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📰 Question: A philosopher of science considers a function $ k(x) $ modeling the "distance" from theory to observation, satisfying $ k(x + y) = k(x) + k(y) - 2k(xy) $. If $ k(1) = 1 $, find $ k(2) $. 📰 Solution: Set $ x = y = 1 $: $ k(2) = k(1) + k(1) - 2k(1 \cdot 1) = 1 + 1 - 2(1) = 0 $. Verify consistency: Try $ x = 2, y = 1 $: $ k(3) = k(2) + k(1) - 2k(2) = 0 + 1 - 0 = 1 $. Try $ x = y = 2 $: $ k(4) = k(2) + k(2) - 2k(4) $ → $ k(4) + 2k(4) = 0 + 0 $ → $ 3k(4) = 0 $ → $ k(4) = 0 $. Assume $ k(x) = 0 $ for all $ x $, but $ k(1) = 1 📰 eq 0 $. Contradiction? Wait, from $ k(2) = 0 $, check $ x = 1, y = -1 $: $ k(0) = k(1) + k(-1) - 2k(-1) = 1 + k(-1) - 2k(-1) = 1 - k(-1) $. Also $ k(0) = k(0 + 0) = 2k(0) - 2k(0) = 0 $? No: $ k(0) = k(0 + 0) = 2k(0) - 2k(0) = 0 $. So $ k(0) = 0 $. Then $ 0 = 1 - k(-1) $ → $ k(-1) = 1 $. Then $ x = -1, y = -1 $: $ k(-2) = 2k(-1) - 2k(1) = 2(1) - 2(1) = 0 $. $ x = 1, y = -1 $: $ k(0) = k(1) + k(-1) - 2k(-1) = 1 + 1 - 2(1) = 0 $, consistent. Now $ x = 2, y = -1 $: $ k(1) = k(2) + k(-1) - 2k(-2) = 0 + 1 - 0 = 1 $, matches. No contradiction. Thus $ k(2) = 0 $. Final answer: $ oxed{0} $. 📰 Finance Consultancy Near Me 9188226 📰 Why Everyones Hiding Their Name In Booksyou Need To Do It Too 6366290 📰 South Philly Cheesesteaks 680534 📰 American Mcgee Alice 1984220 📰 Banking Promotion 1751735 📰 Unlock Django Magic With Django Samuel Ls Revolutionary Framework Tricks 9117984 📰 Raven 2 Gameplay 7162966 📰 Grand Hyatt Scottsdale Resort 8861501 📰 Osiris And 749560 📰 Brello Revelations The Secret Tool Everyones Talking About Freely Available 5822275 📰 Buenos Tardes 6625280 📰 Amazon Prime Lawsuit Settlement 5001093 📰 How To Find My Bank Routing Number 6568192 📰 Get The Perfect Fit The Ultimate Snowboard Size Chart Everyone Needs 4059481 📰 Unlock Faster Smoother Streaming With The Ultimate Plex App For Ipad 2423845