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A real valued function f(x) satisfies the functional equation f(x –y) = f(x) f(y)–f (a–x) f(a + y) where a is a given constant and f(0)=1, then f(2a –x) is equal to -a)–f(x)b)f(x)c)f(a) + f(a –x)d)f(–x)Correct answer is option 'A'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared
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A real valued function f(x) satisfies the functional equation f(x –y) = f(x) f(y)–f (a–x) f(a + y) where a is a given constant and f(0)=1, then f(2a –x) is equal to -a)–f(x)b)f(x)c)f(a) + f(a –x)d)f(–x)Correct answer is option 'A'. Can you explain this answer?, a detailed solution for A real valued function f(x) satisfies the functional equation f(x –y) = f(x) f(y)–f (a–x) f(a + y) where a is a given constant and f(0)=1, then f(2a –x) is equal to -a)–f(x)b)f(x)c)f(a) + f(a –x)d)f(–x)Correct answer is option 'A'. Can you explain this answer? has been provided alongside types of A real valued function f(x) satisfies the functional equation f(x –y) = f(x) f(y)–f (a–x) f(a + y) where a is a given constant and f(0)=1, then f(2a –x) is equal to -a)–f(x)b)f(x)c)f(a) + f(a –x)d)f(–x)Correct answer is option 'A'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice A real valued function f(x) satisfies the functional equation f(x –y) = f(x) f(y)–f (a–x) f(a + y) where a is a given constant and f(0)=1, then f(2a –x) is equal to -a)–f(x)b)f(x)c)f(a) + f(a –x)d)f(–x)Correct answer is option 'A'. Can you explain this answer? tests, examples and also practice JEE tests.