Let f(x)={0if x is rationalxif x is irrational g(x)={0if x is irrationalxif x is rational Then the function (f−g)x is
odd
even
neither odd nor even
odd as well as even
A
odd as well as even
B
odd
C
neither odd nor even
D
even
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Solution
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Let f−g=h. Hence, from the given data h(x)=−x if x is rational =x if x is irrational, Hence, h(−x)=−(−x)=x if x is rational =−x if x is irrational, Therefore, h(x)=−h(−x). Hence, the given function, f−g, is an odd function.
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