2u Mathematics Marathon v1.0 (2 Viewers)

[Riviet]

I was working on a tree diagram, probably post it later tonight.

 

[Jono_2007]

next question

A bag contains 12 jellybeans. 3 are coloured red, 4 are black and 5 are green. Peter eats 3 jellybeans chosen ramdomly from the bag. Find the probability that:

(i) the first jellybean is black
(ii) all 3 jellybeans eaten are black
(iii) exactly one of the jellybeans eaten is black

Spoiler

Answer:
i)4/12=1/3
ii)4/12*3/11*2/10=1/55
iii)4/12 * 8/11 * 7/10 + 8/12 * 4/11 * 7/10 + 8/12 * 7/11 * 4/10
= 84/165
= 28/55

Soulseacher helped me out with the last one.

 

[Jono_2007]

This one is one step further from the 2-unit couse but you should find it faily easy, as its not very hard. What i mean by this, is that the 2 unit couse only covers quadratic polynomials, this is a cubic. But like Riviet said in this thread or another you need to chalange yourselves.

Next question:
When the polynomial 2x^3+ax^2+bx+1 is divided by x-1 the remainder is 4; when divided by x-2 the remainder is 21. Find the values of a and b.

 

[Riviet]

A bag contains 12 jellybeans. 3 are coloured red, 4 are black and 5 are green. Peter eats 3 jellybeans chosen ramdomly from the bag. Find the probability that:

(i) the first jellybean is black
(ii) all 3 jellybeans eaten are black
(iii) exactly one of the jellybeans eaten is black

Spoiler

i) From the diagram (or mentally), P(black 1st)=4/12

ii) Using the diagram, P(black & black & black)= 4/12 x 3/11 x 2/10 = 1/55

iii) This is why I drew a diagram.
Using the diagram, I've highlighted all the possiblilities for eating exactly 1 black bean only. We calculate the probability of each possible outcome (circled in green) and add them all up.

From top to bottom in diagram:

P(exactly 1 black)= (3/12x2/11x4/10)+(3/12x4/11x2/10)+(3/12x4/11x5/10)+(3/12x5/11x4/10)+(4/12x3/11x2/10)+(4/12x3/11x5/10)+(4/12x5/11x3/10)+(4/12x5/11x4/10)+(5/12x3/11x4/10)+(5/12x4/11x3/10)+(5/12x4/11x4/10)+(5/12x4/11x4/10)
= 28/55

 

[Jono_2007]

Next question:
When the polynomial 2x^3+ax^2+bx+1 is divided by x-1 the remainder is 4; when divided by x-2 the remainder is 21. Find the values of a and b.

hint:

Spoiler

For each (x-1) and (x-2) sub in the absolute values of the constants (i.e. 1 and 2) to get the basic equation the solve simultaniously.

 

[YBK]

No need for the hint

Spoiler

P(1)= 2 + a + b +1 = 4
.: a = 1 - b

P(2) = 2 (2)^3 + a(2)^2 + 2b + 1 = 21
4a + 2b = 4

Now substitute a = 1 - b into 4a + 2b = 4

We get 4 - 4b + 2b = 4
.: - 2b = 0
b = 0
Replace back into a = 1 - b to get a
.: a =1

Should be right

 

[Jono_2007]

Correct. Didn't see your last edit. sorry your right to post the ext question!

 

[YBK]

Unfortunately it's not correct.

lmao are u serious?

*was/still am in a rush, I think I just found the stupid little error* (it should be fixed, unless if i have another)

 

[Riviet]

That looks about right YBK. You can post up the next question.

 

[Jono_2007]

The answer is actually:

Spoiler

Let p(x)=2x^3+ax^2+bx+1
for (x-1);
P(1)=2(1)^3+a(1)2+bx+1=4
2+a+b+1=4
a+b=1 i

and for (x-2);
P(2)=2(2)^3+a(2)^2+b(2)+1=21
16+4a+2a+1=21
4a+2a=4
2a+b=2 ii
Now,ii-i
a=1
sub into i b=0
:. a=1 and b=0

Sorry! Didn't see your edit your right!

 

[YBK]

That looks about right YBK. You can post up the next question.

lets see:


Easy one
derrive from first principles:

y = 2x^5 + 3x^3 + x + 3


Numbers are kinda exaggerated I guess

 

[Jono_2007]

derrive from first principles:

y = 2x^5 + 3x^3 + x + 3


Numbers are kinda exaggerated I guess

What does that mean, 'derive from basic principles'?

 

[Riviet]

Differentiating from first principles would mean to use:

lim
h->0 [f(x+h)-f(x)]/h

which would be too hard.

So I think he meant just to differentiate it.

 

[Jono_2007]

I think thats ok!

 

[YBK]

Differentiating from first principles would mean to use:

lim
h->0 [f(x+h)-f(x)]/h

which would be too hard.

So I think he meant just to differentiate it.

You could use Pascal's triangle... but that's not 2unit anymore... and it'd get really annoying with the large numbers
okay then, I'll change the question.


y = 2x^2 + 3x + 3
That should be simple enough now

 

[Jono_2007]

does 2h^2+4ch+3h+3c+3 factorise.

 

[YBK]

does 2h^2+4ch+3h+3c+3 factorise.

It shouldn't matter (are u doing my question???)

 

[Jono_2007]

yes and i came to this part and got stuck.

 

[Jono_2007]

y = 2x^2 + 3x + 3
That should be simple enough now

Spoiler

Answer:
f'(c)=f(c+h)-f(c)/h lim h-->0
=2(c+h)^2-3(c+h)-(2x^2+3x+3)/h
=2(c^2+2ch+h^2)-3c+3h-2c^2+3c+3/h
=2c^2+4ch+2h^2+3h-2c^2+3c+3/h
=2h^2+4ch+3h+3c+3/h
=Got stuck here

 

[Riviet]

Spoiler

Answer:
f'(c)=f(c+h)-f(c)/h lim h-->0
=2(c+h)^2-3(c+h)-(2x^2+3x+3)/h
=2(c^2+2ch+h^2)-3c+3h-2c^2+3c+3/h
=2c^2+4ch+2h^2+3h-2c^2+3c+3/h
=2h^2+4ch+3h+3c+3/h
=Got stuck here

Spoiler

When substituting, don't forget to include any constants, in this case it would be the +3 that you left out. Also in your 3rd line, be careful when expanding and watch the sign.