Cambridge Prelim MX1 Textbook Marathon/Q&A (2 Viewers)

appleibeats · Aug 7, 2015

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

12a)

Give the two functions f(x) = (x + 1)(x - 1)(x - 3) and g(x) = (x - 1)(x + 1), for what value of x is f(x) > g(x)?

b) sketch a graph of the two funcitons on the same number plane and find the magnitude of the area between them.

I can do a) and have sketch part b)..

I found the area of the left hand region to be 13/2 units^2

and the region to the right to be 63/4 units^2

Adding these together I get 89/4 = 22 1/4 units^2

Which is wrong apparently as the answer says 21 1/12 uints^2

Not sure what I have done wrong??

rand_althor · Aug 7, 2015

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

The region on the right you have calculated correctly, but your left region is wrong. f(x) is above g(x) for the left region (x=-1 to x=1), so the working is as follows:

$\begin{align*}A&=\int^1_{-1}(x-3)(x^2-1)-(x^2-1)\, \mathrm{d}x \\&=\int^1_{-1}x^3-x-3x^2+3-x^2+1\, \mathrm{d}x \\&=\int^1_{-1}x^3-4x^2-x+4\, \mathrm{d}x \\&=\left [\frac{x^4}{4}-\frac{4x^3}{3} -\frac{x^2}{2}+4x\right ]^1_{-1} \\&= \left (\frac{1}{4}-\frac{4}{3}-\frac{1}{2}+4 \right ) - \left (\frac{1}{4}+\frac{4}{3}-\frac{1}{2}-4 \right ) \\&=\frac{29}{12}+\frac{35}{12} \\&=\frac{16}{3} \textup{ units}^2\end{align*}$

$$The total area is$:\frac{16}{3}+\frac{63}{4}=21\frac{1}{12} $ units$^2$

appleibeats · Aug 8, 2015

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

A bowl is formed by rotating the rea bounded by y = x^4 and the line y = h about the y axis.

i) FInd the volume of the bowl in terms of h

ii) Hence, find the height of a bowl made to hold a maximum of 2.

Is the answer to i)

V = pi integral from o to h of y^1/2 dy

Is that in terms of h?? Or do I need to replace the y with h??

Unsure about the answer to ii) is it h = cube root of 9/ pi^2 ??

InteGrand · Aug 8, 2015

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

appleibeats said:
A bowl is formed by rotating the rea bounded by y = x^4 and the line y = h about the y axis.

i) FInd the volume of the bowl in terms of h

ii) Hence, find the height of a bowl made to hold a maximum of 2.

Is the answer to i)

V = pi integral from o to h of y^1/2 dy

Is that in terms of h??

Yes, that's the right integral (then, evaluate that integral and the answer will be in terms of h). Remember, the variable of integration is not present in the final answer. I.e. $\int _a ^b f(x)\text{ d}x$ is the same as $\int _a ^b f(y)\text{ d}y$ or $\int _a ^b f(\psi)\text{ d}\psi$ or $\int _a ^b f(\Omega)\text{ d}\Omega$ etc. (it's just a dummy variable). Just avoid using a variable of integration that's a variable that is also present in the limits of integration, as this can cause confusion.

For part (ii), use your answer to part (i) and equate it to 2, then solve for h. This will tell you the value of h that gives a bowl who's capacity is 2 cubic units, which is exactly what the Q is asking for.

appleibeats · Aug 10, 2015

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

A cone is generated by rotating the line y = 2x about the x - axis between x = 0 and x = 3 Using the trapezoidal rule with four funciton values, eestimate the volume of the cone.

I know how to use trapezoidal rule with Integrals. but how do you go about with Volumes.
The answer is 38 pi

InteGrand · Aug 10, 2015

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

appleibeats said:
I know how to use trapezoidal rule with Integrals. but how do you go about with Volumes.

It's the same. Set up an integral for the volume, and apply the trapezoidal rule to this integral.

appleibeats · Aug 10, 2015

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

InteGrand said:
It's the same. Set up an integral for the volume, and apply the trapezoidal rule to this integral.

I have

V = pi Integral from 0 to 3 4x^2 dx

h = 1

when x = 0, 1 , 2, 3
y^2 = 0 , 4, 16, 36

V = 1/2 (0 + 16 + 32+ 36) = 42

Answer is meant to be 38 pi..

Where did I go wrong??

appleibeats · Aug 10, 2015

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

I got it . I was confusing my self with Simpsons rule

It should be instead of above:

V = 1/2 ( 0 + 8 + 32 + 36) pi

= 38 pi

appleibeats · Aug 10, 2015

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

Not sure ewhy I am getting evaluate Ingtegral from -3 to 3 (9 - x^2)^1/2 correct to three decimal places wrong.

I get the answer 36

but the book say 14.137

Isn't it a normal integration

so

The integral = 2 Integral from 0 to 3 (9 - x^2)^1/2 dx

= 2[ - 2/3 (9 - x^2)^3/2 ] 0 to 3

= 2 [ 0 - (-18) ]

= 36

Where have I gone wrong??

InteGrand · Aug 10, 2015

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

appleibeats said:
Not sure ewhy I am getting evaluate Ingtegral from -3 to 3 (9 - x^2)^1/2 correct to three decimal places wrong.

I get the answer 36

but the book say 14.137

Isn't it a normal integration

so

The integral = 2 Integral from 0 to 3 (9 - x^2)^1/2 dx

= 2[ - 2/3 (9 - x^2)^3/2 ] 0 to 3

= 2 [ 0 - (-18) ]

= 36

Where have I gone wrong??

No, because the function being raised to the power 1/2 is not a linear function.

To evaluate that integral, note that the integral $\int _0 ^3 (9-x^2)^{\frac{1}{2}}\text{ d}x$ is the area of a quadrant of a circle of radius 3 (or the original integral is the area of the semicircle).

rand_althor · Aug 10, 2015

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

It isn't a normal integration like that. To do the question, you should realise that $y=\sqrt{9-x^2}$ is the equation of a semicircle with a radius of 3 units and centre (0,0). So the area from x=-3 to x=3, is the area of the semicircle, or half the area of a complete circle with radius 3.

$\begin{align*} A&=\int^3_{-3} \sqrt{9-x^2} \, \mathrm{d}x \\ &=\frac{\pi \times 3^2}{2} \\ &\approx 14.137 \textup{ units}^2 \end{align*}$

appleibeats · Aug 11, 2015

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

An object is moving along the x - axis with values of the velocity v in m/s at time t given in the table below. Given that the distance travelled may be found by calculating the area under a velocity / time graph, use Simpson's rule to estimate the distance travelled by the particle in the first four seconds.

t 0 1 2 3 4
v 1.5, 1.3 , 1.4 , 2.0 , 2.4

rand_althor · Aug 11, 2015

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

$\begin{align*}d&\approx\frac{h}{3} \left [\textup{First} + \textup{Last} + 4(\textup{Odd}) + 2(\textup{Even}) \right ] \\&=\frac{1}{3} \left [1.5+2.4+4(1.3+2)+2(1.4) \right ] \\&=\frac{1}{3} \times \frac{199}{10} \\&=\frac{199}{30} \\&\approx 6.63 \textup{ metres}\end{align*}$

appleibeats · Aug 11, 2015

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

So confused. I thought the number in the denominator is always 6. That what it shows in the cambridge textbook. How come you use 3? I have been using 6 for all the questions and have got them right. That probably the reason I couldn't get the question.

rand_althor · Aug 11, 2015

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

$$Okay let's focus on the part outside the brackets, that is $\frac{h}{3}$. $h$ is the height of one of the intervals. The height is also the average of the first x-coordinate of one interval, and the 2nd x-coordinate from it, which means $h=\frac{b-a}{2}$. If we sub this value for $h$ into $\frac{h}{3}$, we get $\frac{b-a}{6}$. So Simpson's rule can be expressed with either $\frac{h}{3}$ or $\frac{b-a}{6}$. Using your question as an example$:\\d=\frac{b-a}{6} \left [\textup{First} + \textup{Last} + 4(\textup{Odd}) + 2(\textup{Even}) \right ] \\=\frac{2-0}{6} \left [1.5+2.4+4(1.3+2)+2(1.4) \right ] \\=\frac{1}{3}\times \frac{199}{10}$

Also can someone please correct me if I'm wrong.

InteGrand · Aug 11, 2015

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

appleibeats said:
So confused. I thought the number in the denominator is always 6. That what it shows in the cambridge textbook. How come you use 3? I have been using 6 for all the questions and have got them right. That probably the reason I couldn't get the question.

There's a 3 on the denominator simply because it's (2 – 0)/6 (doing it your way with the 6), because each parabola (there are two of them) is over a 2 second interval, i.e. from t=0 to t=2, and from t=2 to t=4. This ((2 – 0)/6) simplifies to 1/3.

(Basically, the 'h' used by rand_althor does not refer to 'b – a'. It refers to half of this. You should just learn one method of Simpson's rule and stick to it, otherwise it can get unnecessarily confusing.)

appleibeats · Aug 12, 2015

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

use log laws to differnentiate :

ln ( (1 + x ) / ( 1 - x) )

I get the answer :

1 / 1 + x - 1/ 1 - x

Answer says Instead of a negative in between the two it its a +.

Not sure why??

InteGrand · Aug 12, 2015

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

appleibeats said:
use log laws to differnentiate :

ln ( (1 + x ) / ( 1 - x) )

I get the answer :

1 / 1 + x - 1/ 1 - x

Answer says Instead of a negative in between the two it its a +.

Not sure why??

$\frac{\mathrm{d}}{\mathrm{d}x}\left(\ln \left(\frac{1+x}{1-x} \right) \right)=\frac{\mathrm{d}}{\mathrm{d}x}\left(\ln (1+x)-\ln (1-x) \right)$

$=\frac{\mathrm{d}}{\mathrm{d}x}\left(\ln (1+x)\right)- \frac{\mathrm{d}}{\mathrm{d}x}\left(\ln (1-x) \right)$

$$=\frac{1}{1+x}- \frac{-1}{1-x}$ (since $\frac{\mathrm{d}}{\mathrm{d}x}\left(\ln (1\pm x)\right)=\frac{\pm 1}{1\pm x}$)$

$$=\frac{1}{1+x}+ \frac{1}{1-x}$.$

Crisium · Aug 12, 2015

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

appleibeats said:
use log laws to differnentiate :

ln ( (1 + x ) / ( 1 - x) )

I get the answer :

1 / 1 + x - 1/ 1 - x

Answer says Instead of a negative in between the two it its a +.

Not sure why??

I think it's because when you separate it into ln(1 + x) - ln(1 - x) you forgot that when you integrate the bolded part it is in the form f'(x) / f(x) and so the numerator should be -1 which should cancel out the negative above to give you a positive

I think markers also prefer if you leave the answer under the one denominator as well (because it can easily be simplified in this case)

appleibeats · Aug 13, 2015

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

Don't understand the graphing of y = log|x|

The answer have the typical log graph and then its reflection in the y axis. Why is that reflection included.

Also Logs can't be can be negative and positive, just not ZERO right?

Cambridge Prelim MX1 Textbook Marathon/Q&A (2 Viewers)

Member

Active Member

Member

Well-Known Member

Member

Well-Known Member

Member

Member

Member

Well-Known Member

Active Member

Member

Active Member

Member

Active Member

Well-Known Member

Member

Well-Known Member

Pew Pew

Member

Users Who Are Viewing This Thread (Users: 0, Guests: 2)