MAT Prep Thread - 2nd November 2016

For 2013 Q1A (lmao) why is it not a>2 ????

Reply 361

Nyathigiz

7

Original post by Mystery.

For 2013 Q1A (lmao) why is it not a>2 ????

Because you use the discrimination equation. Since you want different roots a cant equal 0 because the answer to the discrimination equation will be zero. A can be less than or bigger than 0 since it's being squared so either way the answer will be positive

(edited 7 years ago)

Reply 362

rtc16

1

I've slowly started to get the hang of the multiple choice with the help of the RZC PowerPoint (highly recommend) however I'm still not grasping what first steps to take with the longer questions and was wondering if anybody could recommend any resources to help with the longer questions

Reply 363

1H and J from 2013????

(edited 7 years ago)

Reply 364

Original post by Mystery.

1H and J from 2013????

1H

Square both sides of LHS equation and you get a circle equation (be careful though, don't sketch for x<0 or y<0). the other is a straight line. Draw a quick sketch and work out area using area of sector and area of triangle equations etc

1J

Theres not a hope in hell Ill be able to explain my method for this one and i didnt bother trying to understand the mark scheme answer (i dont understand it either). Sorry but theres at least 1H for you. Maybe someone will come along soon and help you with 1J :biggrin:

Reply 365

Hug_or_Be_Hugged

1

Original post by theaverage

I don't understand how they go about the long division in 2013 question 1, G?

Posted from TSR Mobile

I can help you with that. PM me.

Reply 366

A mad robot sets off towards the North East on a journey from the origin. It travels in stages by moving forward and then rotating on the spot. It follows these pseudo-code instructions:

SUB JOURNEY

DISTANCE = 1000

WHILE (DISTANCE > 0.001)
MOVE DISTANCE
STOP
ROTATE(90, DEGREES, CLOCKWISE)
DISTANCE = DISTANCE / 2
END WHILE

EXPLODE

END SUB

Where does the robot explode?

Someone want to explain the solution...?

It starts off like below:
The forward motion, if it occurs, following the nth turn will be of distance 1000×2−n.This will occur for each n for which

1000×2−n>0.001
Taking logs gives

where does the 2 come from?

Reply 367

Original post by Mystery.

x

Sorry quick question just before I read this. Is this a question 5 from either 2014 or 2015? (i havent done those papers yet and dont want spoilers so just wanna check :biggrin:

)

Reply 368

Original post by DylanJ42

Sorry quick question just before I read this. Is this a question 5 from either 2014 or 2015? (i havent done those papers yet and dont want spoilers so just wanna check :biggrin:

)

No,its from nrich haha

Reply 369

Original post by Mystery.

1H and J from 2013????

also I decided to read through the mark scheme answer for 1J so ill try talk you through this;

Spoiler

Reply 370

Original post by DylanJ42

also I decided to read through the mark scheme answer for 1J so ill try talk you through this;

Spoiler

Ah thankyou so much. I'll read through it later. Currently doing some late night studying on sequences and series ;laugh;

Reply 371

Original post by Mystery.

Ah thankyou so much. I'll read through it later. Currently doing some late night studying on sequences and series ;laugh;

no problem :biggrin:

I was about to go to bed until I seen a post in the MAT thread, I thought id read it and here I am :laugh:

also for your nrich problem is 1000 x (2- n) from the mark scheme? its just im working out the distance travelled after the nth turn to be

\displaystyle \frac{1000}{2^n}

Reply 372

Original post by DylanJ42

no problem :biggrin:

I was about to go to bed until I seen a post in the MAT thread, I thought id read it and here I am :laugh:

also for your nrich problem is 1000 x (2- n) from the mark scheme? its just im working out the distance travelled after the nth turn to be

\displaystyle \frac{1000}{2^n}

Yeah, I copied it from there.

Reply 373

Original post by Mystery.

Yeah, I copied it from there.

hmm you see if the robot hasnt taken any turns n= 0 then the distance is 1000.

If its taken its first turn the distance will be 500

after the 2nd turn itll be 250 etc

this fits; distance after nth turn

\displaystyle = \frac{1000}{2^n}

, whereas

\displaystyle 1000 \times (2-n)

doesnt work, even for n=0 as the robot cant travel 2000 units.

Reply 374

molly221

3

Can someone please explain to me how to do this? It is from 2010. Thank you

Reply 375

Original post by molly221

Can someone please explain to me how to do this? It is from 2010. Thank you

Check option B. Note that option B actually equals to 1.5.

C is less than 1 because its argument is smaller than the base. C eliminated.

5^(1.5) is equal to 5 times sqrt(5). Because sqrt (5) > 2, hence 5^(1.5) is larger than 10. Option D is smaller than 1.5.

By a similar approach we have option A is larger than 1.5. Hence, the answer is A.

Reply 376

molly221

3

Original post by LaserRanger

Check option B. Note that option B actually equals to 1.5.

C is less than 1 because its argument is smaller than the base. C eliminated.

5^(1.5) is equal to 5 times sqrt(5). Because sqrt (5) > 2, hence 5^(1.5) is larger than 10. Option D is smaller than 1.5.

By a similar approach we have option A is larger than 1.5. Hence, the answer is A.

Thanks, helped a lot. :smile:

Reply 377

Original post by danielhx

also for Q4 2014, i dont understand the last part how they can just swap beta and alpha just like that

(Due to convenience I just type alpha as a and beta as b)
Note that originally
0<a<b<pi/2
Now
0<b<a<pi/2

Do you see the whole question actually remains the same except this switch? Hence all steps you have done from i) to v) can be directly applied. You can assume this as a change of the name of variables, ie. somehow you are just giving a new name to a and b respectively.

Reply 378

Original post by KloppOClock

For question 3iv on 2009 I answered it differently but am wondering if it is still correct.

\displaystyle \frac{3n-1}{(4n-1)} \leq An/(B+n)[br][br]1 is negligible for large values of n. Cancel the n's. Ignore B for large values of n as B is a constant.[br]Left with;[br]0.75 \leq A[br][br]

Would this still get the marks?

I don't think so. Check the question, for 3 iv), n can be small numbers as long as it is larger than 1.

Reply 379