Official BOS Trial 2015 Thread (3 Viewers)

Carrotsticks · Sep 19, 2015

Here's a question scrapped from 2U because it was too easy.

$$Solve $ \left | 1 - |x| \right | \geq 1$

Gabriel Moussa · Sep 19, 2015

I shall be attending

Thankyou carrot, PM sent

jono_jonoson · Sep 19, 2015

Carrotsticks said:
Here's a question scrapped from 2U because it was too easy.

$$Solve $ \left | 1 - |x| \right | \geq 1$

Is the answer to this: x < or = -2, x > or = 2

Carrotsticks · Sep 19, 2015

jono_jonoson said:
Is the answer to this: x < or = -2, x > or = 2

Almost.

You feel for the trap I had in mind when making this Q.

jono_jonoson · Sep 19, 2015

Carrotsticks said:
Almost.

You feel for the trap I had in mind when making this Q.

lol.

x < or = -2, x > or = 2, x = 0 ??

Carrotsticks · Sep 19, 2015

jono_jonoson said:
lol.

x < or = -2, x > or = 2, x = 0 ??

There you go =)

I'm curious, did you solve it out algebraically? Or did you perhaps find the solution graphically?

jono_jonoson · Sep 19, 2015

Carrotsticks said:
There you go =)

When I try to do the questions as fast as possible, I always get 'tricked' by those fine details.
I just trust that hopefully when I check my work at the end of my exam, I'll pick up on those little things that may have 'tricked' me.

jono_jonoson · Sep 19, 2015

Carrotsticks said:
There you go =)

I'm curious, did you solve it out algebraically? Or did you perhaps find the solution graphically?

algebraically, I just forgot to include the zero at the end.

I set it equal to zero, and considered both cases (minus and plus). Then did it again (in one of them as the other was equal to zero) as there was an absolute value still.

Then tested a point between -2 and 2 (not zero obviosuly). Didn't work so It had to be on the outside.

Drsoccerball · Sep 19, 2015

ahhaahah i thought you were joking because the latex didnt load... Carrot is question 16 like 2012 question 16 or like others plz dont make it like 2012

Drsoccerball · Sep 19, 2015

Carrotsticks said:
Here's a question that was scrapped from 3U because it was too mark-heavy.

$$By using the substitution $ u=x^2+\frac{1}{x^2}$ , or otherwise, evaluate $ \int \frac{x^2-1}{\left ( x^2+1 \right ) \sqrt{x^4+1}}dx$

Wasnt this a question on the 2015 integration marathon ?

Trebla · Sep 19, 2015

Drsoccerball said:
ahhaahah i thought you were joking because the latex didnt load... Carrot is question 16 like 2012 question 16 or like others plz dont make it like 2012

lol I thought that was the easiest Q16 of the lot

Drsoccerball · Sep 19, 2015

Trebla said:
lol I thought that was the easiest Q16 of the lot

I looked at it and felt i couldnt do any parts...

Drsoccerball · Sep 20, 2015

Carrotsticks said:
Here's a question scrapped from 4U because it was too easy.

$\\ $Two cars $ A $ and $ B $ move around a banked track with radius $ r_1 $ and $ r_2 $ with banking angles $ \alpha $ and $ \beta $ respectively such that $ \alpha + \beta = \frac{\pi}{2} . \\\\ $Let the velocities such that car $ A $ and $ B $ experience no lateral forces be $ V_A $ and $ V_B $ respectively.$ \\\\ $ Show that $ r_1r_2 = \left ( \frac{V_A V_B}{g} \right )^2$

For those who cant do it:

$$ Horizontal: $ \frac{mv_{a}^2}{r_1} = Nsin\alpha$

$$ Vertical: $ mg=Ncos\alpha$

$tan\alpha = \frac{{v_a}^2}{r_1g}$

$$ similarly $ \frac{1}{tan\alpha}= \frac{{v_b}^2}{r_2 g}$

$\therefore r_1 r_2 = (\frac{v_a v_b}{g})^2$

[]

Drsoccerball · Sep 20, 2015

Carrotsticks said:
Here's a question that was scrapped from 3U because it was too mark-heavy.

$$By using the substitution $ u=x^2+\frac{1}{x^2}$ , or otherwise, evaluate $ \int \frac{x^2-1}{\left ( x^2+1 \right ) \sqrt{x^4+1}}dx$

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

$dx= \frac{x^3 du}{2(x^4-1)}$

$I= \int{\frac{x^3 du}{2(x^2+1)^2 \sqrt{x^4+1}}} $ By dividing both top and bottom by $ x^3$

$I= \int{\frac{du}{2\sqrt{u^2+2u}}}$

$I= \frac{1}{2} ln(x^2+ \frac{1}{x^2} +1 + \sqrt{x^4+\frac{1}{x^4} +2 + 2x^2 + \frac{2}{x^2}})$

RealiseNothing · Sep 20, 2015

Will be coming to this again for the fourth year straight

Pretty much acting as a control now to see difficulty over the years.

Drsoccerball · Sep 20, 2015

RealiseNothing said:
Will be coming to this again for the fourth year straight

Pretty much acting as a control now to see difficulty over the years.

Ekman told me that you'll be shouting lunch if we get over 70 ?

RealiseNothing · Sep 20, 2015

Drsoccerball said:
Ekman told me that you'll be shouting lunch if we get over 70 ?

Sure, I made the bet last year and no one got over 70 so why not.

BOS Secret Service · Sep 20, 2015

RealiseNothing said:
Sure, I made the bet last year and no one got over 70 so why not.

getting over 70 in Ext 2?

Drsoccerball · Sep 20, 2015

BOS Secret Service said:
getting over 70 in what, Ext 1 or Ext 2?

Extension 2. Tbh Carrot should troll realisenothing and make the exam super easy so he has to shout 60 people lunch

BOS Secret Service · Sep 20, 2015

RealiseNothing said:
Sure, I made the bet last year and no one got over 70 so why not.

Bad news. This year's cohort is better than last year. You gonna be shouting one person lunch, guaranteed.

Official BOS Trial 2015 Thread (3 Viewers)

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