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Cambridge Prelim MX1 Textbook Marathon/Q&A (3 Viewers)

appleibeats

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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?
 

appleibeats

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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.
 

appleibeats

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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

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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??
 

appleibeats

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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

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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 ......
 

appleibeats

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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

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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

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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
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.
 

appleibeats

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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

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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)
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

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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

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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??
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

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

(Here is a proof)






















 

RealiseNothing

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

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

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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

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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??
 

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