Question

solve inequalities involving absolute values:
1.prove that modulus x < r and modulus x > r.

Answer 1

solution :-  A function f : A --> B is said to be a one - one function or injective mapping, if different elements of A have different f images in B.

If the function f :A → B is such that each element in B (codomain) is the f image of atleast one element in A. then we say that f is a function of A onto B.

given f : R → R, given by f (x) = | x|

We can see that f(-1) = |-1| = 1, f(1) = |1| = 1

⇒ f(-1) = f(1), but -1 ≠ 1.

therefore, f is not one-one.

Now, we consider -1 ∈ R

We know that f(x) = |x| is always positive.

Therefore, there doesn't exist any element x in domain R such that f(x) = |x| = -1

e.g., Range ∈ R⁺

hence, codomain ≠ range

therefore, f is not onto.

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