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

InteGrand · Dec 14, 2015

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

Blitz_N7 said:
Question 9, Chapter 13C of Year 11 Maths 3U Cambridge.
$$Differentiate {xe^x}$. $ Hence find$: \int_0^{2} { xe^x } dx }$

I found the first part being dy/dx=(x+1)e^x but the I have no idea how to do the integration from there :/

$\noindent We rearrange and integrate using the Fundamental Theorem of Calculus, from which we know that $\int _a ^b \frac{\mathrm{d}y}{\mathrm{d}x}\text{ d}x = \Big{[}y(x) \Big{]}_{x=a} ^{x=b}$, where $y(x)$ is the function of $x$, in our case $x\mathrm{e}^x$.$

$\noindent We have$

$\begin{align*}\frac{\mathrm{d}y}{\mathrm{d}x} &= \left(x+1\right) \mathrm{e}^x \quad (\text{from your working}) \\ &= x\mathrm{e}^x + \mathrm{e}^x \\ \Rightarrow x\mathrm{e}^x &= \frac{\mathrm{d}y}{\mathrm{d}x} - \mathrm{e}^x \\ \Rightarrow \int _0 ^2 x \mathrm{e}^x \text{ d}x &= \int_0 ^2 \left( \frac{\mathrm{d}y}{\mathrm{d}x} - \mathrm{e}^x \right) \text{ d}x \quad (\text{integrating both sides from 0 to 2}) \\ &= \Big{[}y(x) - \mathrm{e}^x \Big{]}_{0} ^{2} \quad (\text{using the Fundamental Theorem of Calculus}) \\ &= \Big{[}x\mathrm{e}^x - \mathrm{e}^x \Big{]}_{0} ^{2} \quad (\text{as }y(x) = x\mathrm{e}^x) \\ &= \left(2 \mathrm{e}^2 - \mathrm{e}^2 \right) - \left(0-1 \right) \\ &= \mathrm{e}^2 +1. \end{align*}$

Blitz_N7 · Dec 14, 2015

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

Thank you InteGrand!

Blitz_N7 · Dec 22, 2015

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

Question 21b. Chapter 14 set B
21. The diameters of two circular pulleys are 6cm and 12cm and their centres are 10cm apart.
(a) Calculate the angle α in radians, correct to four decimal places.
(b) Hence find, in centimetres correct to one decimal place, the length of a taut belt that goes round the two pulleys.

My working out is this
http://i.imgur.com/BNZP1GT.jpg

However at the end the answer is supposed to be 49.2cm which is 2sqrt91 + 30.1 . Is the 8.511 not supposed to be there or is the answer from the textbook incorrect?

Ambility · Dec 22, 2015

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

Blitz_N7 said:
Question 21b. Chapter 14 set B
21. The diameters of two circular pulleys are 6cm and 12cm and their centres are 10cm apart.
(a) Calculate the angle α in radians, correct to four decimal places.
(b) Hence find, in centimetres correct to one decimal place, the length of a taut belt that goes round the two pulleys.

My working out is this
http://i.imgur.com/BNZP1GT.jpg

However at the end the answer is supposed to be 49.2cm which is 2sqrt91 + 30.1 . Is the 8.511 not supposed to be there or is the answer from the textbook incorrect?

$\\$I've had a look at your working, and I believe the mistake is in working out the angle both arcs. The first arc (left circle) should be $2\pi-2(1.2661)$ instead of $2\pi-1.2661$. The second arc (right circle) should be $2\pi-2(\frac{\pi}{2}-1.2661)-\pi$ instead of $2\pi-(\frac{\pi}{2}-1.2661)-\pi$.\\Hope this helps.$

Blitz_N7 · Dec 22, 2015

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

Ambility said:
$\\$I've had a look at your working, and I believe the mistake is in working out the angle both arcs. The first arc (left circle) should be $2\pi-2(1.2661)$ instead of $2\pi-1.2661$. The second arc (right circle) should be $2\pi-2(\frac{\pi}{2}-1.2661)-\pi$ instead of $2\pi-(\frac{\pi}{2}-1.2661)-\pi$.\\Hope this helps.$

Thank you Ambility.

hedgehog_7 · Dec 24, 2015

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

(Induction) P.g. 235 Question 4 (a) (i) For even n: n^3 + 2n is divisible by 12.
(B) (ii) n^2 + 2n is a multiple of 8.

In the final step of the induction process for 4(a) i end up with 6k^2 +12k +12 +12m. Am i allowed you just factorise it like this 12(k^2/2 + 2k +2 +2m) or have i done something wrong in the process?

InteGrand · Dec 24, 2015

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

hedgehog_7 said:
(Induction) P.g. 235 Question 4 (a) (i) For even n: n^3 + 2n is divisible by 12.
(B) (ii) n^2 + 2n is a multiple of 8.

In the final step of the induction process for 4(a) i end up with 6k^2 +12k +12 +12m. Am i allowed you just factorise it like this 12(k^2/2 + 2k +2 +2m) or have i done something wrong in the process?

$\noindent You can factorise it, but you'd need to justify that the term in the brackets is an integer. The only term in the brackets that you'd need to talk about for this justification is $\frac{k^2}{2}$ (you'd need to explain why this is an integer; the other terms in the brackets are all obviously integers, and so if we can justify that $\frac{k^2}{2}$ is an integer, we know that the sum of the stuff in the brackets will be an integer, since the sum of integers is an integer). The justification for this is that $k$ is even since the question is about even integers, so $k^2$ is even (to show this, we can write $k=2N$ for some integer $N$, as $k$ is even, so $k^2 = 4N^2 = 2\left(2N^2\right)$, which is even). So as $k^2$ is even, $\frac{k^2}{2}$ is an integer.$

hedgehog_7 · Dec 24, 2015

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

1) Prove n^2 > 10n + 7 using induction, n>=11
2) Prove 2^n > 3n^2 for n>=8

InteGrand · Dec 24, 2015

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

hedgehog_7 said:
1) Prove n^2 > 10n + 7 using induction, n>=11
2) Prove 2^n > 3n^2 for n>=8

$I assume you can do the base case and conclusion for each question, so I'll just show the induction step.$

$1) Assume $n^2 > 10n + 7$ for some integer $n\geq 11$. We need to show that $(n+1)^2 > 10 (n+1)+7$.$

$We have$

$\begin{align*}(n+1)^2 &= n^2 + 2n + 1 \\ &> 10n+2n+1 \quad (\text{by the inductive hypothesis}) \\ &= 10(n+1)+7-16+2n \quad (\text{algebraic manipulation}) \\ &> 10(n+1)+7 \quad (\text{as }-16+2n>0 \text{ since }n\geq 11),\end{align*}$

$as required.$

$2) Assume $2^n > 3n^2$ for some integer $n\geq 8$. We need to show that $2^{n+1}> 3(n+1)^2$.$

$We have$

$\begin{align*}2^{n+1}&=2\times 2^n \\ &> 2\times 3 n^2 \quad (\text{by the inductive hypothesis}) \\ &= 6n^2 \\ &= 3(n+1)^2 + 3n^2 -6n -3 \\ &= 3(n+1)^2 + 3 (n^2 -2n -1) \\ &= 3(n+1)^2 + 3 \left[ (n-1)^2 -2\right] \\ &> 3(n+1)^2 \quad (\text{as }n\geq 8\text{ so } (n-1)^2 >2 \Rightarrow (n-1)^2 -2 >0), \end{align*}$

$as required.$

hedgehog_7 · Dec 24, 2015

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

Silly question which is why im stuck. Why is (n+1)^2 > 10n + 2n +1? by using the inductive hypothesis.

InteGrand · Dec 24, 2015

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

hedgehog_7 said:
Silly question which is why im stuck. Why is (n+1)^2 > 10n + 2n +1? by using the inductive hypothesis.

$Because $(n+1)^2 = n^2 + 2n+1$ and the $n^2$ here is greater than $10n +7>10n$ by the inductive hypothesis. I realise I forgot that the inductive hypothesis was $n^2 > 10n+7$ and was thinking $n^2 > 10n$. So you can rewrite the proof using $10n+7$ instead.$

Blitz_N7 · Jan 5, 2016

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

Excercise 14H, Question 19.
In part d they ask to show that the area of the rectangle is maximised when tana=1/2. However when I find the nature of the stationary point it gives me a minimum. :/

InteGrand · Jan 5, 2016

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

Blitz_N7 said:
Excercise 14H, Question 19.
In part d they ask to show that the area of the rectangle is maximised when tana=1/2. However when I find the nature of the stationary point it gives me a minimum. :/

$What were your functions you got?$

Blitz_N7 · Jan 5, 2016

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

InteGrand said:
$What were your functions you got?$

A=(2tana+1)^2/2tana
dA/da=sec^2a(4tan^2a-1)/2tan^2a
d^2a/da^2=(sec^2a(4tan^4a−tan^2a+sec^2a))/tan^3(x)

InteGrand · Jan 5, 2016

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

Blitz_N7 said:
A=(2tana+1)^2/2tana
dA/da=sec^2a(4tan^2a-1)/2tan^2a
d^2a/da^2=(sec^2a(4tan^4a−tan^2a+sec^2a))/tan^3(x)

$By the way, to simplify this, we should consider things as a function of $\tan \alpha$ instead of $\alpha$. I.e. the area is $A=f(\tau)=\frac{\left(2\tau +1 \right)^2}{2\tau}$, (assuming your functions are all correct), where $\tau \equiv \tan \alpha$, and $\tau >0$ because I suppose $\alpha$ is acute (haven't seen the question, but it should be positive in order for the area $A$ to be positive). Then just consider $\frac{\mathrm{d}A}{\mathrm{d}\tau}$, which is much easier to do.$

$Simplfying, $A=\frac{4\tau^2 + 4\tau + 1}{2\tau}=2\tau + 2 + \frac{1}{2\tau}$, which implies that $\frac{\mathrm{d}A}{\mathrm{d}\tau} = f^\prime (\tau) = 2 + \frac{1}{2}\cdot \left(-\frac{1}{\tau^2}\right)=2-\frac{1}{2} \tau ^{-2}$. So $f^\prime (\tau) = 0 \Longleftrightarrow \frac{1}{2}\cdot \frac{1}{\tau^2} = 2$, so $\frac{1}{\tau^2} = 4$, i.e. $\tau = \frac{1}{2}$ (assuming $\tau$ is positive in the question). Now, $f'' (\tau) =-\frac{1}{2}\times (-2)\tau ^{-3} = \tau ^{-3}$, which is positive for all $\tau>0$, so local minimum occurs, as you said. Maybe post the problem?$

Blitz_N7 · Jan 5, 2016

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

InteGrand · Jan 5, 2016

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

Blitz_N7 said:

It must be a typo, they meant minimised. It is clear by drawing a quick sketch that the area can be made arbitrarily large by making the line near-vertical or near-horizontal. (This shows why it is a good idea to quickly sketch the situation, to help get a feel for the question, if nothing else.)

Blitz_N7 · Jan 5, 2016

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

InteGrand said:
It must be a typo, they meant minimised. It is clear by drawing a quick sketch that the area can be made arbitrarily large by making the line near-vertical or near-horizontal. (This shows why it is a good idea to quickly sketch the situation, to help get a feel for the question, if nothing else.)

That's what I thought. Thank you InteGrand. Btw is it allowed to test nature of stationary points by using a table of values or is it always have to be done by finding second derivative?

InteGrand · Jan 5, 2016

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

Blitz_N7 said:
That's what I thought. Thank you InteGrand. Btw is it allowed to test nature of stationary points by using a table of values or is it always have to be done by finding second derivative?

$You can do it using a first derivative table of values, yes. This is a good idea when the second derivative is very messy to compute. In this question though (after realising the trick of considering the area $A$ as a function of $\tau \equiv \tan \alpha$ rather than just $\alpha$), it is easy to find the second derivative $\frac{\mathrm{d}^2 A}{\mathrm{d}\tau ^2}$.$

axwe7 · Jan 5, 2016

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

Pg 234, Q 1 (a)...

I saw the example, but I'm just soo confused.

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