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

InteGrand · Jun 9, 2015

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

appleibeats said:
I understand your answer and working. But could you explain how you get your second set of points:

$(a_1, a_2)$ , $(a_2,a_1)$$

because the equation is both symmetric. I know it has something to do with the circle being centred at (0,0) but am not entirely sure.

Thanks

$If we interchange $y$ and $x$ in each equation, the equations remain unchanged (e.g. the equation $x^2 + y^2 = 13$ would become $y^2 + x^2 = 13$, which is still the same). This means if $x=a_1,y=a_2$ satisfies both equations, so must $x=a_2,y=a_1$ (because this is the same as interchanging $y$ and $x$).$

appleibeats · Jun 10, 2015

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

InteGrand said:
Without doing any calculations, my guess is that it's a square.

Thanks for that previous question. I understand now how y and x are interchangeable.

Could you please expand on how to get a square in this question.

Thanks.

appleibeats · Jun 11, 2015

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

Desperately need help with Question 26 from 8H.

if the equations mx^2 + 2x + 1 = 0 and x^2 + 2x + m = 0 have a common root, find the possible values of m and the value of the common root in each case.

The answer in the book is:

When m = 1, x = -1, and when m = -3 , x = 1

Thank you for helping me answer this question!

calebwhitaker199 · Jun 12, 2015

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

Could somebody please post a picture of the cover of the 2015 year 11 extension 1 cambridge textbook ??

lollipopstupid · Jun 22, 2015

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

6N 18. [A rather difficult proof] CAMBRIDGE

appleibeats · Jun 24, 2015

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

Need help with question 12 from 10B

the function y = ax^3 + bx^2 + cx + d has a relative maximum at the point (-2,27) and a relative minimum at the point (1,0). Find the values of a,b,c,d.

Thank you for your help.

leehuan · Jun 24, 2015

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

We have several simultaneous equations here:

From the fact that the points (-2,27) and (1,0) pass through the graph of y=ax^3+bx^2+cx+d
27 = a(-2)^3 + b(-2)^2 + c(-2) + d ---------eqn 1
0 = a(1)^3 + b(1)^2 + c(1) + d ----------eqn 2

dy/dx = 3ax^2 + 2bx + c
Since (-2,27) and (1,0) are relative minima and maxima respectively, we know that when x=-2 and x=1 we have stationary points.
At a stationary point, dy/dx = 0
So we also have:
0 = 3a(-2)^2 + 2b(-2) + c ----------eqn 3
0 = 3a(1)^2 + 2b(1) + c -----------eqn 4

My advice is to do eqn 1 - eqn 2 first. Then you have another equation in terms of a, b, and c. Call that eqn 5.

Then solve eqn 3, 4 and 5 to get a, b and c. Which you can sub in to 2 to get d.

appleibeats · Jun 25, 2015

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

Thanks for that help.

Do you also know how to solve Q 13 part b from 10B.

b)

Show that y = x^a(1-x)^b has a turning point whose x coordinate divides the interval between the points (0,0) and (1,0) in the ration a:b

It is obviously related to part a) which I have done but am unsure how to go about this question.

Thank you for your help.

InteGrand · Jun 25, 2015

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

appleibeats said:
Thanks for that help.

Do you also know how to solve Q 13 part b from 10B.

b)

Show that y = x^a(1-x)^b has a turning point whose x coordinate divides the interval between the points (0,0) and (1,0) in the ration a:b

It is obviously related to part a) which I have done but am unsure how to go about this question.

Thank you for your help.

$One way to do it is to find the $x$ value of the point that divides (0,0) and (1,0) in the ratio $a:b$ (call this $X$), and show that the function as a turning point at this $x$ value.$

$Using the division of a line in a ratio formula (or you could work it out intuitively), $X=\frac{b\times 0 + a\times 1}{a+b}=\frac{a}{a+b}$.$

$Now, $y=f(x)=x^a (1-x)^b$, so $f^\prime (x) = ax^{a-1}(1-x)^b -bx^a(1-x)^{b-1}$$

$\Rightarrow f^\prime (x) =x^{a-1}(1-x)^{b-1} (a(1-x)-bx)$

$$\Rightarrow f^\prime (x) =x^{a-1}(1-x)^{b-1} (a-(a+b)x)$.$

$\therefore f^\prime (X)=\left ( \frac{a}{a+b} \right )^{a-1}\left ( 1-\frac{a}{a+b} \right )^{b-1}\left ( a-(a+b)\frac{a}{a+b} \right )$

$=\left ( \frac{a}{a+b} \right )^{a-1}\left ( \frac{b}{a+b} \right )^{b-1}\left ( a-a\right )$

$$=0\Rightarrow $ stationary point at $x=X$.$

$This must be a turning point because $f(0)=f(1)(=0)$ and it is the only point where $f^\prime (x)=0$ in the interval $(0,1)$, as seen from the equation for $f^\prime (x)$, and since $f$ and $f^\prime$ are continuous.$

appleibeats · Jul 3, 2015

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

Need help with 10C.

Question 8:

y = |x| +3

a) find dy/dx when x < 0 and when x > 0

I know what the curve look like, and that when < 0 it has a negative gradient and when x > 0 it is positive gradient.

But how do you show that working out with the absolute value. Isn't any number ( positive or negative) inside the absolute value become positive??

Also Q10)

Part a) can do : differentiate f(x) = ( x - 2) ^ 1/5

Part b) show that there are no stationary points, but that a critical value occurs at x = 2 .

I thought stationary points is when the derivate is 0. But that is wrong as that would mean 2 is a stationary point but it isn't. It is instead a critical point. Please help clarify the meanings.

Thanks for helping and explaining.

Cheers.

InteGrand · Jul 3, 2015

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

appleibeats said:
Need help with 10C.

Question 8:

y = |x| +3

a) find dy/dx when x < 0 and when x > 0

I know what the curve look like, and that when < 0 it has a negative gradient and when x > 0 it is positive gradient.

But how do you show that working out with the absolute value. Isn't any number ( positive or negative) inside the absolute value become positive??

Also Q10)

Part a) can do : differentiate f(x) = ( x - 2) ^ 1/5

Part b) show that there are no stationary points, but that a critical value occurs at x = 2 .

I thought stationary points is when the derivate is 0. But that is wrong as that would mean 2 is a stationary point but it isn't. It is instead a critical point. Please help clarify the meanings.

Thanks for helping and explaining.

Cheers.

$Q.8 (a) Note that $|x|=x$ when $x>0$ and $|x|=-x$ when $x<0$ (e.g. $|-4|=4=-(-4)$, as $-4<0$). So for $x>0$, $y=x+3\Rightarrow \frac{\mathrm{d}y}{\mathrm{d}x}=1$. And for $x<0$, $y=-x+3\Rightarrow \frac{\mathrm{d}y}{\mathrm{d}x}=-1$.$

$Q.10 (b) $f(x)=(x-2)^{\frac{1}{5}}\Rightarrow f^\prime (x) = \frac{1}{5}\frac{1}{(x-2)^\frac{4}{5}}$. Hence $f^\prime (x)\neq 0$ for $x\in \mathbb{R}$, so there are no stationary points. However, at the point where $x=2$ (which is a point where the function is defined), we can see that the derivative is \emph{un}defined. Hence the point where $x=2$ is a critical point. (A critical point for a function is defined as ``any value in its domain where its derivative is 0 or undefined''; see: \url{https://en.wikipedia.org/wiki/Critical\_point\_(mathematics)}.)$

Graphically, the function's slope tends to plus or minus infinity as you approach x = 2: http://www.wolframalpha.com/input/?i=plot+(x-2)^(1/5)

appleibeats · Jul 3, 2015

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

Thanks for your help. Really appreciate it. It now makes more sense.

Also do you know how to do question 10b from 10D.

If y = (2x - 1)^ 4, prove that d/dx(y dy/dx) = y d^2y/dx^2 + (dy/dx)^2.

What is most troubling me is how to find d/dx of y dy/dx which equals i believe 8y(2x -1)^6 ... what do you do with the y at the front??

Thanks for your help in advance.

Cheers

InteGrand · Jul 3, 2015

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

appleibeats said:
Thanks for your help. Really appreciate it. It now makes more sense.

Also do you know how to do question 10b from 10D.

If y = (2x - 1)^ 4, prove that d/dx(y dy/dx) = y d^2y/dx^2 + (dy/dx)^2.

What is most troubling me is how to find d/dx of y dy/dx which equals i believe 8y(2x -1)^6 ... what do you do with the y at the front??

Thanks for your help in advance.

Cheers

$To find $\frac{\mathrm{d}}{\mathrm{d}x}\left(y \frac{\mathrm{d}y}{\mathrm{d}x} \right)$, we just have to use the product rule. Since $y$ and $\frac{\mathrm{d}y}{\mathrm{d}x}$ are both functions of $x$, using the product rule,$

$\frac{\mathrm{d}}{\mathrm{d}x}\left(y\frac{ \mathrm{d}y}{\mathrm{d}x} \right)=\frac{\mathrm{d}y}{\mathrm{d}x}\times \frac{\mathrm{d}y}{\mathrm{d}x}+y\times \frac{\mathrm{d}}{\mathrm{d}x}\left( \frac{\mathrm{d}y}{\mathrm{d}x} \right)$

$$= \left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2+ y\frac{\mathrm{d}^2y}{\mathrm{d}x^2}$, as required. It did not matter what the expression for $y=f(x)$ was (as long as it was twice differentiable).$

appleibeats · Jul 3, 2015

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

could you explain how you used the product rule? I don't quite seem to get that. Thanks.

appleibeats · Jul 3, 2015

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

Disregard my last question, I figured it out. But could you explain what you mean by 'both functions of x' in your working out. Thank you.

InteGrand · Jul 3, 2015

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

appleibeats said:
could you explain how you used the product rule? I don't quite seem to get that. Thanks.

$By the product rule, if we have two functions of $x$, say $u\equiv u(x)$ and $v\equiv v(x)$, then$

$$\frac{\mathrm{d}}{\mathrm{d}x}(uv)=\frac{\mathrm{d}u}{\mathrm{d}x}v+u\frac{\mathrm{d}v}{\mathrm{d}x}$.$

$So finding $\frac{\mathrm{d}}{\mathrm{d}x}\left(y \frac{\mathrm{d}y}{\mathrm{d}x} \right)$ just requires us to apply the product rule using $u(x)=y$ and $v(x)=\frac{\mathrm{d} y}{\mathrm{d}x}$, and hence use $\frac{\mathrm{d}u}{\mathrm{d}x}=\frac{\mathrm{d}y}{\mathrm{d}x}$ (since $u=y$) and similarly $\frac{\mathrm{d}v}{\mathrm{d}x}=\frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{\mathrm{d}y}{\mathrm{d}x} \right)$ (which is just $\frac{\mathrm{d}^2 y}{\mathrm{d}x^2}$) in the product rule formula.$

appleibeats · Jul 3, 2015

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

Also, do you know how to answer question 13 from 10D. Thank you.

InteGrand · Jul 3, 2015

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

appleibeats said:
Also, do you know how to answer question 13 from 10D. Thank you.

What is the question?

appleibeats · Jul 3, 2015

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

Sorry, the question is:

Find positive integers a and b such that

x^2y'' + 2xy' = 12y, where y = x^a + x^-b.

Thank you.

InteGrand · Jul 3, 2015

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

appleibeats said:
Sorry, the question is:

Find positive integers a and b such that

x^2y'' + 2xy' = 12y, where y = x^a + x^-b.

Thank you.

$$y=x^a + x^{-b}$, $a,b$ positive integers.$

$So $y^\prime = ax^{a-1} -bx^{-b-1}=ax^{a-1} -bx^{-(b+1)}$ and $y''=a(a-1)x^{a-2}+b(b+1)x^{-b-2}=a(a-1)x^{a-2}+b(b+1)x^{-(b+2)}$.$

$Since we want $x^2y'' + 2xy' = 12y$, we will substitute the above expressions into this equation and try and solve for $a$ and $b$. Upon substitution into this equation,$

$$x^2 \left(a(a-1)x^{a-2}+b(b+1)x^{-(b+2)}\right)+2x\left(ax^{a-1} -bx^{-(b+1)}\right)=12\left(x^a + x^{-b}\right)$.$

$Expanding out and using index laws etc. yields$

$$a(a-1)x^a + b(b+1)x^{-b}+2ax^a - 2b x^{-b}=12x^a + 12x^{-b}$.$

$Collecting like terms on the left-hand side yields$

$$\left( a(a-1)+2a\right)x^a + \left(b(b+1)-2b\right)x^{-b}=12x^a + 12x^{-b}$.$

$Equating coefficients of like powers of $x$ on the left- and right-hand sides, we can see we require the following:$

$$a(a-1)+2a=12$ (A)$

$and$

$$b(b+1)-2b=12$ (B).$

$Equations (A) and (B) are both quadratic equations that I am sure you can now easily solve to find the suitable $a$ and $b$, making sure to take the \emph{positive} integer solutions (you should obtain $a=3$ and $b=4$).$

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

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