Solve the equation 2 – x = 2b – 2ax, where a and b are real parameters. Find for which values of a the equation has for solution a natural number, if b = 7
Choose one:
A) 0 B) 1 C) 2 D) 3 E) 4

Rationale: solution : we write the given equation like this ( 2 a - 1 ) x = 2 ( b - 1 ) the following cases are possible : if 2 a - 1 ≠ 0 , i . e . a ≠ 1 / 2 , the equation has single solution x = 2 ( b - 1 ) / ( 2 a - 1 ) if a = 1 / 2 and b = 1 , the equation gets the kind 0 . x = 0 and every x is solution if a = 1 / 2 and b ≠ 1 , we get 0 . x = 2 ( b - 1 ) , where 2 ( b - 1 ) ≠ 0 in this case the equation has no solution . if b = 7 and a ≠ 1 / 2 the single solution is x = 2 ( 7 - 1 ) / ( 2 a - 1 ) = 12 / ( 2 a - 1 ) if a is whole number , then 2 a - 1 is also whole number and the solution x = 12 / ( 2 a - 1 ) is natural number when 2 a - 1 is positive divisor of 12 . to have a as whole number , the divisor of 12 mu be odd . but the only whole positive odd numbers divisible by 12 are 1 and 3 . therefore 2 a - 1 = 3 < = > a = 2 or 2 a - 1 = 1 < = > a = 1 a = 2 or 2 a - 1 = 1 < = > a = 1 answer b
Answer: B