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

appleibeats

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

I realised I made a silly mistake, I should have converted the initial volume V(0) to cubic cm, since the dV/dt is given in cubic cm per second.
What was the initial volume V(0) in? And what does that now make V(0)? Do I multiply by 1000.
 

InteGrand

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

What was the initial volume V(0) in? And what does that now make V(0)? Do I multiply by 1000.
The initial volume was given as 2.5 L if I recall correctly. And yes, we'd multiply by 1000 because 1 L = 1000 mL = 1000 cm3. So V(0) = 2500.
 

appleibeats

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

Find the area bounded by the graph of the given function and the y axis between the specified values. Draw a diagram and see if the area is to the right or left of the axis.

a) x = y - 5, y = 0 and y = 6

So I have drawn the graph which is just a line.

I get ; Area = Integration from 6 to 0 of y -5 dx

Eventually giving my the answer 1/2

Not sure what I have done wrong.

The answer is 13
 

VBN2470

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

That's just for the upper triangle (draw a diagram), for the lower triangle bounded between the given curves, the area is , so the total area is sq. units. (No need to use integration, just use standard formula for area of a right-angled triangle)
 

appleibeats

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

a) Show that for n < -1, The integral from N to 1 (x^n) dx converges as N --> infinity, and find the limit.

b) Show that for n > -1, the integral from 1 to e (x^n) dx converges as e --> infinity, and find the limit.

c) Interpret these two results as areas.
 

rand_althor

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



For b) the question in the textbook is different to what you have posted. I'll do the textbook one as your one doesn't actually converge for n>-1.

 

appleibeats

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



For b) the question in the textbook is different to what you have posted. I'll do the textbook one as your one doesn't actually converge for n>-1.

I understand how you found the limit and the answer. What does it mean when it says converges as ...? Also what did you do with the n domain for each question? Are they significant in your answer?
 

rand_althor

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

Converge basically means it approaches a certain value. For example, converges to 0 as .

For a), n<-1, thus n+1<0. So is really just where . As , the curve converges to 0.

For b) n>-1, thus n+1>0. So for , the curve does not converge as - it just keeps bigger when you input a larger value of x. But for , the curve converges to 0.

They are significant as they determine whether the curve actually converges or not.
 

InteGrand

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

Converge basically means it approaches a certain value. For example, converges to 0 as .

For a), n<-1, thus n+1<0. So is really just where . As , the curve converges to 0.

For b) n>-1, thus n+1>0. So for , the curve does not converge as - it just keeps bigger when you input a larger value of x. But for , the curve converges to 0.

They are significant as they determine whether the curve actually converges or not.




 
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appleibeats

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

a) Find the area bounded by the curve y = ax^2 + bx + c and the x - axis between x = h and x = - h, where y > 0 for - h < and equal to x < and equal to h.

b) Hence show that if y = y0, y1, y2 when x = -h, 0 and h respectively, then the area is given by 1/3 h (y0 + 4y1 + y2) .

I can do part a), need help with b).
 

braintic

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

a) Find the area bounded by the curve y = ax^2 + bx + c and the x - axis between x = h and x = - h, where y > 0 for - h < and equal to x < and equal to h.

b) Hence show that if y = y0, y1, y2 when x = -h, 0 and h respectively, then the area is given by 1/3 h (y0 + 4y1 + y2) .

I can do part a), need help with b).
If you substitute the point (0,y1) into the equation of the parabola, you get y1 = c.

So y = ax^2 + bx + y1

Substituting the other points (-h, y0) and (h, y2) gives:
y0 = ah^2 - bh + y1
y2 = ah^2 + bh + y1

By alternately adding and subtracting these equations, you can get expressions for a and b in terms of y0, y1, y2.

Substitute these expressions into your answer to part (i) to get the answer.
 

appleibeats

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

Consider the function G(x) = Integral from x to 0 of g(u) du, where g(u) = 4 - 4/3 u , for 0 < equal to u < 6,
u - 10, for 6 < equal to u < equal to 12,

a) Sketch a graph of g(u)
b) Find the stationary points of the function y = G(x), and determine their nature.
c) Find those values of x for which G(x) = 0
d) Sketch the curve y = G(x), indicating all important features.
e) Find the area bounded by the curve y = G(x) and the x axis for 0 < equal to x < equal to 6.


For part a) Do you just work out the intercepts with the axis in the two different sections and also the endpoints x = 0, 12 and also where the two parts meet, ie, at x = 6

Have no clue about part b) as there are two different parts. Also find stationary points means when first derivative is 0. So is that the derivative of the integral???
 
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InteGrand

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

Consider the function G(x) = Integral from x to 0 of g(u) du
You mean from 0 to x, right? (i.e. 0 on the bottom and x on the top, which is technically read as from '0 to x')
 

appleibeats

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

You mean from 0 to x, right? (i.e. 0 on the bottom and x on the top, which is technically read as from '0 to x')
Yes that is what I meant. Sorry for the confusion.
 

InteGrand

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

Consider the function G(x) = Integral from x to 0 of g(u) du, where g(u) = 4 - 4/3 u , for 0 < equal to u < 6,
u - 10, for 6 < equal to u < equal to 12,

a) Sketch a graph of g(u)
b) Find the stationary points of the function y = G(x), and determine their nature.
c) Find those values of x for which G(x) = 0
d) Sketch the curve y = G(x), indicating all important features.
e) Find the area bounded by the curve y = G(x) and the x axis for 0 < equal to x < equal to 6.


For part a) Do you just work out the intercepts with the axis in the two different sections and also the endpoints x = 0, 12 and also where the two parts meet, ie, at x = 6

Have no clue about part b) as there are two different parts. Also find stationary points means when first derivative is 0. So is that the derivative of the integral???






 

appleibeats

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

I understand that there is a maximum at x = 3 and minimum at x = 10 and I have used a table to show that. Now which formula do I sub in these values to find the y intercept of the coordinates.

Also, still confused on part c), 'Find the values of x for which G(x) = 0 , why is it you only use 4 - 4/3 u, for 0 < equal to u < 6 and not the other section??

Furthermore, in your second post, where G(6) comes from and hence (24 - 24)
 

rand_althor

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

For the y-coordinates sub into one of these formulas:
You should use both those formulas to find when G(x)=0, so you get 2 y-intercepts - y=0 for the first and y=6 for the second.
 
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