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

InteGrand · Aug 13, 2015

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

appleibeats said:
Don't understand the graphing of y = log|x|

The answer have the typical log graph and then its reflection in the y axis. Why is that reflection included.

Also Logs can't be can be negative and positive, just not ZERO right?

$For $x>0$, $|x|=x$, so the graph of $y=\ln |x|$ is just that of $\ln x$ for the right-half (i.e. for $x>0$). For $x<0$, $|x|=-x$ (which is positive, since $x<0)$, so for the left-half of the graph, its the graph of $y=\ln (-x),x<0$, which is why there is a reflection about the $y$-axis, since if you input any negative $x$-value, it will give the same result as inputting the `corresponding' positive $x$ value, e.g. $\ln |-5|=\ln 5$.$

$Logs can be negative, but their argument cannot be negative (\small{assuming we are talking about real-valued functions}).$

Drongoski · Aug 13, 2015

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

appleibeats said:
Don't understand the graphing of y = log|x|

The answer have the typical log graph and then its reflection in the y axis. Why is that reflection included.

Also Logs can't be can be negative and positive, just not ZERO right?

When you take |x|, negative x values become positive.

If f(x) = log |x| then

f(-x) = log |-x| = log |x| = f(x)

That means log|x| is an even function and its graph is symmetric about the y-axis.

Crisium · Aug 13, 2015

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

appleibeats said:
Don't understand the graphing of y = log|x|

The answer have the typical log graph and then its reflection in the y axis. Why is that reflection included.

Also Logs can't be can be negative and positive, just not ZERO right?

ln(1) is zero

Something interesting to note with negative logs though, whenever you're solving an inequality such as

(x)ln(0.4) < 5

When you divide both sides by ln(0.4) remember to swap around the inequality because it is a negative number

I've rarely seen them do something like this but they often do it in 2 unit to bring out the better students

appleibeats · Aug 17, 2015

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

I understand how the log base b b^x = x but am unsure about b^log base b x = x

Also could you help with this question:

Use the identity: a = e^log a to write each expression as a power of e. THus differnetiate it.

x2^x

InteGrand · Aug 17, 2015

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

appleibeats said:
I understand how the log base b b^x = x but am unsure about b^log base b x = x

Also could you help with this question:

Use the identity: a = e^log a to write each expression as a power of e. THus differnetiate it.

x2^x

$For the first one, we know that $b^r= x \Longleftrightarrow r = \log_b x$, by our definition of logs as inverse to exponentiation.$

$For the second one, $2= e^{\ln 2}$, so $2^x =\left( e^{\ln 2}\right)^x = e^{(\ln 2)x}$, by index laws. Now just use the product rule and standard differentiation of an exponential function to differentiate the required function.$

appleibeats · Aug 17, 2015

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

I have sketched y = log ( 1 + x^2)

Now I need to sketch y = log { (1 + x^2) / 2 }

I meant to use log law. I am not sure which one. I answer show it is a shift log 2 but am not sure why.

InteGrand · Aug 17, 2015

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

appleibeats said:
I have sketched y = log ( 1 + x^2)

Now I need to sketch y = log { (1 + x^2) / 2 }

I meant to use log law. I am not sure which one. I answer show it is a shift log 2 but am not sure why.

$$\ln \left( \frac{1+x^2}{2}\right)=\ln \left(1 + x^2\right) -\ln 2$, using the log. law $\ln \left(\frac{A}{B} \right)= \ln A - \ln B$. So the curve shifts down by $\ln 2$ units.$

appleibeats · Aug 17, 2015

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

InteGrand said:
$For the first one, we know that $b^r= x \Longleftrightarrow r = \log_b x$, by our definition of logs as inverse to exponentiation.$

$For the second one, $2= e^{\ln 2}$, so $2^x =\left( e^{\ln 2}\right)^x = e^{(\ln 2)x}$, by index laws. Now just use the product rule and standard differentiation of an exponential function to differentiate the required function.$

I get the answer = 2x ( xlog2 + 1 )

But the answer in the book says = 2^x ( xlog2 + 1)

Not sure why it is to the power and not times 'x' . Is 2x meant to = e^ln2^x instead of what I used 2x = e^(ln2)x ??

InteGrand · Aug 17, 2015

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

appleibeats said:
I get the answer = 2x ( xlog2 + 1 )

But the answer in the book says = 2^x ( xlog2 + 1)

Not sure why it is to the power and not times 'x' . Is 2x meant to = e^ln2^x instead of what I used 2x = e^(ln2)x ??

$Remember, $e^{(\ln 2)x}=2^x$ (this is what we proved in the earlier post), not $2\times x$.$

appleibeats · Aug 17, 2015

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

Show that y = Ae^kx + C is a solution of dy/dx = k(y - C)

Do I just sub in y for LHS and RHS ???

InteGrand · Aug 17, 2015

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

appleibeats said:
Show that y = Ae^kx + C is a solution of dy/dx = k(y - C)

Do I just sub in y for LHS and RHS ???

$y=Ae^{kx}+C$

$\therefore \frac{\mathrm{d}y}{\mathrm{d}x}=Ake^{kx}$

$=k\left(Ae^{kx} \right)$

$=k\left(Ae^{kx}+C -C \right)$

$$=k\left(y -C \right),$ as $y=Ae^{kx}+C $.$

$Hence $y=Ae^{kx}+C$ is a solution of $\frac{\mathrm{d}y}{\mathrm{d}x}=k(y-C)$ (such an equation is called a \textit{differential equation}, an equation involving derivatives of a function. Solutions to these equations are not numbers like solutions to equations like $x^2-x=2$, but rather, \textit{functions}. If a function satisfies a given differential equation (i.e. makes the L.H.S. equal the R.H.S.), then that function is called a \textit{solution} to the differential equation).$

rand_althor · Aug 17, 2015

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

appleibeats said:
Show that y = Ae^kx + C is a solution of dy/dx = k(y - C)

Do I just sub in y for LHS and RHS ???

You differentiate y= Ae^{kx} + C to get dy/dx = k(y-C).

InteGrand · Aug 17, 2015

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

rand_althor said:
Yep, that's correct. You can go from RHS to LHS or vice versa.

$\begin{align*}\frac{\mathrm{d}y}{\mathrm{d}x}&=k(y-C) \\&=k(Ae^{x}+C-C) \\&=kAe^{kx} \\y&=\int kAe^{kx} \, \mathrm{d}x \\&=Ae^{kx} + C\end{align*}$

$Not sure what you did here, you already assumed $y=Ae^{kx}+C$ in the second line, so why are you trying to prove it again (if that's what you did)?$

rand_althor · Aug 17, 2015

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

InteGrand said:
$Not sure what you did here, you already assumed $y=Ae^{kx}+C$ in the second line, so why are you trying to prove it again (if that's what you did)?$

So I can't assume that y=Ae^{kx}+C, and then use that assumption to show that it works? My thinking was, if y is equal to that, and we sub it in to dy/dx and then integrate, and end up with y=Ae^{kx}+C again, then it is a solution.

InteGrand · Aug 17, 2015

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

rand_althor said:
So I can't assume that y=Ae^{kx}+C, and then use that assumption to show that it works? My thinking was, if y is equal to that, and we sub it in to dy/dx and then integrate, and end up with y=Ae^{kx}+C again, then it is a solution.

$Well the reasoning seems a bit circular, it looks like you assumed $y=Ae^{kx}+C$ is a solution and used this to show that $y=Ae^{kx}+C$. If you let $y=Ae^{kx}+C$, you should show that this means that $\frac{\mathrm{d}y}{\mathrm{d}x}=k(y-C)$ (in this way, you don't assume it's a solution \textemdash \text{ } rather, you show it is). Basically, your first three lines are conditioning on $y=Ae^{kx}+C$ being a solution, which wasn't proved yet. Instead, it would be better to simply move your $\frac{\mathrm{d}y}{\mathrm{d}x}$ at the start to the end, in a line under the $kAe^{kx}$ (this would be going from R.H.S. to L.H.S. (the method I showed was going from L.H.S. to R.H.S.)).$

rand_althor · Aug 17, 2015

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

InteGrand said:
$Well the reasoning seems a bit circular, it looks like you assumed $y=Ae^{kx}+C$ is a solution and used this to show that $y=Ae^{kx}+C$. If you let $y=Ae^{kx}+C$, you should show that this means that $\frac{\mathrm{d}y}{\mathrm{d}x}=k(y-C)$ (in this way, you don't assume it's a solution). Basically, your first three lines are conditioning on $y=Ae^{kx}+C$ being a solution, which wasn't proved yet. Instead, it would be better to simply move your $\frac{\mathrm{d}y}{\mathrm{d}x}$ at the start to the end, in a line under the $kAe^{kx}$ (this would be going from R.H.S. to L.H.S. (the method I showed was going from L.H.S. to R.H.S.)).$

Thanks for the clarifications!

appleibeats · Aug 25, 2015

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

What is the natural domain of y = log(log x)?

VBN2470 · Aug 25, 2015

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

appleibeats said:
What is the natural domain of y = log(log x)?

The domain would be $\left\{x\in\mathbb{R}|x>1\right\}$ i.e. the set of all positive real numbers strictly greater than 1.

appleibeats · Aug 25, 2015

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

why is that? Why is it not the usual x > 0 ??

VBN2470 · Aug 25, 2015

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

Note that $\log{1}=0$ , so taking the logarithm of this will give us an undefined output. We need to take values of $x$ such that $\log{x}>0$ which happens only when $x>1$ (since log is an increasing function). Sketch a graph of this function to verify what I have said.

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

Well-Known Member

Well-Known Member

Pew Pew

Member

Well-Known Member

Member

Well-Known Member

Member

Well-Known Member

Member

Well-Known Member

Active Member

Well-Known Member

Active Member

Well-Known Member

Active Member

Member

Well-Known Member

Member

Well-Known Member

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