Cambridge Prelim MX1 Textbook Marathon/Q&A (2 Viewers)

InteGrand · Jan 6, 2016

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

axwe7 said:
Pg 234, Q 1 (a)...

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

What's the question?

axwe7 · Jan 6, 2016

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

Sorry for the late response,

The question is;

Use mathematical induction to prove that for all positive integers 'n':
(a) 1^2 + 2^2 + 3^2 + ... ... .. ... + n^2 = 1/6*n(n+1)(2n+1)

Cheers.

InteGrand · Jan 6, 2016

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

axwe7 said:
Sorry for the late response,

The question is;

Use mathematical induction to prove that for all positive integers 'n':
(a) 1^2 + 2^2 + 3^2 + ... ... .. ... + n^2 = 1/6*n(n+1)(2n+1)

Cheers.

$This is a pretty typical one. For the induction step, assume that $1^2 + 2^2 + 3^2 + \cdots + n^2 = \frac{1}{6}n(n+1)(2n+1)$, and use this to show that $1^2 + 2^2 + 3^2 + \cdots + n^2 + (n+1)^2 = \frac{1}{6}(n+1)((n+1)+1)(2(n+1)+1)$, by noting that $1^2 + 2^2 + 3^2 + \cdots + n^2 + (n+1)^2 =\frac{1}{6}n(n+1)(2n+1) + (n+1)^2 $ (by inductive hypothesis), and then using factorisation and simplification etc. to get it into $\frac{1}{6}(n+1)((n+1)+1)(2(n+1)+1)=\frac{1}{6}(n+1)(n+2)(2n+3)$.$

appleibeats · Jan 7, 2016

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

The variable complex number z satisfies |z - z0 | = r . Find the max and min values of: |z|

I got the answer:

</= represents less than or equal to

|z0| - r </= |z| </= |z0| + r

But the answer has something different.

Instead of my |z0| - r

they have

| |z0| - r |

Can someone explain why there is an extra modulus sign around my entire answer that I got.

InteGrand · Jan 7, 2016

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

appleibeats said:
The variable complex number z satisfies |z - z0 | = r . Find the max and min values of: |z|

I got the answer:

</= represents less than or equal to

|z0| - r </= |z| </= |z0| + r

But the answer has something different.

Instead of my |z0| - r

they have

| |z0| - r |

Can someone explain why there is an extra modulus sign around my entire answer that I got.

You need the absolute value sign to make sure answer is non-negative, since |z| is non-negative. (Draw a diagram.)

appleibeats · Jan 7, 2016

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

find the locus of z if the value of z - 1 / z - i is a) real, b) imaginary

for a) I got x + y = 1
b) I got (x -1/2)^2 + (y - 1/2)^2 = 1/2

The solutions are

a) The line through 1 and i, omitting i

b) The circle with diameter joining 1 and i, omitting these two points.

Could someone explain the answer please.

appleibeats · Jan 7, 2016

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

find the locus of z if the value of z - 1 / z - i is a) real, b) imaginary

for a) I got x + y = 1
b) I got (x -1/2)^2 + (y - 1/2)^2 = 1/2

The solutions are

a) The line through 1 and i, omitting i

b) The circle with diameter joining 1 and i, omitting these two points.

Could someone explain the answer please.

InteGrand · Jan 7, 2016

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

appleibeats said:
find the locus of z if the value of z - 1 / z - i is a) real, b) imaginary

for a) I got x + y = 1
b) I got (x -1/2)^2 + (y - 1/2)^2 = 1/2

The solutions are

a) The line through 1 and i, omitting i

b) The circle with diameter joining 1 and i, omitting these two points.

Could someone explain the answer please.

I haven't bothered doing the question, but notice that your answers match the answers given apart from some omissions of points in the answers. E.g. your x + y = 1 line is precisely the line through 1 (the point (1,0)) and i (the point (0,1)). To see why some points should be omitted, try subbing them in to the expression and see what happens.

InteGrand · Jan 7, 2016

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

appleibeats said:
find the locus of z if the value of z - 1 / z - i is a) real, b) imaginary

for a) I got x + y = 1
b) I got (x -1/2)^2 + (y - 1/2)^2 = 1/2

The solutions are

a) The line through 1 and i, omitting i

b) The circle with diameter joining 1 and i, omitting these two points.

Could someone explain the answer please.

Not sure why they omitted those points for b) though, since they do give imaginary results if you sub. them in to the expression.

appleibeats · Jan 7, 2016

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

THe function f(x) is defined by f(x) = x - 1/x , for x > 0

a) by considering the graphs of y = x and y = 1/x for x > 0, sketch y = f(x)

InteGrand · Jan 7, 2016

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

appleibeats said:
THe function f(x) is defined by f(x) = x - 1/x , for x > 0

a) by considering the graphs of y = x and y = 1/x for x > 0, sketch y = f(x)

$\noindent We sketch the curve $y=f(x) = x - \frac{1}{x},x>0$ using subtraction of ordinates (i.e. subtract $y$-values of the second graph they told us to consider from the first one). Consider the graphs of $y=x$ and $y=\frac{1}{x}$ for $x>0$. Clearly they intersect at and only at $x=1$, so their difference here is 0. So the one and only $x$-intercept of $y=f(x)$ is at $x=1$. To left of this (i.e. where $0<x<1$), the graph of $y=\frac{1}{x}$ is always above that of $y=x$, so the graph of $y=f(x)$ is below the $x$-axis. As $x\to 0^+$, $f(x) \to -\infty$. For $x>1$, the graph of $y=x$ is above that of $y=\frac{1}{x}$, so the graph of $y=f(x)$ is above the $x$-axis. Also, as $x\to \infty$, $f(x) \to \infty$. Moreover, for large $x$, the value of $\frac{1}{x}$ is negligibly small, so the graph of $y=f(x)$ will look like that of $y=x$ and has $y=x$ as an oblique asymptote. Using this information and subtraction of ordinates, we should obtain a sketch like this:$

http://www.graphsketch.com/?eqn1_co..._lines=1&line_width=4&image_w=850&image_h=525 .

$\noindent Here is a link that includes the asymptote in red:$

http://www.graphsketch.com/?eqn1_co..._lines=1&line_width=4&image_w=850&image_h=525 .

appleibeats · Jan 8, 2016

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

I now need to sketch the inverse function:

I plugged it into wolfram and this is what it shows:

http://www.wolframalpha.com/input/?i=x+=+y+-+1/y+for+y+>+0+

I get how when you sub large y values it tends to negative infinity and so approaches as well the asymptote.

But shouldn't the inverse function pass through the point ( 0, 1)

Why does this not occur?

Also why does it as y approaches infinity approach the y axis and not y = x

If the inverse is a reflection across y = x, shouldn't the inverse approach y = x ??

Thanks

InteGrand · Jan 8, 2016

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

appleibeats said:
I now need to sketch the inverse function:

I plugged it into wolfram and this is what it shows:

http://www.wolframalpha.com/input/?i=x+=+y+-+1/y+for+y+>+0+

I get how when you sub large y values it tends to negative infinity and so approaches as well the asymptote.

But shouldn't the inverse function pass through the point ( 0, 1)

Why does this not occur?

Also why does it as y approaches infinity approach the y axis and not y = x

If the inverse is a reflection across y = x, shouldn't the inverse approach y = x ??

Thanks

It actually does do all these things. Observe that in the WolframAlpha plot, the "x"-axis is actually the vertical axis. So they simply plotted the same graph but switched the axes around. So as you can see, the horizontal intercept on the graph is 1. This corresponds to the point where x = 0 and y = 1 (since the ''y''-axis here is actually horizontal). So it does go through the point (0, 1), since by (0, 1), we mean x = 0, y = 1.

InteGrand · Jan 8, 2016

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

appleibeats said:
I get how when you sub large y values it tends to negative infinity

$Actually, this is when we sub. in \emph{small} $y$-values, so that $\frac{1}{y}$ is large.$

appleibeats · Jan 8, 2016

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

Now by conpleting the square or using the quadratic formula, show that

f^-1 x = 1/2 ( x + ( 4 + x^2)^1/2 )

I used the quadratic formula and got the result

But instead of x + (4 + ....

I got x +/ - ( 4 + ....

Why does the answer disregard the negative option??

InteGrand · Jan 8, 2016

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

appleibeats said:
Now by conpleting the square or using the quadratic formula, show that

f^-1 x = 1/2 ( x + ( 4 + x^2)^1/2 )

I used the quadratic formula and got the result

But instead of x + (4 + ....

I got x +/ - ( 4 + ....

Why does the answer disregard the negative option??

$This is because for our original function $f(x)$, it's the function $x-\frac{1}{x}$ \textsl{restricted to }$x>0$. In other words, we only took one branch of the curve $y=x-\frac{1}{x}$, namely the branch to the right of the $y$-axis. So for the inverse, we need to take the corresponding branch when we flip it about the line $y=x$, and it turns out this branch is obtained by taking the positive square root.$

$The reason why we take the positive root is that, since the domain for the function $f$ had $x>0$, when finding the function $y=f^{-1}(x)$, we write $x=y-\frac{1}{y}$, and $y>0$ (since the roles of $y$ and $x$ are reversed, and before $x$ was positive, so this time, $y$ is positive). So since $y$ has to be positive, we just take the positive square root (if we took the negative square root, $y$ would take on negative values, so this is the wrong branch to take). In other words, when we flip our desired branch of $y=x-\frac{1}{x}$ about the line $y=x$, it'll land entirely in places where $y>0$. So we need to have $y>0$.$

appleibeats · Jan 8, 2016

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

Screen Shot 2016-01-08 at 1.16.20 pm.png

Can you please explain part e) .

Paradoxica · Jan 8, 2016

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

$$\noindent The minimum of the quadratic parabola occurs at $x=2$. The inverse of the quadratic is required to be positive, and by of the graph of the quadratic, the inverse is always positive if $x \ge 2$. Solving for the inverse, we have $y = 2+\sqrt{6x-20}$. $N$ is a negative number. $f^{-1}(f(N))=2+\sqrt{6\left(\frac{1}{6}\left((N-2)^2+20\right)\right)-20}=2+\sqrt{(N-2)^2+20-20}=2+|N-2|=2+2-N=4-N$

appleibeats · Jan 8, 2016

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

You lost me at solving for the inverse.......

Paradoxica · Jan 8, 2016

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

appleibeats said:
You lost me at solving for the inverse.......

Have you not learned how to invert a function?

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