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Cambridge Prelim MX1 Textbook Marathon/Q&A (4 Viewers)

Thread starter appleibeats
Start date Mar 5, 2015

A

appleibeats

Member

#41

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

Q 10 from 7H:

Sand being poured from a conveyor belt forms a cone with heigh h and semivertical angle 60 degrees. Show that the volume of the pile is V = pih^3 ( can do this), and differentiate with respect to t. (can do this)

a) Suppose that the sand is being poured at a constant rate of 0.3 m^3 /min , and let A be the area of the base. Find the rate at which the height is increasing:

i) when the height is 4 metres, ( can do this), ii) when the radius is 4 metres ( CANT DO THIS)

b) (CANT DO THIS) Show that dA/dt = 6pih dh/dt, and find the rate of increase o the base area at these times.

c) (CANT DO THIS) At what rate mist the sand be poured if it is required that the height increase at 8 cm/min, when the height is 4m?

ALSO

Q11)

An upturned cone of semivertical angle 45 degrees is being filled with water at a constnat rate of 20cm^3 /s. Find the rate at which the height, the area of the water surface, and the area of the cone wetted by the water, are increasing when the height is 50 cm.

Thanks.

A

appleibeats

Member

#42

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

Hi, worked out the previous question.

Have another one though.

Question 14 from 7I.

Find zeroes and discontinuities of:

a) y = cosx + sinx / cosx - sinx

b) y = cosx - sinx / cosx + sinx

Thanks

P

princesssuku99

New Member

#43

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

hey, does anyone know how to do question 17 and 20 from 2G?

A

Ambility

Active Member

#44

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

appleibeats said:
Hi, worked out the previous question.

Have another one though.

Question 14 from 7I.

Find zeroes and discontinuities of:

a) y = cosx + sinx / cosx - sinx

b) y = cosx - sinx / cosx + sinx

Thanks

Seeing as the problem doesn't give a domain, and the functions go on forever, we need to represent the answer as a general solution because there is going to be infinitely many possible solutions. These fractions will have zeros when the numerator is equal to zero ("zero divided by anything is zero") and will have discontinuities when the denominator is equal to zero ("dividing by zero is undefined"). With that in mind, let's work on 14a:

$\text{Where is the numerator equal to zero?}\\\cos x+\sin x=0\\\text{Divide by }\cos x\\\tan x=-1\\x=135^\circ, 315^\circ, 495^\circ, 675^\circ...\\x=135^\circ+180n^\circ, \text{where }n\in \mathbb{Z}\\\\\text{Where is the denominator equal to zero?}\\\cos x-\sin x=0\\\text{Divide by }\cos x\\\tan x-1=0\\\tan x=1\\x=45^\circ, 225^\circ, 405^\circ, 585^\circ...\\x=45^\circ+180n^\circ, \text{where }n\in \mathbb{Z}$

You can use the answers from 14a. to work out 14b. 14b's denominator is the same as 14a's numerator, and 14b's numerator is the same as 14a's denominator.

Last edited: May 29, 2015

A

appleibeats

Member

#45

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

Thanks for that help.

Also I don't quite understand how to get the answers to the questions in 4 in 7J.

I know how to sketch them and their continuity, but am unsure about how to find where it is not differentiable.

eg, question 4a)

y = |x + 2 |

Also, question 7d from chapter 8A.

write down the general form of a monic quadratic for which one of the zeroes is x = 1. (I know how to do this.) Then find the equation of such a quadratic in which :

d) the curve passes through (3,9).

Thank you.

A

Ambility

Active Member

#46

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

appleibeats said:
Thanks for that help.

Also I don't quite understand how to get the answers to the questions in 4 in 7J.

I know how to sketch them and their continuity, but am unsure about how to find where it is not differentiable.

eg, question 4a)

y = |x + 2 |

Also, question 7d from chapter 8A.

write down the general form of a monic quadratic for which one of the zeroes is x = 1. (I know how to do this.) Then find the equation of such a quadratic in which :

d) the curve passes through (3,9).

Thank you.

I have a method to solve 8A 7d, but it's hardly ideal. I haven't studied that chapter so I recommend asking a teacher.

As for 7J 4a, the graph is not differentiable at a point where there is a sharp turn. This is because there are multiple tangent lines which can be drawn to the graph, multiple slopes of those tangent lines, multiples derivatives, no defined derivative. If you look at the graph of $y=|x+2|$ , there is a point which takes a sudden turn at x=-2. So it's not differentiable at x=-2.

I

InteGrand

Well-Known Member

#47

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

appleibeats said:
Thanks for that help.

Also I don't quite understand how to get the answers to the questions in 4 in 7J.

I know how to sketch them and their continuity, but am unsure about how to find where it is not differentiable.

eg, question 4a)

y = |x + 2 |

Also, question 7d from chapter 8A.

write down the general form of a monic quadratic for which one of the zeroes is x = 1. (I know how to do this.) Then find the equation of such a quadratic in which :

d) the curve passes through (3,9).

Thank you.

Q 7d) from Chapter 8A:

$The general form of the monic quadratic is $Q(x) = A(x-1)(x-a)$. If we want such a quadratic to go through $(3,9)$, since we just need a particular quadratic, we can just let $a$ be some convenient value like 0 (so $Q(x) = Ax(x-1)$), and then we solve for $A$ by substituting in $x=3$ and $Q(3)=9$: $9 = A\times 3 \times (3-1) \Rightarrow 9 = 6A\Rightarrow A = \frac{3}{2}$. So a suitable quadratic is $Q(x)=\frac{3}{2}x(x-1)$.$

Last edited: Jun 2, 2015

A

Ambility

Active Member

#48

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

InteGrand said:
Q 7d) from Chapter 8A:

$The general form of the monic quadratic is $Q(x) = A(x-1)(x-a)$. If we want such a quadratic to go through $(3,9)$, since we just need a particular quadratic, we can just let $a$ be some convenient value like 0 (so $Q(x) = Ax(x-1)$), and then we solve for $A$ by substituting in $x=3$ and $Q(3)=9$: $9 = A\times 3 \times (3-1) \Rightarrow 9 = 6A\Rightarrow A = \frac{3}{2}$. So a suitable quadratic is $Q(x)=\frac{3}{2}x(x-1)$.$

There does exist other suitable quadratics though, right? I came up with $y=-\frac{9}{4}(x-5)(x-1)$ .

A

appleibeats

Member

#49

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

The answer in the text book for that question is y = 1/2 (2x + 3)(x - 1)

Still unsure about how to get this answer.

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 .

Also Question 27 from the same chapter.

A rectangle is inscribed in an isosceles triangle with one of the sides of the rectangle on the base of the triangle. Prove that the rectangle of greatest area occupies half the area of the triangle.

Thanks.

I

InteGrand

Well-Known Member

#50

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

Ambility said:
There does exist other suitable quadratics though, right? I came up with $y=-\frac{9}{4}(x-5)(x-1)$ .

Yep there are infinitely many possible quadratics we could use.

The general form was $Q(x) = A(x-a)(x-1), A \neq 0$ . In order for the point (3, 9) to lie on the curve, we just need to choose a and A (with $A\neq 0$ ) satisfying $9=A(3-a)(3-1)\Longleftrightarrow A = \frac{\frac{9}{2}}{3-a}, a\neq 3$ , and there are clearly an infinite number of pairs $(a,A)$ satisfying this condition.

A

appleibeats

Member

#51

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.

I

InteGrand

Well-Known Member

#52

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

appleibeats said:
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 $\alpha \beta = \frac{-4}{3}$.$

Last edited: Jun 6, 2015

I

InteGrand

Well-Known Member

#53

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

appleibeats said:
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 $-\frac{b}{2a}$).$

A

appleibeats

Member

#54

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.

Thank for all your help.

I

InteGrand

Well-Known Member

#55

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

appleibeats said:
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.

Thank for all your help.

$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: $9x^2 + 18x + 29 =0$.$

Last edited: Jun 8, 2015

A

appleibeats

Member

#56

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.

I

InteGrand

Well-Known Member

#57

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

appleibeats said:
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.$

A

appleibeats

Member

#58

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.

I

InteGrand

Well-Known Member

#59

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

appleibeats said:
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.

A

appleibeats

Member

#60

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

InteGrand said:
$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

Last edited: Jun 9, 2015

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