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3u Mathematics Marathon V 1.1 (2 Viewers)

Riviet

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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

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Riviet said:
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
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

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Sober said:
Next Question

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

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f(x)=ex + e2x
f-1(x) : x=ey + e2y
i.e. (ey)2+ey-x=0

let ey=a

=> a2+a-x=0
.'. a= (-1±√1+4x)/2
.'. ey=(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
 
Last edited:

Riviet

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

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

SoulSearcher

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

Question:
Show that sec2x + tan2x = (cosx + sinx)/(cosx - sinx)
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)/(2cosx-1)
= (sin2x+2sinxcosx+cos2x)/(cos2x-sin2x)
= (sinx+cosx)2/(cosx-sinx)(cosx+sinx)
= (cosx+sinx)/(cosx-sinx)
= RHS, as required
 

followme

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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<sup>-1</sup>(x)
 

Riviet

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followme said:
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<sup>-1</sup>(x)
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<sup>-1</sup>(x) is 2.
Will post up a new question tomorrow / soon. :p
 

Riviet

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Next Question:

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

.ben

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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)
 
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.ben

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8x7x6x5x4x3=20160

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

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.ben said:
But basically isn't it the same because it just depends on perspective?
Yep, but the question wanted different ways. :p

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
2nCr = 22n-1 + (2n)!/2(n!)2
r=0
 

followme

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Riviet said:

Another question:

By considering the value of (1 + x)2n when x=1, prove that:
n
2nCr = 22n-1 + (2n)!/2(n!)2
r=0
(1+x)<sup>2n </sup>let x=1

=<sup>2n</sup>C<sub>0</sub>+<sup>2n</sup>C<sub>1</sub>+<sup>2n</sup>C<sub>2</sub>+<sup>2n</sup>C<sub>3</sub>+...+<sup>2n</sup>C<sub>n</sub>+<sup>2n</sup>C<sub>n</sub>+<sup>2n</sup>C<sub>n+1</sub>+<sup>2n</sup>C<sub>n+2</sub>+...+<sup>2n</sup>C<sub>2n</sub>-<sup>2n</sup>C<sub>n

</sub>
(since <sup>n</sup>C<sub>r</sub>=<sup>n</sup>C<sub>n-r</sub>)

2<sup>2n</sup>=2[<sup>2n</sup>C<sub>0</sub>+<sup>2n</sup>C<sub>1</sub>+<sup>2n</sup>C<sub>2</sub>+<sup>2n</sup>C<sub>3</sub>+...+<sup>2n</sup>C<sub>n</sub>]-<sup>2n</sup>C<sub>n

</sub>so, <sup>2n</sup>C<sub>0</sub>+<sup>2n</sup>C<sub>1</sub>+<sup>2n</sup>C<sub>2</sub>+<sup>2n</sup>C<sub>3</sub>+...+<sup>2n</sup>C<sub>n</sub>= (2<sup>2n</sup>+<sup>2n</sup>C<sub>n</sub>)/2
n
∑ <sup>2n</sup>C<sub>r</sub> =2<sup>2n-1</sup>+(2n)!/2n!(2n-n)!
r
=2<sup>2n-1</sup>+(2n)!/2(n!)<sup>2</sup>

next question: suppose the roots of the equation x<sup>3</sup>+px<sup>2</sup>+qx+r=0 are real, show that the roots are in geometric progression if q<sup>3</sup>=p<sup>3</sup>r.

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

pLuvia

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followme said:
Next question: suppose the roots of the equation x<sup>3</sup>+px<sup>2</sup>+qx+r=0 are real, show that the roots are in geometric progression if q<sup>3</sup>=p<sup>3</sup>r.
Hint: let the roots be a/b, a and ab

x<sup>3</sup>+px<sup>2</sup>+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
 
P

pLuvia

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Ok, umm.

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

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