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

InteGrand · Nov 17, 2015

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

appleibeats said:
how would you check for restrictions??

$Depends on the example. For example, if the parameterisation of the point $P$ was $x(t)=t^2$ and $y(t)=2$, $t\in \mathbb{R}$, then since the range of the function $x(t) = t^2$ is the set of non-negative real numbers, the locus of $P$ will be the portion of the line $y=2$ where $x\geq 0$ (in other words, the right-half of the line, including the point $(0,2)$. In fact, $P$ would trace out this portion of the line twice as $t$ varies through the reals.$

appleibeats · Nov 17, 2015

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

P (2ap, ap^2) and Q (2aq,aq^2) are points on x^2 = 4ay , and he chord PQ subtends a right angle at the vertex O.

Show that pq = -4

It's probably really simple but I can't seem to work it out!

InteGrand · Nov 17, 2015

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

appleibeats said:
P (2ap, ap^2) and Q (2aq,aq^2) are points on x^2 = 4ay , and he chord PQ subtends a right angle at the vertex O.

Show that pq = -4

It's probably really simple but I can't seem to work it out!

$Note that the slope of $OP$ is $m_{OP}=\frac{ap^2}{2ap}=\frac{p}{2}$. Similarly, the slope of $OQ$ is $m_{OQ}=\frac{q}{2}$. Since $\angle POQ$ is a right angle, we have $OP \perp OQ$, so the two lines' slopes must multiply to give $-1$. So $m_{OP}\times m_{OQ}=-1\Rightarrow \frac{p}{2}\times \frac{q}{2} = -1 \Rightarrow pq = -4$ (rearranging). Q.E.D.$

appleibeats · Nov 17, 2015

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

Two points P and Q where p > q, move along the parabola x^2 = 4ay . At all times the x cordinates of P and Q differ by 2a.

Find the midpoint of the chord PQ.

M ( a(p + q), 1/2 a(p^2 + q^2))

Now how do i find the cartesian equation of its locus.

I have put

x = a ( p + q)

y = 1/2 a (p^2 + q^2)

But am stumped on where to go.

MilkyCat_ · Nov 17, 2015

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

Square the x value

kawaiipotato · Nov 18, 2015

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

appleibeats said:
Two points P and Q where p > q, move along the parabola x^2 = 4ay . At all times the x cordinates of P and Q differ by 2a.

Find the midpoint of the chord PQ.

M ( a(p + q), 1/2 a(p^2 + q^2))

Now how do i find the cartesian equation of its locus.

I have put

x = a ( p + q)

y = 1/2 a (p^2 + q^2)

But am stumped on where to go.

$$ Use the given condition that the x-values always differ by 2a. I.e. $ 2ap - 2aq = 2a \Rightarrow p - q = 1$

$$ Squaring the x-coordinate, $ \frac{x^2}{a^2} = (p+q)^2 = p^2 + q^2 + 2pq = (p-q)^2 + 4pq = 1 + 4pq \boxed{1}$

$\frac{2y}{a} = p^2 + q^2 = (p-q)^2 + 2pq = 1 + 2pq \boxed{2}$

$$Multiply$ \boxed{2} $by$ 2 $and subtract the new equation$ $from ${1}$

$\frac{x^2}{a^2} - \frac{4y}{a} = -1 . $is independent of p,q and is the locus of M$$

appleibeats · Nov 25, 2015

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

determine the arcs specified by the following eqn. Sketch each one, showing the centre and radius of the associated circle.

arg ( z - 1 + i) / (z - 1 - i ) = pi/2

So i know it a circle. there are endpoints at (1, -1 ) and ( 1, 1) and that there are not included , hollow dot.

The angle between these two vectors when they meet is pi /2 .

The answers say it is a right hand side semicircle. Why is it a semicircle and not a full circle??

InteGrand · Nov 25, 2015

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

appleibeats said:
determine the arcs specified by the following eqn. Sketch each one, showing the centre and radius of the associated circle.

arg ( z - 1 + i) / (z - 1 - i ) = pi/2

So i know it a circle. there are endpoints at (1, -1 ) and ( 1, 1) and that there are not included , hollow dot.

The angle between these two vectors when they meet is pi /2 .

The answers say it is a right hand side semicircle. Why is it a semicircle and not a full circle??

$The rule for $\arg \left(\frac{z -z_1}{z -z_2}\right)=\alpha$ for $\alpha \in \left(0,\pi\right)$ is to start at the point $z_1$ and go counter-clockwise in a circular arc to the point $z_2$, with an angle of $\alpha$ being subtended in this arc (with open circles at $z_1$ and $z_2$, that is, these points are excluded from the locus). If you have the \textsl{other} arc, this corresponds to the locus $\arg \left(\frac{z -z_2}{z -z_1}\right)=\alpha$.$

appleibeats · Nov 28, 2015

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

how do you sketch y = -1/2 ( x^2 + 1) ??

InteGrand · Nov 28, 2015

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

appleibeats said:
how do you sketch y = -1/2 ( x^2 + 1) ??

$I assume you mean $y = -\frac{1}{2\left( x^2 + 1\right)}$? This is an upside down bell-shaped sort of curve. If you meant $y = -\frac{1}{2} \left( x^2 + 1\right)$ (which is what what you wrote actually means with the way you have the brackets there), this is an easy-to-sketch parabola.$

InteGrand · Nov 28, 2015

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

$Here is a sketch of $y = -\frac{1}{2\left( x^2 + 1\right)}$, via WolframAlpha:$

http://wolframalpha.com/input/?i=plot+y+=+-1/(2(x^2+++1))&x=0&y=0

appleibeats · Nov 28, 2015

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

How do you shade the region defined by the inequality | (z + 1)/ (z - i ) | greater than or equal to 1

Am i right in saying:

1) let z = x + iy
2) rationalise the denominator
3) then get it in the form of x + iy
4) Use sqrt x^2 + y^2
5)get a circle

Is this correct. Seems to long of a method for this question. Is there a faster and shorter way I don't know of??

leehuan · Nov 28, 2015

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

appleibeats said:
How do you shade the region defined by the inequality | (z + 1)/ (z - i ) | greater than or equal to 1

Am i right in saying:

1) let z = x + iy
2) rationalise the denominator
3) then get it in the form of x + iy
4) Use sqrt x^2 + y^2
5)get a circle

Is this correct. Seems to long of a method for this question. Is there a faster and shorter way I don't know of??

Since |a/b|=|a|/|b| (in words, the modulus of the quotient is the quotient of the moduli)

And we also know |z| is positive

We can multiply through to get

|z+1|≥|z-1|

That avoids the need to realise (not rationalise) the denominator

InteGrand · Nov 28, 2015

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

appleibeats said:
How do you shade the region defined by the inequality | (z + 1)/ (z - i ) | greater than or equal to 1

Am i right in saying:

1) let z = x + iy
2) rationalise the denominator
3) then get it in the form of x + iy
4) Use sqrt x^2 + y^2
5)get a circle

Is this correct. Seems to long of a method for this question. Is there a faster and shorter way I don't know of??

$As leehuan has said, don't bother doing it by letting $z=x+iy$, as this is an unnecessarily tedious method. Using leehuan's suggested method (which is the best way to go about this one), we have$

$\begin{align*} \left|\frac{z+1}{z-i} \right| &\geq 1 \quad (z \neq i) \\ \Longleftrightarrow \left|z+1\right| &\geq |z-i|, z\neq i \quad (\text{multiplying through by } \left| z-i \right|).\end{align*}$

$This is a region that you are expected to be able to sketch easily. The way to do it is to notice that the locus of points with $|z+1|= \left| z-(-1)\right|$ being \textsl{equal} to $|z-i|$ is the perpendicular bisector of the line interval joining the point representing $-1$ (i.e. $(-1,0)$) to the point representing $i$ (i.e. the point $(0,1)$). So the region where $|z+1|= \left| z-(-1)\right|$ is \textsl{greater than or equal to} $|z-i|$ is all the points on the perpendicular bisector and those points on the side of the perpendicular bisector that are closer to $i$ than $-1$. So for our region, we should shade in the perpendicular bisector and the half-plane that is on the same side of this bisector as $i$ (as these are all the points closer to $i$ than to $-1$). However, we need to exclude the point $i$ (so draw an open circle here), because we if we had $z=i$ be part of the region, we would have 0 on the denominator of the given inequality, which we cannot have.$

appleibeats · Nov 28, 2015

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

I have sketched the locus of arg( z - i ) = pi/3

Now how do I find the cartesian equation of the locus, stating any restrictions on the domain.

Unsure how to get the equation. Do i sub in x + iy for z ??

InteGrand · Nov 28, 2015

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

appleibeats said:
I have sketched the locus of arg( z - i ) = pi/3

Now how do I find the cartesian equation of the locus, stating any restrictions on the domain.

Unsure how to get the equation. Do i sub in x + iy for z ??

$The locus is a ray starting at $i$ (excluding the point $i$) and making an angle of $\frac{\pi}{3}$ with the positive $x$-axis. The equation of the entire \textsl{line} can be written as $y=mx+b$. Clearly $b=1$, since the $y$-intercept is $1$, as the line goes through $i$. Also, $m=\tan \theta = \tan \frac{\pi}{3} = \sqrt{3}$. So the line's equation is $y=\sqrt{3}x+1$. But for this ray, we include and only include the parts where $x>0$. So the equation of the locus is $y=\sqrt{3}x+1,x>0$.$

appleibeats · Nov 28, 2015

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

For Sketch the locus represented by the equation |z^2 - z -2 | = | z + 1|

Do i just use z = x + iy and simplify

I am half way through this method and its a huge mess of expansion.

InteGrand · Nov 28, 2015

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

appleibeats said:
For Sketch the locus represented by the equation |z^2 - z -2 | = | z + 1|

Do i just use z = x + iy and simplify

I am half way through this method and its a huge mess of expansion.

$Note that $z^2 - z -2=\left(z-2 \right)\left(z+1\right)$, so the given equation is $\left| \left(z-2 \right)\left(z+1\right) \right| = |z+1| \Longleftrightarrow |z-2||z+1| = |z+1|$. So if $z\neq -1$, we can divide through by $|z+1|\neq 0$ to obtain $|z-2|=1$, which is a circle centred at 2 (i.e. the point $(2,0)$) with radius 1. We also need to include the single point $z=-1$ in our locus to account for the case of $z=-1$, as this also satisfies the locus equation. So the locus is the aforementioned circle along with the point $(-1,0)$.$

kyyzhng · Nov 29, 2015

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

I'm not the best at maths but could i get help on Chapter 10: The Geometry of the Derivative 10A: Increasing, Decreasing and Stationary at a Point (Page 359, Exercise 10A, Question 2)

2.a) Show that y = -5x + 2 is decreasing for all x
b) Show that y = x + 7 is increasing for all x

There arent any answers to ths question in the textbook. Any help would be much appreciated ^^

InteGrand · Nov 29, 2015

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

kyyzhng said:
I'm not the best at maths but could i get help on Chapter 10: The Geometry of the Derivative 10A: Increasing, Decreasing and Stationary at a Point (Page 359, Exercise 10A, Question 2)

2.a) Show that y = -5x + 2 is decreasing for all x
b) Show that y = x + 7 is increasing for all x

There arent any answers to ths question in the textbook. Any help would be much appreciated ^^

$(a) $y = -5x +2 \Rightarrow y^\prime = -5<0$, so the derivative is always negative, so the function is decreasing for all $x$.$

$(b) $y = x+7 \Rightarrow y^\prime = 1$, so the derivative is always positive (being equal to $1>0$ always), so the function is increasing for all $x$.$

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