Let f, g a: N ® N such that f (n + 1) = f (n) +f (1) $\forall n \in N$ and g be any arbitrary function. Which of the following statements is NOT true?

Option 1 - If fog is one – one, then g is one – one

Option 2 - f is one – one

Option 3 - If g is onto, then fog is one – one

Option 4 - If f is onto, then f (n) =   <math> <mrow> <mi>n</mi> <mtext> </mtext> <mtext> </mtext> <mo>∀</mo> <mi>n</mi> <mo>∈</mo> <mi>N</mi> </mrow> </math>

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Alok Kumar Singh | Contributor-Level 10

7 months ago

Correct Option - 4

Detailed Solution:

f : N ->N

f (n + 1) = f (n) + f (1)

Let f (1) = a, a $\in$ N

f (2) = f (1) + f (1) = 2a

f (3) = f (2) + f (1) = 3a

and so on

->f (m) = ma, m, a $\in$ N

->f is one – one, Þ option (2) is true.

Suppose f (g (x) is one-one

then f (g (x₁) $\neq$ f (g (x₂) for x₁ $\neq$ x₂

->g (x₁) $\neq$ g (x₂) (as f is one-one)

->g is one – one

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