3u Mathematics Marathon V 1.1 (1 Viewer)

[STx]

^ Check this solution.

 

[Riviet]

Next Question

Consider n distinct points on the circumference of a circle in which an n-sided polygon is formed.

i) If n=4, how many diagonals can be drawn in the quadrilateral?
ii) If n=8, how many diagonals can be drawn in the octagon?
iii) How many diagonals can be drawn in an n-sided polygon?
iv) How many diagonals can be drawn in a 64 sided polygon?
v) Show that the expression in iii) is equal to n(n-3)/2 for n>4

 

[Sober]

Next Question

i) If n=4, how many diagonals can be drawn in the quadrilateral?
ii) If n=8, how many diagonals can be drawn in the octagon?
iii) How many diagonals can be drawn in an n-sided polygon?
iv) How many diagonals can be drawn in a 64 sided polygon?
v) Show that the expression in iii) is equal to n(n-3)/2 for n>4

Spoiler

i) 2
ii) 20
iii) the number of diagonals will equal the number of possible line connections minus the number of edges = ((n-1)+...+2+1)-n which is an arithmetic series = n(n-1)/2 - n = n(n-3)/2
iv) 1952
v) already done

Next Question

Prove that 13*6ⁿ+2 is divisible by 5 for integers n>0

 

[followme]

Next Question

Prove that 13*6ⁿ+2 is divisible by 5 for integers n>0

Spoiler

let n=0
ie 13*1+2=15 = 3*5
therfor it is divisible by 5
it is true for n=0

Assume n=k is true
ie 13*6^k+2 is dividible by 5
ie 13*6^k+2 = 5M whre M is an integer
let n=k+1
[13*6^(k+1)]+2
=[13*6*6^k]+2
=[13*6*(5M-12)/13]+2
=6*5M-12*6+2
=5(6M-14) where 6M-14 is an interger
so it is divisible by 5
therfore if n=k is true, n=k+1 is true

so n=0 is ture, hence n=1 is true, hence n=2 is true and so on.
therefore it is true for all integers n>0

Next Question
let f(x)=e^x + e^(2x)

find:
f inverse (x)
state the domain and range.

 

[STx]

Spoiler

f(x)=e^x + e^2x
f^-1(x) : x=e^y + e^2y
i.e. (e^y)²+e^y-x=0

let e^y=a

=> a²+a-x=0
.'. a= (-1±√1+4x)/2
.'. e^y=(1/2).[(√1+4x)-1)] or (-1/2).[(√1+4x)+1)
.'. y= ln[{(√1+4x)-1}/2] (since y>0 ??)

Now, D: All Positive Reals, x>0
R: All y≥0

 

[Riviet]

STx: feel free to post up a new question, I'm just putting one here to keep things rolling.

Question:
Show that sec2x + tan2x = (cosx + sinx)/(cosx - sinx)

 

[SoulSearcher]

STx: feel free to post up a new question, I'm just putting one here to keep things rolling.

Question:
Show that sec2x + tan2x = (cosx + sinx)/(cosx - sinx)

Spoiler

LHS = sec2x + tan2x
= 1/cos2x + 2tanx/(1-tan²x)
= 1/(2cos²x-1) + 2tanx/(2-sec²x)
= 1/(2cos²x-1) + 2tanx/{(2cos²x-1)/(cos²)}
= 1/(2cos²x-1) + 2sinxcosx/(2cos²x-1)
= (1+2sinxcosx)/(2cos^x-1)
= (sin²x+2sinxcosx+cos²x)/(cos²x-sin²x)
= (sinx+cosx)²/(cosx-sinx)(cosx+sinx)
= (cosx+sinx)/(cosx-sinx)
= RHS, as required

 

[followme]

next question:

y=f(x) is a linear function with slope 0.5
i) find an expression of the inverse function of y=f(x)
ii) hence find the slope of y=f^-1(x)

 

[Riviet]

next question:

y=f(x) is a linear function with slope 0.5
i) find an expression of the inverse function of y=f(x)
ii) hence find the slope of y=f^-1(x)

Spoiler

i) y=x/2 + b, where b is a constant.
ii) x=y/2 + b
y/2=x-b
y=2x-2b
.'. gradient of f^-1(x) is 2.

Will post up a new question tomorrow / soon.

 

[Riviet]

Next Question:

How many different ways of painting 8 different colours on a cube?

 

[.ben]

Spoiler

8x7x6x5x4x3=20160

Next Question

A batsman hits a cricket ball 'off his toes' toward a fieldsman who is 65m away. The ball reaches a maxiumum height of 4.9m and the horizontal component of its velocity is 28m/s. Find the constant speed with which the fieldsman must run forward, starting at the instant the ball is hit, in order to catch the ball at a height of 1.3m above the ground. (g=9.8)

 

[Riviet]

Spoiler

8x7x6x5x4x3=20160

Spoiler

The cube can be flipped and turned over.

 

[.ben]

Spoiler

8x7x6x5x4x3=20160

But basically isn't it the same because it just depends on perspective?

 

[Riviet]

But basically isn't it the same because it just depends on perspective?

Yep, but the question wanted different ways.

Other examples where you would need to be careful with different arrangements are necklaces and round tables.

Another question:
By considering the value of (1 + x)²ⁿ when x=1, prove that:
n
∑ ²ⁿC_r = 2^2n-1 + (2n)!/2(n!)²
r=0

 

[.ben]

So what was the answer?

 

[followme]

Another question:
By considering the value of (1 + x)²ⁿ when x=1, prove that:
n
∑ ²ⁿC_r = 2^2n-1 + (2n)!/2(n!)²
r=0

Spoiler

(1+x)²ⁿlet x=1

=²ⁿC₀+²ⁿC₁+²ⁿC₂+²ⁿC₃+...+²ⁿC_n+²ⁿC_n+²ⁿC_n+1+²ⁿC_n+2+...+²ⁿC_2n-²ⁿC<sub>n

</sub> (since ⁿC_r=ⁿC_n-r)

2²ⁿ=2[²ⁿC₀+²ⁿC₁+²ⁿC₂+²ⁿC₃+...+²ⁿC_n]-²ⁿC<sub>n

</sub>so, ²ⁿC₀+²ⁿC₁+²ⁿC₂+²ⁿC₃+...+²ⁿC_n= (2²ⁿ+²ⁿC_n)/2
n
∑ ²ⁿC_r =2^2n-1+(2n)!/2n!(2n-n)!
r
=2^2n-1+(2n)!/2(n!)²

next question: suppose the roots of the equation x³+px²+qx+r=0 are real, show that the roots are in geometric progression if q³=p³r.

Hint: let the roots be a/b, a and ab

 

[pLuvia]

Next question: suppose the roots of the equation x³+px²+qx+r=0 are real, show that the roots are in geometric progression if q³=p³r.

Hint: let the roots be a/b, a and ab

Spoiler

x³+px²+qx+r=0
Let the roots be a/b, a, ab
a(1/b+1+b)=-p (1)
a²(b+1+1/b)=q (2)
a³=-r (3)

From (1)
(1/b+1+b)=-p/a
Sub into (2)
-ap=q (4)
Sub (3) into (4)
-(-r)^1/3p=q
rp³=q³

 

[Riviet]

New question please pLuvia.

 

[pLuvia]

Ok, umm.

New question
State the domain and range of this inverse function, y=sin^-1sqrt{1/36-x²}.

 

[.ben]

Anyone