O-Level Additional Math

[FrekiWang]

red rose:

Would appreciate help with the following Qs:


1) Arrange the following numbers in ascending order: 2^3333, 3^2222, 6^1111, 9^555
2) Which is bigger, 3^3^3^3 or 4^4^4
3) Solve (x^2-5x+5)^x+2000 = 1

Thanks in advance!

If you are referring to these basic olympiad questions, I guess it is useless to know only the solution, you may need more systematic way of learning them.

1)
2^3333 = (2^3)^1111=8^1111
3^2222=(3^2)^1111=9^1111
9^555=(3^2)^555=3^1110
therefore 3^2222>2^3333>6^1111>9^555

2) Assuming there is no bracket(without any bracket the upper indice will be evaluated first)
3^3^3^3=3^3^9=3^(3 x 3^8)=9^(3^8)
4^4^4=4^(4^4)

since 9>4 and 3^8=9^4>4^4, 3^3^3^3>4^4^4

3) I assume you are asking (x^2-5x+5)^(x+2000)=1
There are three possibilities:
a) x+2000=0 and x^2-5x+5>0
x=-2000, check x^2-5x+5 is positive when x=-2000
b) x^2-5x+5=1
solve this simple quadratic equation we have x=-1 or x=-4
c) x^2-5x+5=-1 and x+2000 is even
solve this simple quadratic equation we have x=-2 or x=-3
reject x=-3 and (-3)+2000 is odd.

Therefore x=-2000, -1,-2 or -4.

[heutistmeintag]

Need help on Sec 3 A maths Trigo question:

Given that cosec A + cot A = 3, evaluate cosec A - cot A and cos A.

Thanks in advance!

[FrekiWang]

heutistmeintag:

Need help on Sec 3 A maths Trigo question:

Given that cosec A + cot A = 3, evaluate cosec A - cot A and cos A.

Thanks in advance!

Sorry I feel lazy to solve it in a standard approach.
1/sinA + cosA/sinA = 3
1 + cosA = 3sinA ...(1)
Let cosecA - cotA = k
1/sinA - cosA/sinA = k
1 - cosA = ksinA ...(2)
(1)*(2) we have,
1 - (cosA)^2 = 3k(sinA)^2
(sinA)^2 = 3k(sinA)^2
3k = 1
k = 1/3 (this is the value of cosecA - cotA)

subst k = 1/3 into (2), we have
1 - cosA = 1/3sinA ... (3)
(1) - (3)*9, we have
(1+cosA) - 9(1-cosA) = 0
-8 + 10cosA = 0
cosA = 0.8

[red rose]

Need help to solve this.


x - 2/x + 3 = 2x^2 + x - 10/x^2 + 6x + 9

I only managed to get x = -2, but the answer key says the answer should be x = 2 or x = -2. Please advise.

Thank you in advance.

[jtoh]

Hi red rose,


I'm a little confused with the presentation of the question - limitations of typing out the question. Could you please clarify if it's :

x - 2/x + 3 or (x-2)/(x+3)

Similarly for the right hand side? Thanks.

[red rose]

jtoh:

Hi red rose,


I'm a little confused with the presentation of the question - limitations of typing out the question. Could you please clarify if it's :

x - 2/x + 3 or (x-2)/(x+3)

Similarly for the right hand side? Thanks.

Hi! It's with brackets. Eg. (x-2) divided by (x+3)

Thanks!

[jtoh]

Thanks red rose. So the right hand side is (2x^2 + x - 10)/(x^2 + 6x + 9)?

[red rose]

jtoh:

Thanks red rose. So the right hand side is (2x^2 + x - 10)/(x^2 + 6x + 9)?

Yes! :thankyou:

[jtoh]

Hi red rose, I only managed to get x = -2 too. Could the answer sheet be wrong? Is this question from a workbook?

[red rose]

jtoh:

Hi red rose, I only managed to get x = -2 too. Could the answer sheet be wrong? Is this question from a workbook?

Nope. It's from a Sec 2 IP school revision worksheet. Thanks for trying anyway. The curious thing is, +2 works too when substituted into the equation.