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Ex 1.2, 5 - Show Signum Function is neither one-one nor onto - To prove injective/ surjective/ bijective (one-one & onto)

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  1. Chapter 1 Class 12 Relation and Functions
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Ex 1.2, 5 Show that the Signum Function f: R → R, given by f(x) = 1 for 𝑥 >0﷮ 0 for 𝑥=0﷮−1 for 𝑥<0﷯﷯ is neither one-one nor onto. f(x) = 1 for 𝑥 >0﷮ 0 for 𝑥=0﷮−1 for 𝑥<0﷯﷯ For example: f(0) = 0 f(-1) = -1 f(1) = 1 f(2) = 1 f(3) = 1 Check onto f: R → R f(x) = 1 for 𝑥 >0﷮ 0 for 𝑥=0﷮−1 for 𝑥<0﷯﷯ Value of f(x) is defined only if x is 1, 0, –1 For other real numbers(eg: y = 2, y = 100) there is no corresponding element x Hence f is not onto Thus, f is neither one-one nor onto

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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He provides courses for Mathematics from Class 9 to 12. You can ask questions here.
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