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

appleibeats · Aug 29, 2015

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

In your second last line of working, does the integral and derivate sign in front of xlnx just cancel each other out?

InteGrand · Aug 29, 2015

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

appleibeats said:
In your second last line of working, does the integral and derivate sign in front of xlnx just cancel each other out?

$Basically, yes. The meaning of just an integral sign with no limits is that we are asking for a function whose derivative is the thing being integrated. I.e. the meaning of $\int f(x)\text{ d}x$ is functions whose derivatives are $f$. So the meaning of $\int \frac{\mathrm{d}}{\mathrm{d}x}\left(x\ln x\right)\text{ d}x$ is a function whose derivative is $\frac{\mathrm{d}}{\mathrm{d}x}\left(x\ln x\right)$ (i.e. a function whose derivative is: the derivative of $x\ln x$). So that function is just $x\ln x$! (because the function $x\ln x$ has derivative $\frac{\mathrm{d}}{\mathrm{d}x}\left(x\ln x\right)$)$

$(You use this all the time subconsciously, e.g. $\int \sin x \text{ d}x=-\cos x +c$, because $-\cos x +c$ has derivative $\sin x$. So integration without any limits (i.e. definite integration) is just anti-differentiation, because instead of differentiating, we are finding something who has a given function as its derivative.)$

$So indefinite integration (anti-differentiation) and differentiation are ``inverse'' processes and cancel each other out. So the following identities hold true:$

$\bullet$ $\frac{\mathrm{d}}{\mathrm{d}x}\int f(x)\text{ d}x=f(x)$

$$\bullet$ $\int \frac{\mathrm{d}}{\mathrm{d}x} \left(f(x)\right) \text{ d}x=f(x) +C$, where $C$ is an arbitrary constant.$

appleibeats · Aug 29, 2015

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

In an experiement in which bacteria are grown on a ptetri dish, it is found that the area A in cm^2 covered by the bacteria increases from 0.5 cm^2 to 1 cm^2 in a period of 3 hours

so

A = 1/2 e^kt

k = 0.23104.....

d) if the diameter of the petri dish is 10 cm how long will it take for the bacteria to cover the dish? Answer to the nearest 10 minutes.

I get the answer 53 hours ,10 minutes

The correct answer is 21 hours and 50 minutes.

rand_althor · Aug 29, 2015

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

$$If the diameter is 10cm, then the area is $25\pi$ cm$^2$, since $A=\pi r^2$. Subbing this into the formula, we have $25\pi=\frac{1}{2}e^{kt}$, where $k\approx0.23104$. To solve for $t$:$

$\begin{align*}25\pi&=\frac{1}{2}e^{kt} \\e^{kt}&=50\pi \\kt&=\ln{(50\pi)} \textup{ (Taking the natural log of both sides)} \\t&=\frac{\ln{(50\pi)}}{k} \\\textup{Subbing in }&k \textup{ gives you: } \\t&\approx21.89 \textup{ hours} \\&\approx 21\textup{ hours } 50 \textup{ minutes}\end{align*}$

appleibeats · Aug 29, 2015

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

Use the discriminant to find any tangents to the curve y = x^3 + 2x^2 - 3x that pass through the origin.

So I used simultaneous eqn with y = mx and the cubic above.

Then got

x( x^2 + 2x - m - 3) = 0

Using the discriminant

Disc = 4( m + 4)

SO tangent when m = -4

But answer also says tangent at m = - 3 because x (x^2 + 2x) = 0 has a repeated root x = 0

This I am unclear about .... not sure what it mean??

InteGrand · Aug 29, 2015

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

appleibeats said:
Use the discriminant to find any tangents to the curve y = x^3 + 2x^2 - 3x that pass through the origin.

So I used simultaneous eqn with y = mx and the cubic above.

Then got

x( x^2 + 2x - m - 3) = 0

Using the discriminant

Disc = 4( m + 4)

SO tangent when m = -4

But answer also says tangent at m = - 3 because x (x^2 + 2x) = 0 has a repeated root x = 0

This I am unclear about .... not sure what it mean??

$The line through the origin with slope $m=-3$ is clearly also a tangent to that curve, because if you differentiate that curve, you see that its slope at $x=0$ is $-3$. And at $x=0$, that curve passes through the origin, so the line through the origin with slope $-3$ is a tangent to that curve.$

appleibeats · Sep 8, 2015

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

cos^ -1 ( -1) is 0, 360 .... using the cosine graph

so when answer a question which gets

x = pi - cos^-1 ( -1)

why is the answer only pi as you let cos... be equal to 0 . But it can also equal to 360... 720 ......

InteGrand · Sep 8, 2015

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

appleibeats said:
cos^ -1 ( -1) is 0, 360 .... using the cosine graph

so when answer a question which gets

x = pi - cos^-1 ( -1)

why is the answer only pi as you let cos... be equal to 0 . But it can also equal to 360... 720 ......

$Firstly, $\cos^{-1} (-1)\neq 0$, since $\cos 0 = 1$. Secondly, $\cos ^{-1} (-1)=\pi$ and yes, there are (infinitely) many values of $x$ such that $\cos x = -1$, but the definition of the function ``$\cos^{-1}$'' is that it will give you only the angle between $0$ and $\pi$ that will have a cosine of something (in this case $-1$). This is because we restrict the domain of the inverse relation of the cosine function in order to make this relation a function (otherwise it would be multivalued). It's like the reason that $\sqrt{4}=2$, but not $-2$, despite the fact that $(-2)^2$ is also 4, is that we have had to restrict the inverse relation of the square function in order to make sure that it is single-valued. So we chose the convention that $\sqrt{x}$ will be the \textit{non-negative} square root of $x$, hence obtaining a unique value for each $x\geq 0$.$

appleibeats · Sep 8, 2015

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

So for the quetstion:

Solve for x over the domain 0 < equal to x < equal to 2pi

cos x + 1 = 0

So 2nd and 3rd Quad

related angle is cos^-1 ( -1) = 180 = pi

so x = pi - pi or x = pi + pi

so x = 0 or 2pi

Where did i go wrong as the answer says x = pi

appleibeats · Sep 8, 2015

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

Also unsure about this question

Solve the eqn for 0 < equal to x < equal to 2pi

tan x = 2^1/2 - 1

rand_althor · Sep 8, 2015

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

appleibeats said:
So for the quetstion:

Solve for x over the domain 0 < equal to x < equal to 2pi

cos x + 1 = 0

So 2nd and 3rd Quad

related angle is cos^-1 ( -1) = 180 = pi

so x = pi - pi or x = pi + pi

so x = 0 or 2pi

Where did i go wrong as the answer says x = pi

Not sure what you have done here. If you draw a graph of y=cos(x) you'll see that the only solution in that domain is x=pi.

InteGrand · Sep 8, 2015

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

appleibeats said:
Also unsure about this question

Solve the eqn for 0 < equal to x < equal to 2pi

tan x = 2^1/2 - 1

$Hint: one solution is $x=22.5^\circ$, i.e. $x=\frac{\pi}{8}$.$

appleibeats · Sep 8, 2015

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

Do I first solve for x in degrees and then convert to radians. Is that the best way to answer the question.

Cause I got this and It is the correct answer but the answer in the book had exact values and I didn't know how to convert it to it.

tanx = 2^1/2 - 1
1st and 3rd quadrant

related angle is tan^-1 (2^1/2 - 1)

x = tan^-1 (2^1/2 - 1) or pi + tan^-1 (2^1/2 - 1)

InteGrand · Sep 8, 2015

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

appleibeats said:
Do I first solve for x in degrees and then convert to radians. Is that the best way to answer the question.

Cause I got this and It is the correct answer but the answer in the book had exact values and I didn't know how to convert it to it.

tanx = 2^1/2 - 1
1st and 3rd quadrant

related angle is tan^-1 (2^1/2 - 1)

x = tan^-1 (2^1/2 - 1) or pi + tan^-1 (2^1/2 - 1)

Need to convert the inverse tan's to pi's (and would need to prove your result, since it's not one of the 'standard' trig. values).

appleibeats · Sep 8, 2015

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

So pretty much say;

tan^-1 (2^1/2 - 1) = 22.5 degrees

22.5 degrees = .5 x 45 degrees

45 degrees = pi / 4 rad
so 22.5 degrees = pi/4 x .5 = pi / 8 rad

Is this correct proof of results??

InteGrand · Sep 8, 2015

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

appleibeats said:
So pretty much say;

tan^-1 (2^1/2 - 1) = 22.5 degrees

22.5 degrees = .5 x 45 degrees

45 degrees = pi / 4 rad
so 22.5 degrees = pi/4 x .5 = pi / 8 rad

Is this correct proof of results??

No, you'd need to prove the first line, not just state it. Once you've proved it, you can just convert it straight to radians (or prove it in radians in the first place rather than degrees).

InteGrand · Sep 8, 2015

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

(Here is a proof)

$Original equation: $\tan x = \sqrt{2}-1$$

$Our calculator says that $\tan^{-1} \left(\sqrt{2}-1 \right)=15^\circ$, so $\frac{\pi}{8}$ should be a solution, but we need to prove this formally, as proof by calculator is not acceptable for such questions.$

$We have$

$$\tan \frac{\pi}{8}=\frac{\sin \frac{\pi}{8}}{\cos \frac{\pi}{8}}$ (definition of tan)$

$$=\frac{2\sin \frac{\pi}{8}\cos \frac{\pi}{8}}{2\cos^2 \frac{\pi}{8}}$ (multiplying top and bottom by $2\cos \frac{\pi}{8}$)$

$$=\frac{\sin \frac{\pi}{4}}{2\times \frac{1}{2}\left(1+\cos \frac{\pi}{4} \right)}$ (double angle formula on numerator and half-angle formula on denominator)$

$$=\frac{\frac{\sqrt{2}}{2}}{1+ \frac{ \sqrt{2}}{2} }$ (using standard trig. values)$

$$=\frac{\sqrt{2}}{2+\sqrt{2}}$ (multiplying top and bottom by 2)$

$$=\frac{\sqrt{2}}{2+\sqrt{2}}\times \frac{2-\sqrt{2}}{2-\sqrt{2}}$ (to rationalise the denominator)$

$=\frac{2\sqrt{2}-2}{4-2}$

$$=\sqrt{2}-1$.$

RealiseNothing · Sep 9, 2015

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

Could also just use compound angle formula for tan to prove it.

appleibeats · Sep 10, 2015

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

OP and OQ are radii of length r centimetres of a circle centered at O. The arc PQ of the circle subtends an angle of G radians at O and the perimeter of the sector OPQ is 12 cm.

a) Show that the area A cm^2 of the sector is given by A = 72G / ( 2 + G)^2

DID PART A

b) Hence find the maximum area of the sector.

NOT SURE ABOUT PART B

Usually I would find the vertex of a downward parabola to get the x values but not sure what to do here??

InteGrand · Sep 10, 2015

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

appleibeats said:
OP and OQ are radii of length r centimetres of a circle centered at O. The arc PQ of the circle subtends an angle of G radians at O and the perimeter of the sector OPQ is 12 cm.

a) Show that the area A cm^2 of the sector is given by A = 72G / ( 2 + G)^2

DID PART A

b) Hence find the maximum area of the sector.

NOT SURE ABOUT PART B

Usually I would find the vertex of a downward parabola to get the x values but not sure what to do here??

$Maximise your area function in part (a) for $G$ in the range 0 to $2\pi$ (using calculus) (I assume you know about how to optimise functions like this).$

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