2 unit maths revision (1 Viewer)

blackops23

Member
did i do it wrong?

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

OmmU

★ BoS Deity ★
I've seen similar questions in 2U exams but it would start off with:

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

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

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

$\bg_white \therefore$ = RHS

Question:

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

Using 4 ordinates.

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ohexploitable

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

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

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

$\bg_white \therefore$ = RHS
lols, now try this:

$\bg_white \int\frac{x+2}{x+1}\,dx$

OmmU

★ BoS Deity ★
lols, now try this:
Oh haha xD I thought we were skipping it Dam not lucky enough:

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

Question:

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

Using 5 ordinates.

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Drongoski

Well-Known Member
Question:

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

Using 4 ordinates.

4 ordinates - is this a trick question?

OmmU

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

Squishxmishyx

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

Using 5 ordinates.
$\bg_white \frac {1}{3}[0+68+2(10)+4(2+30)] = 72$

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

jamesfirst

Active Member
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

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

rawrence

Member
Question
12. Show $\bg_white \frac {d^{2}y}{dx^{2}} = b^{2}y$ for $\bg_white y=ae^{bx}$.
$\bg_white y = ae^{bx}\\\frac{dy}{dx}=bae^{bx}\\\frac{d^2y}{d^2x}=b²ae^{bx}\\\\ \text{Since } y = ae^{bx}\\\frac {d^{2}y}{dx^{2}} = b^{2}y$

AAEldar

Premium Member
$\bg_white y = ae^{bx}\\\frac{dy}{dx}=bae^{bx}\\\frac{d^2y}{d^2x}=b²ae^{bx}\\\\ \text{Since } y = ae^{bx}\\\frac {d^{2}y}{dx^{2}} = b^{2}y$
You left the squared off the b in the 3rd line.

NewiJapper

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

rawrence

Member
You left the squared off the b in the 3rd line.
Oops, just beginning to latex, I tend to forget that

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

Squishxmishyx

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

Drongoski

Well-Known Member
Solve $\bg_white \sqrt {3} tan \theta +1$ for $\bg_white 0\leq \theta \leq2\pi$
you can't

Iruka

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

Squishxmishyx

Olive You.
What you wrote wasn't an equation, hence it does not have a solution.
Woops. Fixed hopefully.

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

SpiralFlex

Well-Known Member
$\bg_white \tan \theta = -\frac{1}{\sqrt{3}}$

Second and last quadrants.

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

Next question taken from 1995 CSSA trials.

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

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

ii. Find the coordinates of any points of inflexion.

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hscishard

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