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

appleibeats · Feb 3, 2016

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

I got the correct answers x = 0, pi, 2pi, pi/4, and 5pi/4 as the textbook answers says.

However shouldn't the answers pi and 5pi/4 not be included as if you sub them into cosx then cosx< 0 which is wrong as you stated above ??

InteGrand · Feb 3, 2016

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

appleibeats said:
I got the correct answers x = 0, pi, 2pi, pi/4, and 5pi/4 as the textbook answers says.

However shouldn't the answers pi and 5pi/4 not be included as if you sub them into cosx then cosx< 0 which is wrong as you stated above ??

$\noindent Yeah, we shouldn't include $\pi$ and $\frac{5\pi}{4}$, as they are not in the domain of our function.$

drsabz101 · Feb 3, 2016

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

How do I find the domain the natural domain of

InteGrand · Feb 3, 2016

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

drsabz101 said:
How do I find the domain the natural domain of View attachment 32853

$\noindent Is that $\left(\sqrt{9-x}\right)^2$?$

appleibeats · Feb 3, 2016

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

Screen Shot 2016-02-03 at 7.53.34 pm.png

Having troubles with part b.

I brought everything to the left and got

seca ( seca - 2) = 0

then made seca = 0 or seca = 2

so a = 90, 270, 60 , 300

But the answers says 60 or 300 only. Why isn't 90 or 270 included??

Shuuya · Feb 3, 2016

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

sec(a) = 0 has no solutions, so 90 and 270 are not solutions.

appleibeats · Feb 3, 2016

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

Screen Shot 2016-02-03 at 8.34.00 pm.png

I did part b) and got up to the part where

t= 0 or t = root 3

so tan x/2 = 0 or tan x/2 = root 3

so new domain is 0 < x/2 < pi

so x/2 = 0 or pi/3

and so x = 0 or 2pi/3

The answer have 0, 2pi/3 and 2pi

Not sure why 2pi is in the answers. I thought it was outside the domain??

InteGrand · Feb 3, 2016

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

appleibeats said:
View attachment 32855

I did part b) and got up to the part where

t= 0 or t = root 3

so tan x/2 = 0 or tan x/2 = root 3

so new domain is 0 < x/2 < pi

so x/2 = 0 or pi/3

and so x = 0 or 2pi/3

The answer have 0, 2pi/3 and 2pi

Not sure why 2pi is in the answers. I thought it was outside the domain??

$\noindent It's in the answers because $2\pi$ just falls inside the domain, as the domain had $0\leq x \leq 2\pi$.$

wu345 · Feb 5, 2016

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

Hey, can someone please do these two?
http://i.imgur.com/fKj2m6q.png (Ex9I)

http://i.imgur.com/Okk5fO1.png (Ex9I)

Blitz_N7 · Feb 6, 2016

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

Not sure how to go about this one :/

Paradoxica · Feb 6, 2016

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

Blitz_N7 said:
Not sure how to go about this one :/

Hahahaha... this one I usually see as an improper integral.

Divide everything by x^2

hedgehog_7 · Feb 6, 2016

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

induction question: 2^n >= 2n^3 (for integer values greater than 12)

InteGrand · Feb 7, 2016

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

hedgehog_7 said:
induction question: 2^n >= 2n^3 (for integer values greater than 12)

$\noindent The base case of $n=13$ clearly holds, as $2^{13} = 8192 > 4394 = 2 \cdot 12^3$. Now suppose the result holds for some particular integer $k\geq 13$, i.e. $2^k \geq 2k^3$. We need to show that $2^{k+1}\geq 2(k+1)^3\Longleftrightarrow 2\cdot 2^k \geq 2 \left(k^3 + 3k^2 + 3k +1\right)\Longleftrightarrow 2^k \geq k^3 + 3k^2 + 3k +1$. We have$

$\begin{align*} 2^k &\geq 2k^3 \quad (\text{by the inductive hypothesis}) \\ &= k^3 + k^3 \\ &\geq k^3 + 3k^2 + 3k+1,\end{align*}$

$\noindent as $k^3 \geq 3k^2 + 3k+1$ for all $k\geq 13$, as required. The result is thus true by the Principle of Mathematical Induction$

$\noindent \textbf{Note.} Here is a proof that $k^3 \geq 3k^2 + 3k+1$ for all $k\geq 13$.$

$\noindent Let $k\geq 13$, then$

$\begin{align*}k^3 &\geq 13k^2 \\ &= 3k^2 + 10k^2 \\ &\geq 3k^2 + 4k \quad (\text{since clearly }10k^2 \geq 4k \text{ whenever }k\geq 13) \\ &\geq 3k^2 + 3k+1 \quad (\text{as clearly }4k \geq 3k+ 1 \text{ whenever }k\geq 13).\end{align*}$

wu345 · Feb 7, 2016

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

why are my questions getting ignored ;_;

InteGrand · Feb 7, 2016

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

wu345 said:
why are my questions getting ignored ;_;

$\noindent I think because they're both just tedious algebra. For the first one, let $P$ be like $\left(2p,p^2\right)$, where $p\in \mathbb{R}$ and $p\neq 0$ ($p$ can't be 0, because then $P$ is the origin, and so the normal at $P$ is the $y$-axis, which doesn't meet the parabola again, so the question would become meaningless).$

$\noindent Then obtain the equation of the normal at $P$. You can do this using point-slope form and the fact that the slope of the \textit{tangent} at $p$ is $p$, so the normal has slope $-\frac{1}{p}$. Then, solve the normal equation simultaneously with the parabola to find the coordinates of $Q$. Once you have the coordinates, to show that $\angle PSQ =90^\circ$, all you need to do is calculate the slopes of $PS$ and $QS$ and show their product is $-1$, so that $PS \perp QS$. (Remember that $S = (0,1)$.)$

$\noindent For part (b), obtain the equations of the tangents at $P$ and $Q$ by tedious algebra, or the fact that the tangent at $(x_1,y_1)$ on that parabola has equation $xx_1 = 2(y+y_1)$ (there's heaps of useful formulas like this in that textbook in the chapter you're doing). Then solve these tangent equations simultaneously to find the coordinates of $T$. Then calculate $PQ$ and $PT$ using the distance formula and show they are equal.$

$\noindent Your second question can be done using similarly tedious methods.$

InteGrand · Feb 7, 2016

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

$\noindent Another way to do your part (b) of first Question (proving that $PQ = PT$) is to just show that $\theta = 45^\circ$, where $\theta = \angle PQT$. This can be done by using the angle between two lines formula: $\tan \theta = \left| \frac{m_{PQ} - m_{QT}}{1+ m_{PQ} m_{QT}}\right |$. (Note that $\theta$ is acute, as it is an angle in a right-angled triangle. Hence this formula actually is the tan of $\theta$. If $\theta = \angle PQT$ were \textit{obtuse} instead, then the R.H.S. of this formula would have given the negative of the tan of $\theta$.)$

$\noindent So you calculate the slopes required and plug them in to that `angle between two lines' formula. This should come out to be $1$, which means that $\tan \theta =1\Rightarrow \theta = 45^\circ$. This means that $\triangle PQT$ is an isosceles right-angled triangle, with $PQ = PT$.$

scienceking · Feb 7, 2016

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

how to do 4I question 13b? i keep getting the wrong answer

InteGrand · Feb 7, 2016

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

scienceking said:
how to do 4I question 13b? i keep getting the wrong answer

What is the question?

wu345 · Feb 7, 2016

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

InteGrand said:
$\noindent I think because they're both just tedious algebra. For the first one, let $P$ be like $\left(2p,p^2\right)$, where $p\in \mathbb{R}$ and $p\neq 0$ ($p$ can't be 0, because then $P$ is the origin, and so the normal at $P$ is the $y$-axis, which doesn't meet the parabola again, so the question would become meaningless).$

$\noindent Then obtain the equation of the normal at $P$. You can do this using point-slope form and the fact that the slope of the \textit{tangent} at $p$ is $p$, so the normal has slope $-\frac{1}{p}$. Then, solve the normal equation simultaneously with the parabola to find the coordinates of $Q$. Once you have the coordinates, to show that $\angle PSQ =90^\circ$, all you need to do is calculate the slopes of $PS$ and $QS$ and show their product is $-1$, so that $PS \perp QS$. (Remember that $S = (0,1)$.)$

$\noindent For part (b), obtain the equations of the tangents at $P$ and $Q$ by tedious algebra, or the fact that the tangent at $(x_1,y_1)$ on that parabola has equation $xx_1 = 2(y+y_1)$ (there's heaps of useful formulas like this in that textbook in the chapter you're doing). Then solve these tangent equations simultaneously to find the coordinates of $T$. Then calculate $PQ$ and $PT$ using the distance formula and show they are equal.$

$\noindent Your second question can be done using similarly tedious methods.$

I have tried the method you suggested and all other methods I can think of at least twice and I still can't get it - could you please show your working if you have time? I get Q (-2(p+(2/p)), (p+(2/p))^2), is that right?

InteGrand · Feb 7, 2016

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

wu345 said:
I have tried the method you suggested and all other methods I can think of at least twice and I still can't get it - could you please show your working if you have time? I get Q (-2(p+(2/p)), (p+(2/p))^2), is that right?

Are you talking about for the first question?

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

Member

Well-Known Member

Member

Well-Known Member

Member

Well-Known Member

Member

Well-Known Member

Member

Member

-insert title here-

Member

Well-Known Member

Member

Well-Known Member

Well-Known Member

New Member

Well-Known Member

Member

Well-Known Member

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