Which of the following functions is injective | vedclass

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(A) function $f: A 	o B$ is injective (one-to-one) if $f(x_1) = f(x_2) \implies x_1 = x_2$. It is surjective (onto) if the range equals the codomain.

Vedclass · Accepted Answer

$f : N 	o N$,$f(x) = 2x + 3$

Vedclass · Answer

(A) function $f: A 	o B$ is injective (one-to-one) if $f(x_1) = f(x_2) \implies x_1 = x_2$. It is surjective (onto) if the range equals the codomain.For option $A$: $f: N 	o N$,$f(x) = 2x + 3$.Injective: $2x_1 + 3 = 2x_2 + 3 \implies 2x_1 = 2x_2 \implies x_1 = x_2$. Thus,it is injective.Surjective: For $y \in N$,$2x + 3 = y \implies x = \frac{y-3}{2}$. If $y=4$,$x = 0.5 
otin N$. Thus,it is not surjective.For option $B$: $f: R 	o R$,$f(x) = \frac{4x+3}{5}$. This is a linear function,which is both injective and surjective (bijective).For option $C$: $f(x) = x^3 - x$. $f(0) = 0$,$f(1) = 0$,$f(-1) = 0$. Since $f(0) = f(1)$,it is not injective.For option $D$: $f(x) = \ln(|x| + 1)$. $f(1) = \ln(2)$,$f(-1) = \ln(2)$. Since $f(1) = f(-1)$,it is not injective.Therefore,the correct option is $A$.

Which of the following functions is injective but not surjective?

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