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

Blitz_N7 · Jan 27, 2016

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

I used 20% of 50 m/s instead of 5sqrt58. :/
Thanks for your help Integrand!

appleibeats · Jan 30, 2016

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

Screen Shot 2016-01-30 at 11.45.16 am.png

I get part a)

But am unsure on how to answer part b) using part a)

part b) ii) answer says root 2 -1 = tan22.5 degrees

InteGrand · Jan 30, 2016

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

appleibeats said:
View attachment 32839

I get part a)

But am unsure on how to answer part b) using part a)

part b) ii) answer says root 2 -1 = tan22.5 degrees

$\noindent (b) (i) We have $t=\tan \frac{\theta}{2}$, where $\theta = 225^\circ,\, \frac{\theta}{2}=112\tfrac{1}{2}^{\circ}$. Note that $t<0$ (using ASTC for example), and our equation is $\frac{2t}{1-t^2}=\tan \theta = 1, \, t<0$. Rearranging that equation gives us $2t = 1-t^2 \Rightarrow t^2 +2t = 1$. Completing the square, we have $\left(t+1 \right)^2 = 2 \Rightarrow t = -1 \pm \sqrt{2}$. Since $t<0$ for us, we have $t=-1-\sqrt{2}$.$

$\noindent (ii) The positive solution to the quadratic represents the value of $t$ such that $\frac{2t}{1-t^2}=1$ and $t>0$. In other words, it is the value of $\tan \frac{\alpha}{2}$, where $\alpha$ is any angle such that $\tan \alpha = 1$ and $\tan \alpha >0$, i.e. $\alpha$ must be $45^\circ$ (or add any multiply of $360^\circ$ to this, so the tan will be the same and still be positive, so $\alpha = 45^\circ + n\cdot 360^\circ$, where $n\in \mathbb{Z}$). Hence the positive solution is the value of $\tan \frac{\alpha}{2}=\tan \left(\frac{45^\circ + n\cdot 360^\circ}{2}\right)=\tan \left(22\tfrac{1}{2}^{\circ}+n\cdot 180^\circ\right)=\tan 22\tfrac{1}{2}^\circ$ (by periodicity of the tan function), where $n\in\mathbb{Z}$. Since the positive solution was found to be $-1+\sqrt{2}$, we have $\tan 22\tfrac{1}{2}^\circ=-1+\sqrt{2}$.$

appleibeats · Jan 30, 2016

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

There is a multiple choice question that says what is the range of g(x) = tan(1/2 inverse cos x)

a) y less than 0
b) y all real
c) 0 < y <1
D) y greater than or equal to 0

I choice b

But the answer says D.

Can someone explain how to get the correct answer.

InteGrand · Jan 30, 2016

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

appleibeats said:
There is a multiple choice question that says what is the range of g(x) = tan(1/2 inverse cos x)

a) y less than 0
b) y all real
c) 0 < y <1
D) y greater than or equal to 0

I choice b

But the answer says D.

Can someone explain how to get the correct answer.

$\noindent Observe that as $x$ varies in its domain $\Big{(}$which is $-1< x\leq 1$, noting that $x\neq -1$, since if $x=-1$, $\frac{1}{2}\cos ^{-1} x = \frac{\pi}{2}$, so $g(x)$ is undefined here$\Big{)}$, $\frac{1}{2}\cos ^{-1}x$ varies between $0$ and $\frac{\pi}{2}$, so $\tan \left(\frac{1}{2}\cos ^{-1} x\right)$ varies between 0 and $\infty$ $\Big{(}$since from the tan graph we see how that this is how tan behaves when its argument varies between $0$ and $\frac{\pi}{2}$$\Big{)}$. Hence the answer is (D).$

appleibeats · Jan 31, 2016

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

i) Sketch the region in the complex plane which simulatenously satisfies:

-pi/4 < or equal to arg(z + i) < or equal to pi/4 and |z - 2| < or equal to 1

ii) Find the maximun value of |z|
iii) Find the minium values of arg(z), given -pi < arg(z) < or equal to pi

I have done and understood parts i and ii

But am unsure about part iii)
the answer says arg(z) = - pi/6

Blitz_N7 · Feb 1, 2016

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

In this question, is it safe to assume that since the motion is parabolic, then the highest height occurs at half its range, i.e: x=4?

InteGrand · Feb 1, 2016

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

Blitz_N7 said:
In this question, is it safe to assume that since the motion is parabolic, then the highest height occurs at half its range, i.e: x=4?

Yes, that's correct.

Blitz_N7 · Feb 1, 2016

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

Thanks!

Drsoccerball · Feb 1, 2016

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

Blitz_N7 said:
In this question, is it safe to assume that since the motion is parabolic, then the highest height occurs at half its range, i.e: x=4?

Just as long as the question says "No air resistance."

si2136 · Feb 2, 2016

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

Given that one root of the equation ax^2 + bx + c = 0 is 3 times the other, show that 3b^2 = 16ac

InteGrand · Feb 2, 2016

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

si2136 said:
Given that one root of the equation ax^2 + bx + c = 0 is 3 times the other, show that 3b^2 = 16ac

$\noindent Let the two roots be $\alpha, \,3\alpha$. By sum of roots, we have $\alpha+3\alpha = 4\alpha = -\frac{b}{a}\Rightarrow \alpha = -\frac{b}{4a}$. Now, by product of roots, $\alpha \cdot 3\alpha = 3\alpha^2 = \frac{c}{a} \Rightarrow 3 \left( -\frac{b}{4a}\right)^2=\frac{c}{a}$. Rearranging and simplifying this yields the desired result. $\blacksquare$.$

drsabz101 · Feb 2, 2016

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

How do o do ex 2B . Q 4 and 5

InteGrand · Feb 2, 2016

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

drsabz101 said:
How do o do ex 2B . Q 4 and 5

What are the questions?

drsabz101 · Feb 2, 2016

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

I'll post it tomorrow actually

Shuuya · Feb 2, 2016

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

drsabz101 said:
I'll post it tomorrow actually

It's this right?

Blitz_N7 · Feb 3, 2016

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

I was able to do parts A and B but I'm stuck in part C i and ii.

InteGrand · Feb 3, 2016

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

Blitz_N7 said:
I was able to do parts A and B but I'm stuck in part C i and ii.

$\noindent For (c), if we differentiate the function, we find using parts (a) and (b) that $y^\prime = -1-(-1)=0$, provided that $\cos x >0$ (i.e. $x$ is in the first or fourth quadrants). So each piece of the graph has 0 derivative, so is a horizontal line. To find the value of the function (which is constant in each piece as it is a horizontal line), we sub. in a convenient value of $x$. For first quadrant, sub. in $x = \frac{\pi}{4}$ say and evaluate the function to find its value in this quadrant. For the fourth quadrant, we can use $x =-\frac{\pi}{4}$.$

appleibeats · Feb 3, 2016

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

Screen Shot 2016-02-03 at 5.02.40 pm.png

I need help with part b)

I get y ' = -sinx/cosx + 1/cos^2 x - 1

Now I make this equal to 0 for stat pts but am unsure on the next step.

I tried multiplying by cos^2 x to everything but there will be a sinxcosx .

InteGrand · Feb 3, 2016

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

appleibeats said:
View attachment 32852

I need help with part b)

I get y ' = -sinx/cosx + 1/cos^2 x - 1

Now I make this equal to 0 for stat pts but am unsure on the next step.

I tried multiplying by cos^2 x to everything but there will be a sinxcosx .

$\noindent Note that $y^\prime = -\tan x + \sec^2 x -1$. Hence $y^\prime =-\tan x + \tan^2 x$, as $\sec^2 x -1 = \tan^2 x$. Now it is easy to solve $y^\prime =0$. Just make sure to only include solutions $x$ such that $\cos x>0$, since the domain of our function is $\{ x\in \mathbb{R}: \cos x>0\}$ (since we have a $\ln (\cos x)$).$

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

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