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Show that f: R- Rdefined by f(x) = x + 2 is a bijection also find f (4) and {f^-1(x):2 |
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Answer» iven, FUNCTION f:R→R such that f(x)=1+x 2 , Let A and B be two sets of REAL numbers.Let x 1 ,x 2 ∈A such that f(x 1 )=f(x 2 ).⇒1+x 12 =1+x 22 ⇒x 12 −x 22 =0⇒(x 1 −x 2 )(x 1 +x 2 )=0⇒x 1 =±x 2 . Thus f(x 1 )=f(x 2 ) does not imply that x 1 =x 2 .For instance, f(1)=f(−1)=2, i.e. , two ELEMENTS (1, -1) of A have the same image in B. So, f is many-one function.Now, y=1+x 2 ⇒x= y−1 ⇒elements < y have no pre-image in A (for instance an element -2 in the codomain has no pre-image in the domain A). So, f is not onto.Hence, f is neither one-one onto. So, it is not bijectionhope it HELPS. |
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