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if \(\ f(x) =\begin {cases}x\,, &\quad \text{if } x \text{ is rational}\\0\,, &\quad \text{if } x \text{ is irrational}\end{cases}\) and \(\ g(x) =\begin {cases}0\,, &\quad \text{if } x \text{ is rational}\\x\,, &\quad \text{if } x \text{ is irrational}\end{cases}\) then f - g is (a) neither one-one nor onto (b) one-one and onto (c) one-one and into (d) many-one and onto |
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Answer» Answer : (b) = one - one and onto . (f – g) (x) = \(\ f(x) - g(x) =\begin {cases}x\,, &\quad \text{if } x \text{ is rational}\\-x\,, &\quad \text{if } x \text{ is irrational}\end{cases}\) Distinct elements of R have distinct images in (f – g). Hence (f – g) is one-one. Also Range of (f – g) = R ⇒ f is onto. ⇒ (f – g) R → R is a one-one onto function |
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