Passage: Let u be a positive integer and f be a real number lying between 0 and 1. Further, ( 2 +1 )^ 10 =u+f and ( 2 -1 )^ 10 =v. Question: What is the value of (v+f) ?

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Correct answer: B. 1

Given ( 2 +1)^ 10 = u+f and ( 2 -1)^ 10 = v. Adding both equations: u+f+v = ( 2 +1)^ 10 + ( 2 -1)^ 10 Expanding using the binomial theorem: u+f+v = 2 [^ 10 C_0 ( 2 )^ 10 + ^ 10 C_2 ( 2 )^8 + + ^ 10 C_ 10 ] The right hand side is an even integer. Let it be 2k. u+f+v = 2k f+v = 2k - u Since 2k and u are integers, f+v must be an integer. We are given 0 Also, 0 Adding the inequalities for f and v: 0 Since f+v is an integer strictly between 0 and 2, the only possible value is 1. Therefore, v+f = 1. Answer: 1

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