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

braintic · Sep 19, 2015

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

appleibeats said:
so I let t = tan ( x/2)

Domain 0< x/2 < 180 ( less than and equal to)

then use t formula

1 + t^2 / 1 - t^2 + 2t / 1 + t^2 = 1 - t^2/ 1 + t^2

1 + t^2 + 2t / 1 - t^4 = 1 - t^2 / 1 + t^2

Then I cross multiply and expand and simplify to get

t^6 - 2t^4 - 2t^3 - 3t^2 - 2t = 0

Not sure where to go from here.

Looks terribly wrong.

When you multiply through by (1-t²) (1+t²), you can't get a t^6 term.

appleibeats · Sep 19, 2015

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

rand_althor said:
$\begin{align*}\sec{x}+\sin{x}&=\cos{x} \\\frac{1+t^2}{1-t^2}+\frac{2t}{1+t^2}&=\frac{1-t^2}{1+t^2} \\\textup{Multiply through by }(1-t^2)&(1+t^2) \\(1+t^2)^2+2t(1-t^2)&=(1-t^2)^2 \\(1+t^2)^2-(1-t^2)^2+2t(1-t^2)&=0 \\\textup{Since }a^2-b^2&=(a-b)(a+b) \\(1+t^2+1-t^2)(1+t^2-(1-t^2))+2t(1-t^2)&=0 \\2(2t^2)+2t(1-t^2)&=0 \\2t(2t+1-t^2)&=0 \\t(t^2-2t-1)&=0 \\\end{align*}$

so t = 0 or t = t^2 - 2t - 1

so

tan x/2 = 0 or (tanx/2)^2 - 2(tanx) - 1 = 0 for 0 < x < 180

x = 0 or not sure how to solve this part.

rand_althor · Sep 19, 2015

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

$\begin{align*}t^2-2t-1&=0 \\\textup{Completing the square:}& \\t^2-2t+1&=2 \\(t-1)^2&=2 \\t-1&=\pm \sqrt{2} \\t&=1 \pm \sqrt{2} \\\end{align*}$

If you solve this using your calculator, you get $\frac{x}{2}=-\frac{\pi}{8},\frac{3\pi}{8}$ . However, since these are not standard values, you need to prove it. Look at this post to see a possible method.

appleibeats · Sep 19, 2015

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

Find an approximation of Integral from 1 to 3 of dx/ x^2 + 1 using the trapezoidal rule with 3 subintervals

I am unsure about the numbers to use as it doesn't evenly split for 3 subintervals.

dan964 · Sep 19, 2015

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

x values of 1, 5/3, 7/3, 3
and their function values.
(gaps of 2/3)

appleibeats · Sep 19, 2015

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

Use the substitution u = x - 2 to evaluate Integral from 2 to 3 of x (x - 2)^1/2 dx giving your answer correct to 2 decimal places

= approx 1.73

ii)

Suppose that the area of a particular function is given by the integral above, Use the simpson rule with 3 functional values and the trapezoidal rule with 5 functional values to estimate the area, and determine which gives the better aprrox.

So using the trapezoidal rule

Integral = approx 1.69358

Simpsons Rule:

Integral = approx 1.67851

Therefore Trapezoidal Rule give the better approx.

I was marked wrong for my trapezoidal rule and my final conclusion but right for the simpsons rule. Not sure where I have went wrong, I checked and I don't seem to see any mistake made.

appleibeats · Sep 19, 2015

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

Find the general solution to the equation cosx = cos3x

rand_althor · Sep 19, 2015

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

$\begin{align*}\cos{x}&=\cos{3x} \\\cos{x}&=4\cos^3{x}-3\cos{x} \, (*)\\4\cos^3{x}-4\cos{x}&=0 \\\cos{x} \left (\cos^2{x}-1 \right )&=0 \\\cos{x}&= 0 \to x=2n\pi \pm \cos^{-1}{0}=2n\pi \pm \frac{\pi}{2} \\\cos^2{x}&=1 \\\cos{x}&=1\to x=2n\pi \pm \cos^{-1}{1}=2n\pi \textup{ or} \\\cos{x}&=-1\to x=2n\pi \pm \cos^{-1}{(-1)}=2n\pi \pm \pi \\\end{align*}$

* Check this post if you are unsure how to get there.

appleibeats · Sep 19, 2015

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

The lines y = sinx and y = 3/5pi intersect at the origin and at A .

i) Show that x = 5pi/ 6 is a root to the equation f(x) = sinx - 3x/5pi

braintic · Sep 19, 2015

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

rand_althor said:
$\begin{align*}\cos{x}&=\cos{3x} \\\cos{x}&=4\cos^3{x}-3\cos{x} \, (*)\\4\cos^3{x}-4\cos{x}&=0 \\\cos{x} \left (\cos^2{x}-1 \right )&=0 \\\cos{x}&= 0 \to x=2n\pi \pm \cos^{-1}{0}=2n\pi \pm \frac{\pi}{2} \\\cos^2{x}&=1 \\\cos{x}&=1\to x=2n\pi \pm \cos^{-1}{1}=2n\pi \textup{ or} \\\cos{x}&=-1\to x=2n\pi \pm \cos^{-1}{(-1)}=2n\pi \pm \pi \\\end{align*}$

* Check this post if you are unsure how to get there.

Why don't you combine those last 2 answers into just x = nπ

braintic · Sep 19, 2015

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

appleibeats said:
Find the general solution to the equation cosx = cos3x

We don't need triple angle formulae for this:

3x = x + 2kπ
2x = 2kπ
x = kπ

OR

3x = (2π - x) + 2kπ
4x = 2(k+1)π
x = (k+1)π/2
which is identical to:
x = kπ/2

This answer includes all answers from the first solution.
So the simplest answer is:
x = kπ/2

appleibeats · Sep 20, 2015

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

The lines y = sinx and y = 3/5pi intersect at the origin and at A .

i) Show that x = 5pi/ 6 is a root to the equation f(x) = sinx - 3x/5pi

braintic · Sep 21, 2015

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

appleibeats said:
The lines y = sinx and y = 3/5pi intersect at the origin and at A .

i) Show that x = 5pi/ 6 is a root to the equation f(x) = sinx - 3x/5pi

just substitute .....

appleibeats · Oct 6, 2015

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

Use a graphical approach to determine the number of positive solutions of sin x = x / 200.

kawaiipotato · Oct 6, 2015

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

sinx = x/200
200sinx = x

Sketch y = 200sinx
Sketch y = x
Both on the same cartesian plane.
The amount of solutions will be evident

appleibeats · Oct 6, 2015

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

kawaiipotato said:
sinx = x/200
200sinx = x

Sketch y = 200sinx
Sketch y = x
Both on the same cartesian plane.
The amount of solutions will be evident

So doing this I get the answer 3. But the answer in the textbook is 63. Still unsure how to do the question.

Zlatman · Oct 6, 2015

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

appleibeats said:
So doing this I get the answer 3. But the answer in the textbook is 63. Still unsure how to do the question.

Are you using radians for sin x?

EDIT: Once y > 200, x and sin x will have no more solutions. Therefore, we're looking at the domain of x < 200 (since y = x, and y must be less than 200). In this domain, sin x completes 31.8 cycles (200 divided by 2pi), and y = x passes through sin x twice in the first half of each cycle. This gives us 64 solutions. But x = 0 is included in this, and that is not positive; hence we have 63 solutions.

I hope that makes some sense.

appleibeats · Oct 6, 2015

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

a) carefully sketch the curve y = sin^2 x for 0 < or equal to x < or equal to 2pi

b) Explain why y = sin^2x has range 0 </= y </= 1

c) Write down the period and amplitude of y = sin^2 x .

InteGrand · Oct 6, 2015

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

appleibeats said:
a) carefully sketch the curve y = sin^2 x for 0 < or equal to x < or equal to 2pi

b) Explain why y = sin^2x has range 0 </= y </= 1

c) Write down the period and amplitude of y = sin^2 x .

a) http://www.graphsketch.com/?eqn1_co..._lines=1&line_width=4&image_w=850&image_h=525

$b) Since $\sin x$ is between -1 and 1, $\sin^2 x$ is between 0 and 1.$

$c) Recall that $\sin^2 x = \frac{1}{2}-\frac{1}{2}\cos 2x$, so the amplitude is $\frac{1}{2}$ and the period is $\pi$.$

appleibeats · Oct 7, 2015

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

I have sketched y = sinx and y = cosx, for 0 </= x </= 2pi

But am unsure how to sketch y = sinx + cosx for 0 </= x </= 2pi.

Then how do you calculate the period and amplitude of y = sinx + cosx

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

Well-Known Member

Member

Active Member

Member

what

Member

Member

Active Member

Member

Well-Known Member

Well-Known Member

Member

Well-Known Member

Member

Well-Known Member

Member

Member

Member

Well-Known Member

Member

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