# SMO JUNIOR SECTION 2012 SOLUTIONS PART 4

Problem 11 :

Let a and b be real numbers such that a > b, 2a + 2b = 75 and 2-a + 2-b = 12-1. Find the value of 2a-b

Solution :

2a + 2b = 75   ---(1)

2-a + 2-b = 12-1   ---(2)

(1) x (2)

(2a + 2b)(2-a + 2-b) = 75 x (1/12)

2a-a + 2a-b + 2b-a + 2b-b = 75/12

1 + 2a-b + 2b-a + 1 = 75/12

2 + 2a-b + 2-(a-b) = 75/12

2a-b + 2-(a-b) = (75/12) - 2

2a-b + 2-(a-b) = (51/12)

2a-b + 2-(a-b) = (17/4)

2a-b + 2-(a-b) = 4 + (1/4)

Then, 2a-b = 4

Problem 12 :

Find the sum of all positive integers x such that

is an integer.

Solution :

The given fraction is an improper fraction. Converting it into mixed fraction, we get

To get integer, it is necessary to be x2 - 1 also be integer.

Factors of 120 are

1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60, 120

The possible values of x2 - 1 are

0, 2, 3, 4, 5, 11

Sum of (0, 2, 3, 4, 5, 11) :

= 0 + 2 + 3 + 4 + 5 + 11

= 25

Problem 13 :

Consider the equation

√(3x2-8x+1) + √(9x2-24x-8) = 3

It is known that the largest root of the equation is - k times the smallest root . Find k.

Solution :

√(3x2-8x+1) + √(9x2-24x-8) = 3

y = 2 and y = -5 (rejected because y should be greater than or equal to 0.)

since k = 2

3x- 8x + 1 = 22

3x- 8x + 1 = 4

3x- 8x - 3 = 0

(x - 3)(3x + 1) = 0

x = 3 and x = -1/3 ==> -3-1

The largest root = 3, which is equal to -k x  the smallest root.

3 = -k(-1/3)

3 = k/3

k = 9

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