Prove that π(π₯) = π₯^2 + 2π₯ is not injective.
We have , "f(x)=x^2+2x"
"Calculate f(x_1):\\\\\n\n\\Rightarrow f(x_1)=x_1^2+2x\\\\\n\nCalculate f(x_2):\\\\\n\n\\Rightarrow f(x_2)=x_2^2+2x"
Suppose the function is injective then "f(x_1)=f(x_2)"
if this condition is not true then function is not injective
Now, "f(x_1)=f(x_2)"
"\\Rightarrow x_1^2+2x=x_2^2\u200b+2x\u200b"
"\\Rightarrow x_1^2-x_2^2\u200b+2x\u200b_1-2x_2=0\\\\\n\n\\Rightarrow (x_1\u2212x_2)(x_1\u200b+x_2)+2x_1\u22122x_2=0\\\\\n\n\\Rightarrow (x_1\u200b\u2212x_2)(x_1\u200b+x_2+2)=0\\\\"
Since, "x_1\\neq x_2" and "x_1\u200b+x_2\u200b+2\\neq0 ," for any "x\u2208N"
So function f(x) is not injective.
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