Answer: Let x,y∈Z such that f(x)=f(y) ⇒x 2 +x=y 2 +y ⇒x 2 −y 2 +x−y=0 ⇒(x−y)(x+y+1)=0 ⇒x=y or x=−y−1 Since f(x)=f(y) does not yield the unique solution x=y but also provides the solution x=−y−1, So it is not a one – one function. For example, if y=1, then x=1 from x=y and also x=−2 from y=−x−1 This means that 1 and −2 have the same image Hence, f is a many-one function Reply
Answer:
Let x,y∈Z such that f(x)=f(y)
⇒x
2
+x=y
2
+y
⇒x
2
−y
2
+x−y=0
⇒(x−y)(x+y+1)=0
⇒x=y or x=−y−1
Since f(x)=f(y) does not yield the unique solution x=y but also provides the solution x=−y−1,
So it is not a one – one function.
For example, if y=1, then x=1 from x=y and also x=−2 from y=−x−1
This means that 1 and −2 have the same image
Hence, f is a many-one function