Cambridge Prelim MX1 Textbook Marathon/Q&A (4 Viewers)

appleibeats · Dec 1, 2015

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

I know what the graph of y = lnx is, but can someone help me through the transformation of this to f(x) = 1 - lnx

isn't the negative sign in front of the log mean that it is reflected in the y axis. The answer says its reflected in the x axis. Why is this??

InteGrand · Dec 1, 2015

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

appleibeats said:
I know what the graph of y = lnx is, but can someone help me through the transformation of this to f(x) = 1 - lnx

isn't the negative sign in front of the log mean that it is reflected in the y axis. The answer says its reflected in the x axis. Why is this??

$In general, if we have the graph of $y=f(x)$, then the graph of $y=-f(x)$ is the graph of $y=f(x)$ reflected about the $x$-axis. This is explained as follows. Consider any point $P\left(a,b\right)$ that lies on the graph of $y=f(x)$, so that means $b=f(a)$. Now at $x=a$ on the graph of $y=-f(x)$, we have $y=-f(a)=-b$. So the point $A$ on the graph of $y=f(x)$ has become $(a,-b)$ on the graph of $y=-f(x)$, i.e. \textsl{flipped about the }$x$-\textsl{axis} (because the $x$-value is the same ($a$), but the $y$-value has been negated (to $-b$)). Since this happens for any point $P$ on the graph of $y=f(x)$, it means that the graph of $y=-f(x)$ is just obtained by flipping every point (i.e. the entire graph) of $y=f(x)$ about the $x$-axis.$

$The transformation that would flip the graph of $y=f(x)$ about the $y$-axis would be sketching $y=f(-x)$. See if you can see why this is, using a similar approach to above.$

appleibeats · Dec 1, 2015

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

Find all the roots to the equation z^3 + 1 = 0 in modulus -argument form , giving the principle argument.

InteGrand · Dec 1, 2015

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

appleibeats said:
Find all the roots to the equation z^3 + 1 = 0 in modulus -argument form , giving the principle argument.

$$We have$

$\begin{align*} z^3 + 1 &= 0 \\ \Rightarrow z^3 &= -1 \\ \Rightarrow z^3 &= \mathrm{cis}\left(\left(2k+1\right)\pi \right), \quad k\in \mathbb{Z} \\ \Rightarrow z &= \mathrm{cis}\left(\left(2k+1\right)\frac{\pi}{3} \right). \end{align*}$

$Since we had a third-degree polynomial, we'll have 3 roots. We can obtain them by subbing in $k=-1,0,1$. We could also do $k=0,1,2$ or other things, but we'd need to convert the argument to principal argument then, whereas if we do $k=-1,0,1$, they'll be the principal argument already.$

$k=-1:$

$$z = \mathrm{cis}\left(-\frac{\pi}{3} \right)$.$

$k=0:$

$$z = \mathrm{cis}\left(\frac{\pi}{3} \right)$.$

$k=1:$

$$z = \mathrm{cis}\left(\frac{3\pi}{3} \right)=\mathrm{cis} \left(\pi\right)\quad (=1)$.$

$So the roots are $z= \mathrm{cis}\left(-\frac{\pi}{3} \right), \mathrm{cis}\left(\frac{\pi}{3} \right), \mathrm{cis}\left(\pi \right)$.$

leehuan · Dec 1, 2015

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

InteGrand said:
$$We have$

$\begin{align*} z^3 + 1 &= 0 \\ \Rightarrow z^3 &= -1 \\ \Rightarrow z^3 &= \mathrm{cis}\left(\left(2k+1\right)\pi \right), \quad k\in \mathbb{Z} \\ \Rightarrow z &= \mathrm{cis}\left(\left(2k+1\right)\frac{\pi}{3} \right). \end{align*}$

$Since we had a third-degree polynomial, we'll have 3 roots. We can obtain them by subbing in $k=-1,0,1$. We could also do $k=0,1,2$ or other things, but we'd need to convert the argument to principal argument then, whereas if we do $k=-1,0,1$, they'll be the principal argument already.$

$k=-1:$

$$z = \mathrm{cis}\left(-\frac{\pi}{3} \right)$.$

$k=0:$

$$z = \mathrm{cis}\left(\frac{\pi}{3} \right)$.$

$k=1:$

$$z = \mathrm{cis}\left(\frac{3\pi}{3} \right)=\mathrm{cis} \left(\pi\right)\quad (=1)$.$

$So the roots are $z= \mathrm{cis}\left(-\frac{\pi}{3} \right), \mathrm{cis}\left(\frac{\pi}{3} \right), \mathrm{cis}\left(\pi \right)$.$

Why (2k+1) and not 2k

Edit, nvm same idea

2kpi+pi was what I was thinking

appleibeats · Dec 5, 2015

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

How do you factorise

y = x^3 - 6x^2 + 5x + 12

InteGrand · Dec 5, 2015

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

appleibeats said:
How do you factorise

y = x^3 - 6x^2 + 5x + 12

$Observe that $-1,3$ and $4$ are roots, so $y = (x+1)(x-3)(x-4)$.$

appleibeats · Dec 5, 2015

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

But if you needed to show working, how would you go about doing so. I have tried grouping but that doesn't work .

InteGrand · Dec 5, 2015

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

appleibeats said:
But if you needed to show working, how would you go about doing so. I have tried grouping but that doesn't work .

$Just do what I did, except actually show they are roots by substituting in those values and saying they give $0$. E.g. to show that $3$ was a root, you could just do this (calling the polynomial $P$):$

$$P(3) = 3^3 -6\times 3^2 + 5\times 3 +12 =0\Rightarrow 3$ is a root.$

appleibeats · Dec 5, 2015

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

Take logs of both sides and differentiate y = e^1/x

I get up to x = 1/ (lny)

but am unsure on how to proceed.

kawaiipotato · Dec 5, 2015

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

appleibeats said:
Take logs of both sides and differentiate y = e^1/x

I get up to x = 1/ (lny)

but am unsure on how to proceed.

dx/dy = -1/(y(lny)^2)
so dy/dx = -y(lny)^2
= -e^(1/x)(1/x)^2

appleibeats · Dec 5, 2015

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

kawaiipotato said:
dx/dy = -1/(y(lny)^2)
so dy/dx = -y(lny)^2
= -e^(1/x)(1/x)^2

How do you get your first line of working??

are you using a log law ??

kawaiipotato · Dec 5, 2015

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

appleibeats said:
How do you get your first line of working??

are you using a log law ??

I'm using the rule
$\left(\frac{1}{v}\right)' = - \frac{v'}{v^2}$

$\left(\frac{1}{\ln y}\right)' = - \frac{ ( \ln y )' }{ (\ln y)^2 } = - \frac{\frac{1}{y}}{(\ln y)^2} = - \frac{1}{y( \ln y)^2}$

appleibeats · Dec 5, 2015

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

I don't remember this log law. Is it a standard one that I should know?? Or is it from somewhere else??

leehuan · Dec 5, 2015

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

appleibeats said:
I don't remember this log law. Is it a standard one that I should know?? Or is it from somewhere else??

It's using the function of a function rule to differentiate something like say y=1/(x^2+1)

*of which that example -> y=(x^2+1)^-1
dy/dx = -2x (x^2+1)^-2
= 2x/(x^2+1)^2

No logarithm laws here.

kawaiipotato · Dec 5, 2015

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

appleibeats said:
I don't remember this log law. Is it a standard one that I should know?? Or is it from somewhere else??

It's not a logarithm law. It's a variant of the quotient rule used in differentiation.
Quotient rule:
$\left(\frac{u}{v}\right)' = \frac{u' v - v' u }{v^2}$

$$ In this case, u = 1 and so u' = 0$

$Which will give you the rule mentioned in the previous post$

$\left(\frac{1}{v}\right)' = - \frac{v' }{v^2}$

leehuan · Dec 5, 2015

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

kawaiipotato said:
It's not a logarithm law. It's a variant of the quotient rule used in differentiation.
Quotient rule:
$\left(\frac{u}{v}\right)' = \frac{u' v - v' u }{v^2}$

$$ In this case, u = 1 and so u' = 0$

$Which will give you the rule mentioned in the previous post$

$\left(\frac{1}{v}\right)' = - \frac{v' }{v^2}$

The prime notation when used in such a manner is easily confusing I've found. Use d/dx instead and then it becomes clearer.

appleibeats · Dec 5, 2015

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

Is it not just the chain rule??

leehuan · Dec 5, 2015

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

appleibeats said:
Is it not just the chain rule??

I said it was, when I said function of a function.

(The chain rule is actually the statement dy/dx=dy/du du/dx)

appleibeats · Dec 5, 2015

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

Find d/dx ( x^2logx) ... = 2xlnx + x

Hence evaluate Integral from 1 to e xlnx dx

I get the answer : (e^2 + 1)/4

Could someone see if they get the same result.

The answer says its (e^2 - e + 1)/ 2

Still can't figure where I have gone wrong.

Cambridge Prelim MX1 Textbook Marathon/Q&A (4 Viewers)

Member

Well-Known Member

Member

Well-Known Member

Well-Known Member

Member

Well-Known Member

Member

Well-Known Member

Member

Well-Known Member

Member

Well-Known Member

Member

Well-Known Member

Well-Known Member

Well-Known Member

Member

Well-Known Member

Member

Users Who Are Viewing This Thread (Users: 0, Guests: 4)