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

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

Does anyone know how to answer Q15 from 8H.

if a and b are the zeros of the function y = 3x^2 - 5x - 4, find the value of a^3 + b^3.

Thanks.

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

Originally Posted by appleibeats
Does anyone know how to answer Q15 from 8H.

if a and b are the zeros of the function y = 3x^2 - 5x - 4, find the value of a^3 + b^3.

Thanks.
$For a given quadratic, we know how to find the values of \alpha + \beta and \alpha \beta, where \alpha,\beta are the roots, by just inspecting the coefficients of the quadratic' terms. So our goal is to express \alpha^3 + \beta^3 in terms of \alpha + \beta and \alpha \beta, which is done as follows: \alpha^3 + \beta^3 = (\alpha + \beta)(\alpha^2 -\alpha \beta + \beta^2) (factorising) and use \alpha ^2 + \beta ^2 = (\alpha + \beta)^2 -2\alpha \beta.$

$So, \alpha^3 + \beta^3 = (\alpha + \beta)(\alpha^2 -\alpha \beta + \beta^2) = (\alpha + \beta)((\alpha + \beta)^2 - 3\alpha \beta)= \left( -\frac{-5}{3}\right)\left( \left( -\frac{-5}{3}\right)^2 -3 \left( \frac{-4}{3}\right)\right)=\frac{305}{27}, since \alpha + \beta = -\frac{-5}{3}, and \alpha \beta = \frac{-4}{3}.$

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

Originally Posted by appleibeats
Also does anyone know how to do Question 23 from 8 F.

A piece of string of length l is bent to form the sector of a circle of radius r. Show that the area of the sector is maximised when r = 1/4l
$If we make the sector of radius r, we have two radii, and one arc. Therefore, this arc must have length \ell - 2r, since 2r units of the available \ell units have been used to make the radii. Since the arc length is \ell - 2r and the radius is r, the area of the sector is A=\frac{\text{arc length}}{\text{full circle's circumference}}\times (\text{area of full circle})=\frac{\ell - 2r}{2\pi r}\times \pi r^2 = \frac{\ell - 2r}{2}\times r. Simplify this to the form ar^2 + br + c and you will see that we have now expressed the area of the sector as a quadratic in r whose coefficient of r^2 is negative (which means it represents a concave down parabola), so I trust you will now be able to find which r makes this a maximum (i.e. the r value of the vertex, which is -\frac{b}{2a}).$

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

Thanks for that help.

Also having troubles with 3c) from 8D.

Solve the following simultaneous equations:

x^2 + y^2 -2x + 6y - 35 = 0 and 2x + 3y = 5

How do you get rid of the horrible fractions??

Also 11c) from 8H

If a and b are the roots of the equation 3x^2 + 2x + 7 = 0 , find a + b and ab. Hence form the equation with integer coefficients having roots:

a +2b and b + 2a.

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

Originally Posted by appleibeats
Also 11c) from 8H

If a and b are the roots of the equation 3x^2 + 2x + 7 = 0 , find a + b and ab. Hence form the equation with integer coefficients having roots:

a +2b and b + 2a.

$We note that \alpha + \beta = -\frac{2}{3} and \alpha \beta = \frac{7}{3}.$
$The quadratic with roots A =\alpha + 2\beta and B=2\alpha + \beta has sum of roots 3\alpha + 3\beta = 3(\alpha + \beta)=-2, and product of roots (\alpha + 2\beta)(2\alpha + \beta)=2\alpha^2 + 5\alpha \beta + 2\beta^2 = 2(\alpha^2 + \beta^2)+5\alpha \beta = 2( \left(\alpha + \beta \right)^2 - 2\alpha \beta) + 5\alpha \beta = 2(\alpha + \beta)^2+\alpha \beta = \frac{29}{9}, by substituting \alpha + \beta = -\frac{2}{3} and \alpha \beta = \frac{7}{3}.$

$Since our required equation has \textit{sum of roots}=-2 and \textit{product of roots}=\frac{29}{9}, it could have been x^2 -(-2)x +\frac{29}{9}=0 \Longleftrightarrow x^2 + 2x + \frac{29}{9} = 0. Since we wanted an equation with integer coefficients though, we multiply through by 9, giving us this as our answer: 9x^2 + 18x + 29 =0.$

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

Thanks for that help.

Do you know how to solve Q24 and Q25 from 8H

24)the line x + y -1 = 0 intersects the circle x^2 + y^2 = 13 at A (a1, a2) and B (b1, b2). Without finding the corordinates of A and B, find the length of the chord AB. ( HINT: form a quadratic equation in x and evaluate |a1 - b1| and similarly find |a2 - b2| )

25) for what values of m are the roots of x^2 + 2x + 3 = m(2m +1) real and positive?

Thanks.

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

Originally Posted by appleibeats
Thanks for that help.

Do you know how to solve Q24 and Q25 from 8H

24)the line x + y -1 = 0 intersects the circle x^2 + y^2 = 13 at A (a1, a2) and B (b1, b2). Without finding the corordinates of A and B, find the length of the chord AB. ( HINT: form a quadratic equation in x and evaluate |a1 - b1| and similarly find |a2 - b2| )

25) for what values of m are the roots of x^2 + 2x + 3 = m(2m +1) real and positive?

Thanks.
$24. First of all, note that if (a_1, a_2) is a solution, then so is (a_2,a_1), because the equations are both symmetric in x and y. So we can say b_1 = a_2 and b_2 = a_1. So the distance AB will be \sqrt{(a_1 - b_1)^2 + (a_2 - b_2)^2}=\sqrt{(a_1 - a_2)^2 + (a_2 - a_1)^2}=\sqrt{2(a_1 - a_2)^2}=\sqrt{2}|a_1 - a_2|.$

$Now, x+y -1 = 0 \Rightarrow y = 1-x. Substitute into the circle's equation, then at the points of intersection, x^2 + (1-x)^2 = 13 \Rightarrow x^2 + x^2 - 2x + 1 -13 = 0 \Rightarrow 2x^2 - 2x -12 = 0. The roots of this equation are a_1 and a_2, since these are the x-values of the points of intersections. So using sum and product of roots, a_1 + a_2 = 1, a_1 a_2 = -6.$

$Therefore, (a_1-a_2)^2 = (a_1 + a_2)^2 - 4a_1 a_2 = 1^2 - 4(-6) = 25 \Rightarrow |a_1 - a_2|=\sqrt{(a_1-a_2)^2} = \sqrt{25}=5. Hence AB = \sqrt{2}\cdot 5 = 5\sqrt{2} units.$

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

Thanks for the help.

Do you know how to to 8E Q 26.

OAB is a triangle in which OA is perpendicular to OB. OA and OB have lengths of 60cm and 80cm respectively. A rectangle inscribed inside the triangle so that one of its sides lies along the base OA of the triangle. By using similar triangles find the size of the rectangle of maximum area that may be inscribed in the triangle.

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

Originally Posted by appleibeats
Thanks for the help.

Do you know how to to 8E Q 26.

OAB is a triangle in which OA is perpendicular to OB. OA and OB have lengths of 60cm and 80cm respectively. A rectangle inscribed inside the triangle so that one of its sides lies along the base OA of the triangle. By using similar triangles find the size of the rectangle of maximum area that may be inscribed in the triangle.
Without doing any calculations, my guess is that it's a square.

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

Originally Posted by InteGrand
$24. First of all, note that if (a_1, a_2) is a solution, then so is (a_2,a_1), because the equations are both symmetric in x and y. So we can say b_1 = a_2 and b_2 = a_1. So the distance AB will be \sqrt{(a_1 - b_1)^2 + (a_2 - b_2)^2}=\sqrt{(a_1 - a_2)^2 + (a_2 - a_1)^2}=\sqrt{2(a_1 - a_2)^2}=\sqrt{2}|a_1 - a_2|.$

$Now, x+y -1 = 0 \Rightarrow y = 1-x. Substitute into the circle's equation, then at the points of intersection, x^2 + (1-x)^2 = 13 \Rightarrow x^2 + x^2 - 2x + 1 -13 = 0 \Rightarrow 2x^2 - 2x -12 = 0. The roots of this equation are a_1 and a_2, since these are the x-values of the points of intersections. So using sum and product of roots, a_1 + a_2 = 1, a_1 a_2 = -6.$

$Therefore, (a_1-a_2)^2 = (a_1 + a_2)^2 - 4a_1 a_2 = 1^2 - 4(-6) = 25 \Rightarrow |a_1 - a_2|=\sqrt{(a_1-a_2)^2} = \sqrt{25}=5. Hence AB = \sqrt{2}\cdot 5 = 5\sqrt{2} units.$
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

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

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

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

Originally Posted by InteGrand
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.

13. ## 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!

14. ## 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 ??

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

6N 18. [A rather difficult proof] CAMBRIDGE

16. ## 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.

17. ## 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.

18. ## 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.

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

Originally Posted by appleibeats
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.

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

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

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

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

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

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

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

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

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

20. ## 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.

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

Originally Posted by appleibeats
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...%5E%281%2F5%29

22. ## 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??

Cheers

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

Originally Posted by appleibeats
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??

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

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

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

24. ## 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.

25. ## 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.

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