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

Usernameismyname · Mar 2, 2016

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

Hey

It'll be greatly appreciated if someone could help me out with question 22 of Exercise 4J.

Thanks

Blitz_N7 · Mar 2, 2016

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

Unsure on how to do this question :/

integral95 · Mar 2, 2016

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

Blitz_N7 said:
Unsure on how to do this question :/

a) it's a geometric series with the ratio r = -3 and first term a = 4, use the sum of GP formula with n terms and simplify

b) let Sn = -728 and solve for n

c) ehh you could always do guess and check, but if not, then it's solving the biggest value of n in the inequality

$4(3)^{n-1} < 1000000$ since you can make r = 3 (absolute values implies that all values are positive)

d) ahhh just do guess and check

davidgoes4wce · Mar 2, 2016

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

AfroNerd said:
Use mathematical induction to prove that for all positive integers n:

a) View attachment 32967

a) View attachment 32968

I answered that EXERCISE 6N Q 1 (e) with the helpful advice from Integrand. Using n=1 for this case.

davidgoes4wce · Mar 3, 2016

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

drsabz101 said:
What are the steps involved in solving this (using the chain rule)

View attachment 32969

With Exercise 7H, Q 9 first try to visualise the image of the parabola. I have included a screenshot of the path of the bullet:

davidgoes4wce · Mar 3, 2016

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

$Exercise 7H Q9$

$y=-20x^2+400x$

$\frac{dy}{dt} = [-40x +400] \ \frac{dx}{dt}$

$(a) (i) $ When x=8 $, \frac{dx}{dt} = \frac{1}{2} , \frac{dy}{dt}=[-40(8)+400] \times \frac{1}{2}=40 ms^{-1}$

$(ii) $ When x=18 $, \frac{dx}{dt} = \frac{1}{2} , \frac{dy}{dt}=[-40(18)+400] \times \frac{1}{2}=-100 ms^{-1}$

$(iii) $ When x=10 $, \frac{dx}{dt} = \frac{1}{2} , \frac{dy}{dt}=[-40(10)+400] \times \frac{1}{2}=0 ms^{-1}$

$(iv) $ When y=1500, by substituting it into the equation y=-20x^2+400x $$

$1500=-20x^2+400x$

$0=-20x^2+400x-1500$

$0=-20(x-15)(x-5)$

$x=5,15$

$$ When x=5 $, \frac{dy}{dt}=[-40(5)+400] \times \frac{1}{2} =100 ms^{-1}$

$$ When x=15,$ \frac{dy}{dt}=[-40(15)+400] \times \frac{1}{2}=-100 ms^{-1}$

$Q9 (b) (i) \frac{dy}{dt}=30ms^{-1}, \frac{dy}{dt}=[-40x+400] \frac{dx}{dt}$

$30=[-40x+400] \times \frac{1}{2}$

$x=8.5 \ metres$

$y=-20(8.50) [8.5-20] =1955 \ metres$

$(ii) \frac{dy}{dt}=-70 ms^{-1}$

$-70 = [-40x+400] \times \frac{1}{2}$

$\therefore x=13.5 \ metres$

$y=-20 (13.5) [13.5-20]= 1755 \ metres$

$9 (c) I tried to again draw a diagram, since we know that the horizontal velocity is constant, and the vertical velocity mentioned is 10 $ ms^{-1} , $ by drawing a Triangle up the angle can be found$

$tan \theta = \frac{10}{\frac{1}{2}} = 87 \degree 8 ^\prime$

$Q9 (d)$

$\frac{dy}{dt}=0$

$0=[-40x+400] \times \frac{1}{2}$

$x=10$

$y=-20 (10) [10-20] =2000 \ metres$

$The range the ball travels, we set y=0.$

$0=-20x(x-20)$

$x=0, x=20$

$\therefore $ The range the bullet travels is x=20 $ \ m$

Blitz_N7 · Mar 3, 2016

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

I need help in this question. I can see that Tn=n/2^(n-1) but I dont know how to find Sn.

integral95 · Mar 3, 2016

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

Blitz_N7 said:
I need help in this question. I can see that Tn=n/2^(n-1) but I dont know how to find Sn.

what's the final answer?

Blitz_N7 · Mar 3, 2016

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

Paradoxica · Mar 3, 2016

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

Blitz_N7 said:

By inspection, I presume?

AfroNerd · Mar 3, 2016

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

can someone please explain how n only = 27 and not ±27 in the equation n^(2/3) = 9

InteGrand · Mar 3, 2016

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

AfroNerd said:
can someone please explain how n only = 27 and not ±27 in the equation n^(2/3) = 9

$\noindent Note that $n^{\frac{2}{3}} = \left(n^{\frac{1}{3}}\right)^2$. So $n^{\frac{2}{3}} = 9\Longleftrightarrow \left(n^{\frac{1}{3}}\right)^2 = 9$, so $n^{\frac{1}{3}} = \pm 3$, so upon cubing, we have $n = \pm 27$.$

AfroNerd · Mar 3, 2016

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

answers say n=27 even though i did those steps

InteGrand · Mar 3, 2016

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

AfroNerd said:
answers say n=27 even though i did those steps

Was it part of a larger question where n had to be positive (e.g. because it was the number of some objects or something)?

InteGrand · Mar 3, 2016

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

Blitz_N7 said:
I need help in this question. I can see that Tn=n/2^(n-1) but I dont know how to find Sn.

$\noindent We have $S_n = \sum_{k=1}^{n} \frac{k}{2^{k-1}}$, for $n=1,2,3,\ldots$. Hence $2S_n = \sum_{k=1}^{n} \frac{2k}{2^{k-2}}= \sum_{k=1}^{n} \frac{k}{2^{k-2}}$. So we have$

$\begin{align*}2S_n - S_n &= \sum_{k=1}^{n} \frac{k}{2^{k-2}} - \sum_{k=1}^{n} \frac{k}{2^{k-1}} \\ \Rightarrow S_n &= \sum_{k=0}^{n-1} \frac{k+1}{2^{k-1}} - \sum_{k=1}^{n} \frac{k}{2^{k-1}} \quad (\text{shifting the index of the first sum}) \\ &= \sum_{k=0}^{n-1} \frac{k}{2^{k-1}} + \sum_{k=0}^{n-1} \frac{1}{2^{k-1}} - \sum_{k=1}^{n} \frac{k}{2^{k-1}} \\ &= \color{red}\sum_{k=1}^{n-1} \frac{k}{2^{k-1}}\color{black} + \sum_{k=0}^{n-1} \frac{1}{2^{k-1}} - \color{red}\sum_{k=1}^{n-1} \frac{k}{2^{k-1}}\color{black} - \frac{n}{2^{n-1}} \quad (\text{removing the last term from the last sum, so that we can cancel the red sums;the reason the first sum can actually start at }k=1 \text{ is that the }k=0 \text{ term is just 0}) \\ &= \sum_{k=0}^{n-1} \frac{1}{2^{k-1}} - \frac{n}{2^{n-1}} \\ &= \frac{2\left(1-\frac{1}{2^n}\right)}{1-\frac{1}{2}}- \frac{n}{2^{n-1}} \quad (\text{using G.P. sum formula})\end{align*}$

$\begin{align*}&= \frac{2\left(1-\frac{1}{2^n}\right)}{\frac{1}{2}}- \frac{n}{2^{n-1}}\\ &= 4\left(1-\frac{1}{2^n}\right)- \frac{n}{2^{n-1}}\\ \Rightarrow S_n &=4 - \frac{1}{2^{n-2}}- \frac{n}{2^{n-1}} \quad (\text{expanding and simplifying}). \end{align*}$

$\noindent For the final part, taking the limits as $n\to \infty$ of both sides, we have$

$\begin{align*}\lim_{n\to \infty} S_n &= \lim_{n\to \infty } \left(4 - \frac{1}{2^{n-2}}- \frac{n}{2^{n-1}}\right) \\ \Rightarrow S_{\infty} &= 4 - 0 -0 \quad (\text{since the terms on the R.H.S. above that involve }n \text{ all go to 0 as }n\to \infty) \\ &= 4. \end{align*}$

AfroNerd · Mar 4, 2016

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

InteGrand said:
Was it part of a larger question where n had to be positive (e.g. because it was the number of some objects or something)?

No, the question just asked to solve. Also, calculators like wolfram alpha only got 27 as well

Paradoxica · Mar 4, 2016

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

AfroNerd said:
No, the question just asked to solve. Also, calculators like wolfram alpha only got 27 as well

I suspect they took the principle valued power.

InteGrand · Mar 4, 2016

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

AfroNerd said:
No, the question just asked to solve. Also, calculators like wolfram alpha only got 27 as well

$\noindent I think it's to do with some calculators computing cube roots as logarithms, which are undefined at negative values, so these calculators only end up working with positive values of $x$ for these power functions. For example, here's how WolframAlpha plots the graph of $y= x^{\frac{2}{3}}$:$

http://wolframalpha.com/input/?i=plot+y+=+x^(2/3)&x=0&y=0.

$\noindent It only has has the right branch of the graph. The actual graph also has a mirror image left branch, and can be found about halfway down this page:$

http://www.sosmath.com/calculus/diff/der09/der09.html.

si2136 · Mar 5, 2016

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

View attachment 32979

Is this correct?

ml125 · Mar 5, 2016

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

si2136 said:
View attachment 32979

Is this correct?

Yeah, it is

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

New Member

Member

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Member

Well-Known Member

Member

-insert title here-

New Member

Well-Known Member

New Member

Well-Known Member

Well-Known Member

New Member

-insert title here-

Well-Known Member

Well-Known Member

Well-Known Member

Users Who Are Viewing This Thread (Users: 0, Guests: 1)