We have , f(x)=x2+2x
Calculatef(x1):⇒f(x1)=x12+2xCalculatef(x2):⇒f(x2)=x22+2x
Suppose the function is injective then f(x1)=f(x2)
if this condition is not true then function is not injective
Now, f(x1)=f(x2)
⇒x12+2x=x22+2x
⇒x12−x22+2x1−2x2=0⇒(x1−x2)(x1+x2)+2x1−2x2=0⇒(x1−x2)(x1+x2+2)=0
Since, x1=x2 and x1+x2+2=0, for any x∈N
So function f(x) is not injective.
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