# Thread: Cambridge Prelim MX1 Textbook Marathon/Q&A

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

Originally Posted by rand_althor
Thanks for the corrections. Is there any difference if it is $x\to0^+$ rather than $x\to0^-$?
$Yes, there is. The case of x\to 0^- is quite complicated, whereas for x\to 0^+ it is much more straightforward because the function is defined for positive x for all n, but not negative x for all n, so we need to consider cases of n for negative x.$

$If f_n(x) = x^n,n>-1, then depending on n, there will be problems for negative x, due to taking roots of negative numbers. E.g. if n=\frac{1}{4}, we have trouble. For integer values of n and n of the form \frac{a}{b}, where a and b are coprime integers and b is an odd number, the function will be real-valued for negative x. In such cases, if -10, then f_n (x)\to 0^+ if a is even and 0^- if a is odd, as x\to 0^-. If n=0, we have f_0 (x) = x^0 = 1\text{ }\forall x \neq 0, and in this case f_0(x) \to 1 as x\to 0^{\pm}.$

$Now, if n is \emph{not} an integer or of the form \frac{a}{b}, where a and b are coprime integers and b is an odd number, that is, if n is irrational or of the form \frac{a}{b}, where a and b are coprime integers and b is \emph{even}, the function f_n will not take on real values for negative x (e.g. things like (-1)^{\pi} or (-3)^{\frac{3}{8}} would occur), so the function is undefined for negative x if we only want a real-valued function.$

2. ## 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).

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

Originally Posted by appleibeats
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.

4. ## 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???

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

Originally Posted by appleibeats
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')

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

Originally Posted by InteGrand
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.

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

Originally Posted by appleibeats
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???
$Given: G(x) = \int_{0}^{x}g(u)\text{ d}u, where$

$g(u)=\begin{cases}4-\frac{4}{3}u & \text{ if } 0\leq u<6 \\ u-10& \text{ if } 6\leq u\leq 12 \\ \end{cases}.$

$a) We just sketch the linear function g(u)=4-\frac{4}{3}u for 0\leq u < 6 and g(u) = u-10 for 6\leq u \leq 12 (no x, rather, u is the variable on the horizontal axis here).$

$b) By the Fundamental Theorem of Calculus (which is what we use to differentiate the integral), G^\prime (x) = g(x). Note that g(x) is continuous for all x\in (0,12), so G^\prime (x) exists and is continuous for all x in this open interval. So stationary points occur when G^\prime (x) = 0 \Rightarrow g(x) = 0 \Rightarrow 4-\frac{4}{3}x=0 (for 0

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

$Also note that$

$\int _0 ^x \left(4 - \frac{4}{3}u \right)\text{ d}u = \left[4u - \frac{2}{3}u^2 \right]_0 ^x$

$= 4x - \frac{2}{3}x^2.$

$For 0\leq x<6 then, we have G(x) = 4x - \frac{2}{3}x^2, and for x\geq 6, we have G(x) = G(6) + \int _6 ^x \left(u-10 \right)\text{ d}u = (24-24)+ \left[\frac{u^2}{2}-10u \right]_6 ^x = \frac{x^2}{2}-10x +42. So we have this formula for calculating values for G(x):$

$G(x)=\begin{cases} 4x - \frac{2}{3}x^2 & \text{ if } 0\leq x<6 \\ \frac{x^2}{2}-10x +42& \text{ if } 6\leq x\leq 12 \\ \end{cases}.$

$(This is clearer if you first sketch the graph of y=g(u) on the domain 0\leq u \leq 12, and then visualise G(x) as an area so far'' function, i.e. we vary the point x\geq 0, and G(x) varies by giving the signed area (integral) of the function sketched from 0 to x.)$

9. ## 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)

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

For the y-coordinates sub into one of these formulas:
Originally Posted by InteGrand
$G(x)=\begin{cases} 4x - \frac{2}{3}x^2 & \text{ if } 0\leq x<6 \\ \frac{x^2}{2}-10x +42& \text{ if } 6\leq x\leq 12 \\ \end{cases}.$
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.

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

Can anyone do this?
Find the exact gradient, with rational denominator, of the normal to the parabola y^2 = 12x at the point where x=4 in the first quadrant.

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

I used implicit differentiation and got the answer, but is there any way I can square root both sides and take the positive answer (y=root12x instead of y=-root12 because it states "at the point x=4 in first quadrant".)

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

LOL nevermind, I got it thanks

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

8b from 11G

Graph of y = 5x and y = 5

Shaded area is the area within the y axis, y = 5x and y = 5

Find the volume of the solid generated by rotating the region about the x axis ..

I get the answer 25 pi / 3

However the answer in the book say 50 pi / 3

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

Take the difference volumes and you should get $V={\pi}\int_{0}^{1} 5^2-(5x)^2 \text{d}x=\frac{50\pi}{3}$

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

Why can't you just take the integral as usual , ie V = pi integral 0 to 1 25x^2 dx

Why do you need to take the differences of volume??

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

Because the region you are rotating is bounded by the y-axis (not x-axis), so you are taking the the difference of volumes when rotating around the x-axis. In this case you are findig the volume of solid of revolution of the rectangle bounded by $[0,1] \times [0,5]$ (which is a cylinder with volume $25{\pi}$) and subtracting the volume of the solid of revolution bounded by the x-axis, x = 1 and y = 5x. A diagram will greatly help you visualise what's going on.

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

We can also do it without using integration by subtracting the volume of a cone (using the cone volume formula) from the cylinder volume.

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

Also similarly for 8c

ii) volume of the solid generated by rotating region about the y axis

region is the area between x axis, x = y^2 and x = 4

My working:

V = pi integral from 0 to 2 8 - y^4 dy

then integrate normally.

What is wrong with my working??

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

Originally Posted by appleibeats
Also similarly for 8c

ii) volume of the solid generated by rotating region about the y axis

region is the area between x axis, x = y^2 and x = 4

My working:

V = pi integral from 0 to 2 8 - y^4 dy

then integrate normally.

What is wrong with my working??
What is the reason for your working?

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

I am trying to find the volume. Not sure where I have gone wrong. The answer is 128 pi / 5 u^3 and I am getting way off.

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

Originally Posted by appleibeats
I am trying to find the volume. Not sure where I have gone wrong. The answer is 128 pi / 5 u^3 and I am getting way off.
What's the reasoning for your working, as in, why do you think you've done it right in the first place?

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

I realised my mistake. Really stupid of me. I finally got the answer.

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

c)i ) A sphere of radius r is generated by rotating the semicircle y = (r^2 - x^2)^ 1/2 about the x asix. Show that the volume of the sphere is given by 4/3 pi r^3

I can do this first part

ii) A spherical cap of radium r and heigh h is formed by rotating the semicircle y = (r^2 - x^2)^1/2 between x = r and x = r - h about the x - axis . Show that the volume of the cap is geven by 1/3 pi h^2( 3r - h)

I am quite certain it has something to do with part i) but am not sure how to use part i) .

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

Originally Posted by appleibeats
c)i ) A sphere of radius r is generated by rotating the semicircle y = (r^2 - x^2)^ 1/2 about the x asix. Show that the volume of the sphere is given by 4/3 pi r^3

I can do this first part

ii) A spherical cap of radium r and heigh h is formed by rotating the semicircle y = (r^2 - x^2)^1/2 between x = r and x = r - h about the x - axis . Show that the volume of the cap is geven by 1/3 pi h^2( 3r - h)

I am quite certain it has something to do with part i) but am not sure how to use part i) .
doesn't need part i) :P

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