2 unit maths revision (3 Viewers)

blackops23 · Mar 10, 2011

did i do it wrong?

Answ: x + ln(x+1) + C

OmmU · Mar 12, 2011

ohexploitable said:
I've seen similar questions in 2U exams but it would start off with:

$\text{Show: }1+\frac{1}{x+1}=\frac{x+2}{x+1}$

RHS:
$= 1+\frac{1}{x+1}$

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

$= \frac{x + 2}{x+1}$

$\therefore$ = RHS

Question:

Use Simpson's rule to find an approximation for:
$\int_{0}^{4} (x^3 + x)dx$

Using 4 ordinates.

ohexploitable · Mar 12, 2011

OmmU said:
RHS:
$= 1+\frac{1}{x+1}$

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

$= \frac{x + 2}{x+1}$

$\therefore$ = RHS

lols, now try this:

ohexploitable said:
$\int\frac{x+2}{x+1}\,dx$

OmmU · Mar 12, 2011

ohexploitable said:
lols, now try this:

Oh haha xD I thought we were skipping it

Dam not lucky enough:

$x + ln(x+1) + c$

Question:

Use Simpson's rule to find an approximation for:
$\int_{0}^{4} (x^3 + x)dx$

Using 5 ordinates.

Drongoski · Mar 12, 2011

OmmU said:
Question:

Use Simpson's rule to find an approximation for:
$\int_{0}^{4} (x^3 + x)dx$

Using 4 ordinates.

4 ordinates - is this a trick question?

OmmU · Mar 13, 2011

Sorry, fixed to 5. Think I would have noticed after the first time xD

Squishxmishyx · Mar 13, 2011

OmmU said:
Use Simpson's rule to find an approximation for:
$\int_{0}^{4} (x^3 + x)dx$

Using 5 ordinates.

$\frac {1}{3}[0+68+2(10)+4(2+30)] = 72$

Question
12. Show $\frac {d^{2}y}{dx^{2}} = b^{2}y$ for $y=ae^{bx}$ .

jamesfirst · Mar 13, 2011

Hey guys, How do you integrate tanx ????

There was this question, it said to use the simpsons rule to find the area for tan x. But part ii) said compared it with the actual integration of tan x... >.>

ohexploitable · Mar 13, 2011

jamesfirst said:
Hey guys, How do you integrate tanx ????

There was this question, it said to use the simpsons rule to find the area for tan x. But part ii) said compared it with the actual integration of tan x... >.>

$\int\tan x\,dx=\int\frac{\sin x}{\cos x}\,dx=-\int\frac{-\sin x}{\cos x}=-\log{(\cos x)}$

rawrence · Mar 14, 2011

Squishxmishyx said:
Question
12. Show $\frac {d^{2}y}{dx^{2}} = b^{2}y$ for $y=ae^{bx}$ .

$y = ae^{bx}\\\frac{dy}{dx}=bae^{bx}\\\frac{d^2y}{d^2x}=b²ae^{bx}\\\\ \text{Since } y = ae^{bx}\\\frac {d^{2}y}{dx^{2}} = b^{2}y$

AAEldar · Mar 15, 2011

rawrence said:
$y = ae^{bx}\\\frac{dy}{dx}=bae^{bx}\\\frac{d^2y}{d^2x}=b²ae^{bx}\\\\ \text{Since } y = ae^{bx}\\\frac {d^{2}y}{dx^{2}} = b^{2}y$

You left the squared off the b in the 3rd line.

NewiJapper · Mar 15, 2011

y=(x-1)/(x+1) Find x. Yawwwwwwwwwwwwwn.

rawrence · Mar 15, 2011

AAEldar said:
You left the squared off the b in the 3rd line.

Oops, just beginning to latex, I tend to forget that

$y = ae^{bx}\\\frac{dy}{dx}=bae^{bx}\\\frac{d^2y}{d^2x}=b^2ae^{bx}\\\\ \text{Since } y = ae^{bx}\\\frac {d^{2}y}{dx^{2}} = b^{2}y$

Squishxmishyx · Apr 22, 2011

Solve $\sqrt {3} tan \theta +1$ for $0\leq \theta \leq2\pi$

Drongoski · Apr 22, 2011

Squishxmishyx said:
Solve $\sqrt {3} tan \theta +1$ for $0\leq \theta \leq2\pi$

you can't

Squishxmishyx · Apr 23, 2011

Drongoski said:
you can't

Why? I got the answer now -.-.

Iruka · Apr 23, 2011

What you wrote wasn't an equation, hence it does not have a solution.

Squishxmishyx · Apr 23, 2011

Iruka said:
What you wrote wasn't an equation, hence it does not have a solution.

Woops. Fixed hopefully.

Squishxmishyx said:
Solve $\sqrt {3} tan \theta +1 =0$ for $0\leq \theta \leq2\pi$

SpiralFlex · Apr 23, 2011

$\tan \theta = -\frac{1}{\sqrt{3}}$

Second and last quadrants.

$\therefore \theta = \frac{5\pi }{6}$ and $\frac{11\pi}{6}$

Next question taken from 1995 CSSA trials.

Question 4: The function $y= x^3 - 3x^2 - 9x + 1$ is defined in the domain $-4 \leq x \leq 5$

i. Find the coordinates of any turning points and determine their nature.

ii. Find the coordinates of any points of inflexion.

hscishard · Apr 23, 2011

^ lol
Imagine a paper with just those types of questions...you'll have blank writing booklets XD

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