos^{-1}x=\frac{\pi}{2}$

yasmineturner · « **Reply #1031 on:** November 12, 2016, 04:02:22 pm »

Hello,
I was just wondering if someone could please help me with this question: find the greatest coefficient in (x+2)^8, using Tk+1/Tk>1?
Thankyou

RuiAce · « **Reply #1032 on:** November 12, 2016, 04:23:01 pm »

Quote from: yasmineturner on November 12, 2016, 04:02:22 pm

Hello,
I was just wondering if someone could please help me with this question: find the greatest coefficient in (x+2)^8, using Tk+1/Tk>1?
Thankyou

$\textit{Take care that depending on what convention (or even what textbook) you use}\\ \textit{Some minor details might not correspond.}\\ \textit{The final answer is the same though.}$
$\text{Note that }T_k=\binom{8}{k}x^{8-k}2^k\text{ and }T_{k+1}=\binom{8}{k+1}x^{8-(k+1)}2^{k+1}$
$\text{We are interested in the coefficients only.}\\ \text{Upon comparing coefficients, we observe that}\\ \begin{align*}\frac{T_{k+1}}{T_k}&=\frac{\binom{8}{k+1}2^{k+1}}{\binom{8}{k}2^k}\\&=2 \times \frac{\frac{8!}{(k+1)!(8-(k+1))!}}{\frac{8!}{k!(8-k)!}}\\ &= 2\times\frac{k!(8-k)!}{(k+1)!(7-k)!}\\ &=\frac{2(8-k)}{k+1} \end{align*}\\ \text{Note the use of the identity }\boxed{n!=n(n-1)!}$
$\text{Now, observe that if }T_{k+1} > T_k\text{ then the next coefficient is larger}\\ \text{If }T_{k+1} < T_k\text{ then the next coefficient is smaller.}\\ \text{As requested, we will consider }\boxed{T_{k+1} > T_k}\text{ or equivalently }\boxed{\frac{T_{k+1}}{T_k} > 1}$
$\begin{align*}\frac{T_{k+1}}{T_k} &> 1\\ \frac{2(8-k)}{k+1} &> 1\\ 16-2k &> k+1\tag{k>0}\\ k& < 5\end{align*}$
$\text{Hence when }k < 5\text{ we have }T_{k+1} > T_k\\ \text{When }k=5\text{ we have }T_6=T_5\text{ (check!!)}\\ \text{When }k > 5\text{ we have }T_{k+1} < T_k$
$\text{Hence the greatest coefficient occurs when }k=5\\ \text{So the answer is }\binom{8}{5}2^5=1792$

samuels1999 · « **Reply #1033 on:** November 16, 2016, 09:33:05 pm »

Hi Everyone,

I just needed a bit of help with doing the second part of this question. I kinda know how it works, but I don't know how to set out my answer.

Thanks,
Samuel

RuiAce · « **Reply #1034 on:** November 16, 2016, 09:45:24 pm »

Quote from: samuels1999 on November 16, 2016, 09:33:05 pm

Hi Everyone,

I just needed a bit of help with doing the second part of this question. I kinda know how it works, but I don't know how to set out my answer.

Thanks,
Samuel

$\text{Since }\alpha,\beta\text{ and }\gamma\text{ are the angles of a triangle}\\ \gamma=180^\circ - (\alpha+\beta)\text{ (angle sum of a triangle)}$
$\text{Hence if }\sin \gamma = 2\sin\alpha\cos\beta \text{, using part a)}\\ \begin{align*}\sin (180^\circ - (\alpha+\beta))&=\sin(\alpha+\beta)+\sin(\alpha-\beta)\\ \therefore \sin(\alpha+\beta)&=\sin (\alpha+\beta) + \sin (\alpha-\beta)\\ \sin (\alpha-\beta)&=0\\ \therefore \alpha-\beta&=0\tag{*}\\ \therefore \alpha&=\beta\end{align*}$
$\text{So since two angles of the triangle are equal}\\ \triangle ABC\text{ is isosceles.}$
$\textit{Note: In line (*)}\\ \text{Technically we should have }\alpha-\beta=0\text{ or }\alpha-\beta = 180^\circ\\ \text{But if the latter were true then }\alpha > 180^\circ\text{ which is not possible}\\ \text{since no angle in a triangle can exceed }180^\circ$

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