Cambridge Prelim MX1 Textbook Marathon/Q&A (1 Viewer)

drsabz101 · Jun 5, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

the term after n

Blitz_N7 · Jun 5, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

What is the best (HSC) method to do part (d)? Thanks in advance.

InteGrand · Jun 5, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

Blitz_N7 said:
What is the best (HSC) method to do part (d)? Thanks in advance.

Recall that a number is divisible by 3 if and only if the sum of its digits is a multiple of 3. So just find how many ways you can form numbers from that so that the digit sum is a multiple of 3.

Blitz_N7 · Jun 5, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

InteGrand said:
Recall that a number is divisible by 3 if and only if the sum of its digits is a multiple of 3. So just find how many ways you can form numbers from that so that the digit sum is a multiple of 3.

Alright, thanks! ^.^

davidgoes4wce · Jun 5, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

appleibeats said:
View attachment 33235

Unsure how to do part b). I get the answer 13/6 but the answer is n = 7

$Decided to have a go at this question as well$

$Q 21 (a) \ Three consecutive coefficients: ^nC_r,^nC_{r+1}, ^nC_{r+2}$

$\frac{^nC_r}{^nC_{r+1}}=\frac{9}{24}$

$\frac{^nC_{r+1}}{^nC_{r+2}}=\frac{24}{42}$

$\frac{^nC_r}{^nC_{r+1}}=\frac{n!}{(n-r)!r!} \div \frac{n!}{(r+1)!(n-r-1)!} =\frac{9}{24}$

$\frac{n!}{(n-r)!r!} \times \frac{(r+1)!(n-r-1)!}{n!} =\frac{9}{24}$

$Cancelling n! leaves us with$

$\frac{(r+1)!(n-r-1)!}{(n-r)!r!}=\frac{9}{24}$

$\frac{r!(r+1)(n-r-1)!}{(n-r)(n-r-1)!r!}=\frac{9}{24}$

$$ Cancelling terms again leaves us with , $ \frac{r+1}{n-r}=\frac{9}{24}$

$24=9n-33r \ \textcircled{1}$

$\frac{^nC_{r+1}}{^nC_{r+2}}=\frac{24}{42}$

$\frac{^nC_{r+1}}{^nC_{r+2}}=\frac{n!}{(n-r-1)!(r+1)!} \div \frac{n!}{(r+2)!(n-r-2)!} =\frac{24}{42}$

$\frac{n!}{(n-r-1)!(r+1)!} \times \frac{(r+2)!(n-r-2)!}{n!} =\frac{24}{42}$

$Cancelling n! leaves us with$

$\frac{(r+2)!(n-r-2)!}{(n-r-1)!(r+1)!}=\frac{24}{42}$

$\frac{(r+1)!(r+2)(n-r-2)!}{(n-r-2)!(n-r-1)(r+1)!}=\frac{24}{42}$

$$ Cancelling terms again leaves us with , $ \frac{r+2}{n-r-1}=\frac{24}{42}$

$108=24n-66r \ \textcircled{2}$

$$ Solving simultaneously $ \textcircled{1} \ and \ \textcircled{2}, n=10$

$Q21(b)$

$$ Knowing $ ^nC_0=1, ^nC_1=n, ^nC_{2}=\frac{n}{2}(n-1), ^nC_3=\frac{n(n-1)(n-2)}{6}$

$$ With an arithmetic difference $ ^nC_3-^nC_2=^nC_2-^nC_1$

$$ Substituting the terms in we get, $ \frac{(n-1)(n-2)}{6}-\frac{n}{2}(n-1)=\frac{n}{2}(n-1)-n$

$\frac{n(n-1)(n-2)}{6}-n(n-1)+n=0$

$$ Knowing that n cannot equal, we can factorize n out, to have the equation, $ \frac{(n-1)(n-2)}{6}-(n-1)+1=0$

$(n-1)(n-2)-6(n-1)+6=0$

$n^2-9n+14=0$

$\therefore n=2,n=7$

$n cannot be 2 for our purposes as it does not contain the term x^3$

$Q21 (c) (i)$

$2 \times ^nC_5=^nC_4+^nC_6$

$Divide $ ^nC_ 5 $ both sides$

$2=\frac{^nC_4}{^nC_5}+\frac{^nC_6}{^nC_5}$

$2=\frac{5}{n-4}+\frac{n-5}{6}$

$Rearranging, we get the quadratic formula , n^2-21n+98=0$

$(ii)$

$Using PSF, factorise 0=(n-14)(n-7)$

$n in this case is : 7,14$

davidgoes4wce · Jun 5, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

I also learnt a little trick today

if you have something like

$\frac{^nC_5}{^nC_6}=\frac{6}{n-5}$

$I'll be honest and admit up until today that I have done it the long way by cancelling terms and using up 3 lines to get to that part. Would that be something the advanced students know in 1 step?$

si2136 · Jun 5, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

davidgoes4wce said:
I also learnt a little trick today

if you have something like

$\frac{^nC_5}{^nC_6}=\frac{6}{n-5}$

$I'll be honest and admit up until today that I have done it the long way by cancelling terms and using up 3 lines to get to that part. Would that be something the advanced students know in 1 step?$

$The more practice you have, the better you'll be able to 'see' the patterns$

appleibeats · Jun 6, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

Screen Shot 2016-06-06 at 8.35.51 pm.png

I found the general formula

Then found T14 and T15

Then made T14/T15 > 1

I then simplified and got sinOcos^9O > 6/15 where O is theta

However I think this is wrong and I don't know where I have gone wrong.

Answers says Theta = 22 degrees

Paradoxica · Jun 6, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

appleibeats said:
View attachment 33248

I found the general formula

Then found T14 and T15

Then made T14/T15 > 1

I then simplified and got sinOcos^9O > 6/15 where O is theta

However I think this is wrong and I don't know where I have gone wrong.

Answers says Theta = 22 degrees

tanθ>2⁄5
θ>tan^-12⁄5 ≈ 21.8°

pasq · Jun 10, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

View attachment 332

davidgoes4wce · Jun 10, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

pasq said:
View attachment 332

davidgoes4wce · Jun 11, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

Blitz_N7 said:
I got 2.863km/h

Here is my worked out solution, I got a different answer.

Blitz_N7 · Jun 11, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

You are right. I misread two as ten. ._.

drsabz101 · Jun 15, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

Can someone show the working out to these questions:

- Q5c and d

Extensions question 16 and 17c

Shuuya · Jun 15, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

drsabz101 said:
Can someone show the working out to these questions:

View attachment 33270 - Q4c and d

Extensions question 16 and 17c

View attachment 33271

View attachment 33272

Do you mean 5c) and d)?

let b² - 4ac=0, then solve for 'l' by factorising/using the quadratic formula

drsabz101 · Jun 15, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

why would u let the discriminant equal to 0?

integral95 · Jun 15, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

drsabz101 said:
why would u let the discriminant equal to 0?

If a quadratic can be factorised in a form of a perfect square,it would only have one root, and letting the discriminant equal to 0 implies there's only one root.

drsabz101 · Jun 15, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

for q.16 am I supposed to find the discriminant let it equal to 0. What is the next step?

InteGrand · Jun 15, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

drsabz101 said:
for q.16 am I supposed to find the discriminant let it equal to 0. What is the next step?

Assume a, b real.

$\noindent Since the roots are real, this is equivalent to saying the discriminant is \emph{greater than or equal to} 0, so $a^2 - 8(b+3)\geq 0$. Thus $a^2 \geq 8(b+3)$. So to minimise $a^2+b^2$, whatever value for $b$ we take, we should take the $a$ that makes $a^2$ as small as possible while still having the condition $a^2 \geq 8(b+3)$ hold. We consider two cases.$

$\noindent \textbf{Case 1.} $b\leq -3$$

$\noindent Here, $b+3 \leq 0$, so any $a$ will make the condition hold. So to minimise $a^2 + b^2$ here, we should clearly make $a=0$ and $b=-3$, giving $a^2 + b^2 =9$ for the minimum.$

$$\noindent \textbf{Case 2.} $b\geq -3$ (There is an overlap with Case 1 but it doesn't matter as these two cases are still exhaustive.)$

$\noindent Here, $b+3$ is non-negative, so the smallest allowable $a^2$ is by taking $a=\sqrt{8(b+3)}$, so $a^2 = 8(b+3)$. Then $a^2 + b^2 = b^2 + 8(b+3) = (b+4)^2 + 8$. This is an increasing function in $b$ when $b\geq -3$, so is minimised by taking $b=-3$. So then $a^2 = 0$ and the minimum for $a^2 + b^2$ is again $9$.$

So in summary, the answer is 9.

davidgoes4wce · Jun 25, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

Exercise 4B Question 10

Would this be a good enough explanation?

$\angle ABP=90-\theta $ (sum of angles in a triangle) $$

$\angle PBC=90-(90-\theta)=\theta $ (Complementary angle) $$

$AQ=AP+PQ$

$= x \cos \theta + y \ sin \theta$

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