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) ?

Step-by-step solution

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 [¹⁰C₀ ( 2 )¹⁰ + ¹⁰C₂ ( 2 )^8 + + ¹⁰C₁₀] 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