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Paragraph: Consider, f(x)= | ( (2 r₁-1 ) 6 ) x |+ | ( r₂ 6 ) x | . Let a fair dice is thrown twice and r₁, r₂ are the numbers obtained on the dice its first and second throw respectively. Question: The probability that f(1) is an integer, is:

Options

  1. A5 9
  2. B4 9
  3. C1 3
  4. D2 3

Correct answer

C. 1 3

Step-by-step solution

aligned & f(x)= | ( (2 r₁-1 ) 6 ) x |+ | ( r₂ 6 ) x | & f(1)= | (2 r₁-1 ) 6 |+ | ( r₂ 6 ) | aligned Possible value of (2 r₁-1 ) 6 = 6 , 3 6 , 5 6 , 7 6 , 9 6 , 11 6 Possible values of r₂ 6 = 6 , 2 6 , 3 6 , 4 6 , 5 6 , 6 6 (i) For f(1) to be an integer aligned & (2 r₁-1 ) 6 _ 3 6 , 9 6 & 6 , 5 6 , 7 6 , 11 6 r₂ 6 _ 3 6 , 6 6 & Required probability = 2 6 2 6 + 4 6 2 6 = 4 36 = 1 3 aligned aligned & (ii) f(2)= | (2 r₁-1 ) 3 |+ | ( r₂ 3 ) | & (2 r₁-1 ) 3 = 3 , 3 3 , 5 3 , 7 3 , 9 3 , 11 3 & r₂ 3 = 3 , 2 3 , 3 3 , 4 3

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