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

appleibeats · Jul 29, 2015

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

InteGrand said:
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 · Jul 29, 2015

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

appleibeats said:
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 · Jul 29, 2015

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 · Jul 29, 2015

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 $\frac{1}{2}\times{5}\times{5}=12.5$ , so the total area is $12.5+0.5=13$ sq. units. (No need to use integration, just use standard formula for area of a right-angled triangle)

appleibeats · Jul 30, 2015

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 · Jul 30, 2015

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

$\begin{align*}\textup{a)} \int^N_1 x^n\, \mathrm{d}x &= \left [\frac{x^{n+1}}{n+1} \right ]^N_1 \\&=\lim_{N\to\infty}\frac{N^{n+1}}{n+1} -\frac{1^{n+1}}{n+1} \\&=\frac{1}{\frac{n+1}{N^{n+1}}}-\frac{1}{n+1} \\&=-\frac{1}{n+1}\end{align*}$

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.

$\begin{align*}\textup{b)} \int^1_\varepsilon x^n\, \mathrm{d}x &= \left [\frac{x^{n+1}}{n+1} \right ]^1_\varepsilon \\&=\frac{1^{n+1}}{n+1} -\lim_{\varepsilon\to0^+}\frac{\varepsilon^{n+1}}{n+1} \\&=\frac{1}{n+1}\\\end{align*}$

appleibeats · Jul 30, 2015

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

rand_althor said:
$\begin{align*}\textup{a)} \int^N_1 x^n\, \mathrm{d}x &= \left [\frac{x^{n+1}}{n+1} \right ]^N_1 \\&=\lim_{N\to\infty}\frac{N^{n+1}}{n+1} -\frac{1^{n+1}}{n+1} \\&=\frac{1}{\frac{n+1}{N^{n+1}}}-\frac{1}{n+1} \\&=-\frac{1}{n+1}\end{align*}$

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.

$\begin{align*}\textup{b)} \int^1_\varepsilon x^n\, \mathrm{d}x &= \left [\frac{x^{n+1}}{n+1} \right ]^1_\varepsilon \\&=\frac{1^{n+1}}{n+1} -\lim_{\varepsilon\to0^+}\frac{\varepsilon^{n+1}}{n+1} \\&=\frac{1}{n+1}\\\end{align*}$

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 · Jul 30, 2015

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

Converge basically means it approaches a certain value. For example, $\frac{1}{x}$ converges to 0 as $x\to\infty$ .

For a), n<-1, thus n+1<0. So $x^{n+1}$ is really just $\frac{1}{x^a}$ where $a=n+1$ . As $x\to\infty$ , the curve converges to 0.

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

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

InteGrand · Jul 30, 2015

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

rand_althor said:
Converge basically means it approaches a certain value. For example, $\frac{1}{x}$ converges to 0 as $x\to\infty$ .

For a), n<-1, thus n+1<0. So $x^{n+1}$ is really just $\frac{1}{x^a}$ where $\textbf{a=n+1}$ . As $x\to\infty$ , the curve converges to 0.

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

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

$It's not asking about whether the function converges, but rather whether the \emph{integral} converges.$

$Also, in a), it should be $a=-(n+1)$.$

$And in b), the function $f(x)=x^{n},n>-1$ may also approach $\infty$ as $x\to 0^+$; this happens precisely when $-1<n<0$. Also in this range of $n$, $f(x)\to 0^+$ as $x\to \infty$.$

rand_althor · Jul 30, 2015

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

InteGrand said:
$It's not asking about whether the function converges, but rather whether the \emph{integral} converges.$

$Also, in a), it should be $a=-(n+1)$.$

$And in b), the function $f(x)=x^{n},n>-1$ may also approach $\infty$ as $x\to 0^+$; this happens precisely when $-1<n<0$. Also in this range of $n$, $f(x)\to 0^+$ as $x\to \infty$.$

Thanks for the corrections. Is there any difference if it is $x\to0^+$ rather than $x\to0^-$ ?

InteGrand · Jul 30, 2015

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

rand_althor said:
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 $-1<n<0$, then we can write $n=-m$, where $0<m<1$, and $f_n (x) = \frac{1}{x^m}$, which, as $x\to 0^-$, approaches $+\infty$ if $a$ is even, and $-\infty$ if $a$ is odd. If $n>0$, 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.$

appleibeats · Jul 31, 2015

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 · Jul 31, 2015

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

appleibeats said:
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 · Jul 31, 2015

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???

InteGrand · Jul 31, 2015

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

appleibeats said:
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 · Jul 31, 2015

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

InteGrand said:
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 · Jul 31, 2015

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

appleibeats said:
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 $0<x<6$) or $x-10=0$ (for $6<x<12$), i.e. $x=3,10$. I'll let you calculate $G(3)$ and $G(10)$. You should be able to see that the stationary point at $x=3$ is a local maximum (because the derivative, which is just equal to $g(x)$, changes from positive to negative as we go from just left of 3 to just right of 3), and the stationary point at $x=10$ is a local minimum (as the derivative goes from negative to positive).$

InteGrand · Jul 31, 2015

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 $x$.)$

appleibeats · Aug 1, 2015

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 · Aug 1, 2015

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

For the y-coordinates sub into one of these formulas:

InteGrand said:
$$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.

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

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