Prove that f(x) = x^2 + x + 1 is a one - one function
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Step-by-step explanation:
ANSWER
We have, f(x)=x
2
+x+1
Calculate f(x
1
):
⇒ f(x
1
)=x
1
2
+x
1
+1
Calculate f(x
1
):
⇒ f(x
2
)=x
2
2
+x
1
+1
Now, f(x
1
)=f(x
2
)
⇒ x
1
2
+x
1
+1=x
2
2
+x
2
+1
⇒ x
1
2
−x
2
2
+x
1
−x
2
=0
⇒ (x
1
−x
2
)(x
1
+x
2
)+x
1
−x
2
=0......... [ Since, (a
2
−b
2
=(a+b)(a−b) ]
⇒ (x
1
−x
2
)(x
1
+x
2
+1)=0
Since, x
1
+x
2
+1
=0 for any x∈N
∴ x
1
=x
2
So, f is one-one function.
Clearly, f(x)=x
2
+x+1≥3 for all x∈N
So, f(x) does not assume values 1 and 2.
∴ f is not an onto function.
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