Extension One Revising Game (1 Viewer)

[Zeber]

12C5.2^7.m^5 = 24057
rearranging, m = 3/4

 

[Trebla]

Solve for x:
ln (1 + 1/x) ≥ ln (3x - 1)

 

[Timothy.Siu]

Solve for x:
ln (1 + 1/x) ≥ ln (3x - 1)

x cant be equal or between 0 and -1 and x cannot be less than or equal to 1/3
(1+1/x)≥(3x-1)
x^2+x≥3x^3-x^2
0≥x(3x^2-2x-1)
0≥x(3x+1)(x-1)

0<=x<=1 or x<=-1/3 but thats not in the domain and stuff....
so 1/3<x<=1
i think....

 

[alez]

ouch
i cant even get to the product rule bit

 

[azureus88]

y' = (x^2)(1-y)

y = (x^2-y')/(x^2)
=(x^2-(x^2)(1-y))/(x^2)
xy = 1-x^2 +yx^2
xy-yx^2 = 1-x^2
y = (1-x^2)/(x(1-x^2)
= 1/x if x is not equal to 1,-1

as x approaches infinite y appraoches 0

 

[lolokay]

i)
by considering the product rule find y in terms of x of:
y'.e^x + y.e^x = x

hence find the value of y as x approaches infinity

ii)
find y in terms of x of:
y' + y.x^2 = x^2

hence find the value of y as x approaches infinity

Spoiler

d(y.e^x)/dx
= y.e^x + y.e^x

Int [y'.e^x + y.e^x].dx = Int [x].dx
y.e^x = 1/2 x² + C
y = 1/2 x²e^-x + Ce^-x
so as x->infinity, y->0

ii) d/dx y.e^1/3x3
= y'.e^1/3x3 + x²y.e^1/3x3
= x².e^1/3x3
integating each side wrt x gives
y.e^1/3x3 = e^1/3x3 + C
y = 1 + Ce^-1/3x3
as x->infinity, y->1

 

[azureus88]

New Q:

Given that a^2 + b^2 = 1, prove that the expression
tan <SUP>-1 </SUP>[(ax)/(1-bx)] - tan <SUP>-1 </SUP>[(x-b)/a] is independant of x.
<SUP><?xml:namespace prefix = o ns = "urn:schemas-microsoft-comfficeffice" /><o></o></SUP>

 

[lolokay]

let: tan -1 [(ax)/(1-bx)] = A, tan -1 [(x-b)/a] = B

calculating tan(A-B) gives: ( a²x - x + b + bx² - b²x)/(a + ax² - 2abx)
subbing in -x = -a²x - b²x gives
(b(x²+1) - 2b²x)/(a(x²+1) - 2ab)
= b/a (x² + 1 - 2b)/(x² + 1 - 2b)
= b/a

so tan -1 [(ax)/(1-bx)] - tan -1 [(x-b)/a] = tan^-1(b/a)
which is independant of x

 

[Trebla]

Prove by induction that for positive integer n, the n-th derivative of xe^x is given by:
dⁿ(xe^x)/dxⁿ = e^x(x + n)

 

[addikaye03]

Prove by induction that for positive integer n, the n-th derivative of xe^x is given by:
dⁿ(xe^x)/dxⁿ = e^x(x + n)

d^n(xe^x)/dx^n = e^x(x + n)

Step 1: prove true for n=1 [d(xe^x)/dx=e^x(x+1)]
LHS=d/dx(xe^x)
u=x, v=e^x
u'=1, v'=e^x
d/dx=e^x+xe^x
=e^x(x+1)
=RHS therefore true for n=1

Step2: Assume true for n=k
d^k(xe^x)/dx^k=e^x(x+k)

Step 3: Prove true for n=k+1
d^k+1(xe^x)/dx^k+1=e^x(x+k+1)
LHS=d/dx(d^k(xe^x)/dx^k)
=d/dx(e^x(x+k)) by assumption.
u=e^x, v=(x+k)
u'=e^x, v'=1
d/dx=e^x(x+k)+e^x
=e^x(x+k+1)
=RHS

Step 4: Since true for n=1 and true for n=k+1, then true for n=2,3,...all positive integer to the nth.

Was bored so thought i would answer it, heres a Q to someone:

If x^2+y^2=a^2+b^2=1. prove that |ax+by|<=1

 

[Timothy.Siu]

wats the s?

 

[addikaye03]

wats the s?

sorry. Fixed it

 

[Trebla]

This looks like an Extension 2 type question lol.
x² + y² = 1
a² + b² = 1
So
(x² + y²)(a² + b²) = 1
a²x² + a²y² + b²x² + b²y² = 1
But (ay - bx)² ≥ 0
=> a²y² + b²x² ≥ 2abxy
Hence
a²x² + a²y² + b²x² + b²y² ≥ a²x² + b²y² + 2abxy
1 ≥ (ax + by)²
Thus
| ax + by | ≤ 1

 

[lyounamu]

sorry. Fixed it

abs(ax+by) = sqrt (a^2x^2 + b^2y^2 + 2abxy)
= sqrt (a^2(a^2+b^2 - y^2) + b^2(a^2+b^2-x^2) + 2abxy)
= sqrt (a^4 + 2a^2b^2 + b^4 - a^2y^2 - b^2x^2 + 2abxy)
= sqrt ( 1 - a^2y^2 - b^2x^2 + 2abxy)
= sqrt ( 1 - (ay-bx)^2)
since (ay-bx)^2 >= 0

therefore, abs (ax+by) <= 1

EDIT: damn yoooo~~~~ I shall refine my typing skills.

 

[addikaye03]

This looks like an Extension 2 type question lol.
x² + y² = 1
a² + b² = 1
So
(x² + y²)(a² + b²) = 1
a²x² + a²y² + b²x² + b²y² = 1
But (ay - bx)² ≥ 0
=> a²y² + b²x² ≥ 2abxy
Hence
a²x² + a²y² + b²x² + b²y² ≥ a²x² + b²y² + 2abxy
1 ≥ (ax + by)²
Thus
| ax + by | ≤ 1

CORRECT.
That was the exact method i used, down to the last line. So i give u 6/4 marks!
And it was classed as harder 3U

 

[lyounamu]

CORRECT.
That was the exact method i used, down to the last line. So i give u 6/4 marks!
And it was classed as harder 3U

Your questions was quite an open-ended question. Any working that would have proved that abs (ax+by) >=1 would be fine.

There isn't much of a correct solution or incorrection solution as long as you can prove it, right?

 

[addikaye03]

I agree, it was a very open ended Q and i guess any proof would be acceptable. I think it could be done diagramatically, u think? Anyone got any Q to share??
Keep this thread alive i say haha

 

[Timothy.Siu]

uhh ok

Prove by induction that 7+77+777+....+777...to n digits=7(10ⁿ⁺¹-9n-10)/81

 

[youngminii]

This looks like an Extension 2 type question lol.
x² + y² = 1
a² + b² = 1
So
(x² + y²)(a² + b²) = 1
a²x² + a²y² + b²x² + b²y² = 1
But (ay - bx)² ≥ 0
=> a²y² + b²x² ≥ 2abxy
Hence
a²x² + a²y² + b²x² + b²y² ≥ a²x² + b²y² + 2abxy
1 ≥ (ax + by)²
Thus
| ax + by | ≤ 1

Don't quite understand the last part :S
And you were meant to type your own question!

By the way, Harder 3u is 4u

 

[youngminii]

uhh ok

Prove by induction that 7+77+777+....+777...to n digits=7/81(10ⁿ⁺¹-9n-10)

This doesn't work.