i took this paper 2 years ago and some of the questions still stump me.
still cant believe such questions are tested in sec1 ._.
Q1
suppose a, b and c are roots of (11-x)^3 + (13-x)^3 = (24-2x)^3
what is the sum of a+b+c?
A)30 B)36 C)40 D)42 E)44
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Q2
suppose that a+x^2=2006, b+x^2=2007 and c+x^2=2008 and abc=3
find the value of (a/bc)+(b/ca)+(c/ab)-(1/a)-(1/b)-(1/c)
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Q3
suppose that x-y=1.
find the value of x^4 - x(y^3) - (x^3)y - 3(x^2)y + 3x(y^2) + y^4
thanks for your help!
Hi,
Let's consider Q3 first.
Since x - y = 1, we will need to represent the expression in terms of x - y.
So x^4 - xy^3 - x^3y - 3x^2y + 3xy^2 + y^4
= x(x^3 - y^3) - y(x^3 - y^3) - 3xy(x - y)
= (x - y)(x^3 - y^3) - 3xy(1)
= (1)[(x - y)^3 + 3x^2y - 3xy^2] - 3xy
= 1^3 + 3xy(x - y) - 3xy
= 1 + 3xy(1) - 3xy
= 1
Nice questions! Thanks for sharing.
Cheers,
Wen Shih
Hi,
For Q1, obtain the coefficients of x^3 and x^2 which are 6 and -216 respectively.
a + b + c = -(-216/6) = 36, based upon the result below:
For ax^3 + bx^2 + cx + d = 0, sum of roots = -b/a.
In addition, sum of product of any 2 roots = c/a
and product of roots = d/a.
Thanks!
Cheers,
Wen Shih
Since Mr. Wee did Q1 and Q3, I shall try Q2
(a/bc)+(b/ca)+(c/ab)-(1/a)-(1/b)-(1/c)
= (a/bc) - (1/c) + (b/ca) - (1/a) + (c/ab) - (1/b)
= (a-b)/bc + (b-c)/ac + (c-a)/ab
from a+x^2=2006, b+x^2=2007 and c+x^2=2008
a = 2006 - x^2, b = 2007 - x^2, c = 2008 - x^2
a - b = -1
b - c = -1
c - a = 2
c - b = 1
Also, abc is given as 3
So, (a-b)/bc + (b-c)/ac + (c-a)/ab
= -1/bc - 1/ac + 2/ab
= (-a -b +2c)/abc
= ([c - a] + [c - b]) / abc
= (2 + 1) / 3
= 1 (answer)
Hi,
Olympiad questions emphasise a lot on algebraic manipulations (e.g. factorisation, rearrangement, taking sums/differences, etc.)
Thanks!
Cheers,
Wen Shih
Originally posted by wee_ws:Hi,
For Q1, obtain the coefficients of x^3 and x^2 which are 6 and -216 respectively.
a + b + c = -(-216/6) = 36, based upon the result below:
For ax^3 + bx^2 + cx + d = 0, sum of roots = -b/a.
In addition, sum of product of any 2 roots = c/a
and product of roots = d/a.
Thanks!
Cheers,
Wen Shih
Just replace 12 - x by u. The answer comes out immediately. Alternatively by observation note that 11, 12, 13 are the roots. Junior olympiad questions do not need that much working.
Originally posted by eagle:Since Mr. Wee did Q1 and Q3, I shall try Q2
(a/bc)+(b/ca)+(c/ab)-(1/a)-(1/b)-(1/c)
= (a/bc) - (1/c) + (b/ca) - (1/a) + (c/ab) - (1/b)
= (a-b)/bc + (b-c)/ac + (c-a)/ab
from a+x^2=2006, b+x^2=2007 and c+x^2=2008
a = 2006 - x^2, b = 2007 - x^2, c = 2008 - x^2
a - b = -1
b - c = -1
c - a = 2
c - b = 1Also, abc is given as 3
So, (a-b)/bc + (b-c)/ac + (c-a)/ab
= -1/bc - 1/ac + 2/ab
= (-a -b +2c)/abc
= ([c - a] + [c - b]) / abc
= (2 + 1) / 3
= 1 (answer)
Alternatively,
(a/bc)+(b/ca)+(c/ab)-(1/a)-(1/b)-(1/c) = (a^2 + b^2 + c^2 - ab - bc - ac)/3 =(-a - b + 2c)/3 = 3/3 = 1
Wow so pro
Anyway, in the smo junior round 1, there are 35 questions. You just need to score more than 14/35 to get silver medal. Gold need to score more questions.
National Average Scores
Junior Section
Gold Award: 20 or more marks for round 1 and special round score
Silver Award: 14 or more questions correct
Bronze Award: 11-13 questions correct
Honourable mention: 10 questions correct
thanks for all the help!
you can still go for smo senior/open you know
really uh. my school now only organises for sec1 to go for the junior comp.