A function from A = {x : -1 leq x leq 1} to i | vedclass

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(C) function is a bijection if it is both injective (one-to-one) and surjective (onto).For the domain and codomain $A = [-1, 1]$:$A) f(x) = x|x|$ is a

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$f(x) = x^2$

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(C) function is a bijection if it is both injective (one-to-one) and surjective (onto).For the domain and codomain $A = [-1, 1]$:$A) f(x) = x|x|$ is a bijection because it is strictly increasing and maps $[-1, 1]$ onto $[-1, 1]$.$B) f(x) = x^3$ is a bijection because it is strictly increasing and maps $[-1, 1]$ onto $[-1, 1]$.$C) f(x) = x^2$ is not a bijection. It is not injective because $f(1) = f(-1) = 1$. It is also not surjective because the range is $[0, 1]$,which is not equal to the codomain $[-1, 1]$.$D) f(x) = \sin \left(\frac{\pi x}{2}\right)$ is a bijection because it is strictly increasing and maps $[-1, 1]$ onto $[-1, 1]$.Thus,the correct option is $C$.

$A$ function from $A = \{x : -1 \leq x \leq 1\}$ to itself which is not a bijection is

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