KCET · Maths · Differential Equations
Let \( f: N \rightarrow N \) defined by \( f(n)=\left\{\begin{array}{cl}\frac{n+1}{2} ; & \text { if } n \text { is odd } \\ \frac{n}{2} ; & \text { if } n \text { is even }\end{array}\right. \)
then \( \mathrm{f} \) is
- A one-one and onto
- B one-one but not onto
- C onto but not one-one
- D neither one-one nor onto
Answer & Solution
Correct Answer
(C) onto but not one-one
Step-by-step Solution
Detailed explanation
Given that, \( f: N \rightarrow N \) defined by
\( f(n)=\left\{\begin{array}{cc}\frac{n+1}{2} & n \text { is odd } \\ \frac{n}{2} & n \text { is even }\end{array}\right. \)
For \( n=1 \), we have \( f(1)=\frac{1+1}{2}=1 \)
and, if \( n=2 \), we have \( f(2)=\frac{2}{2}=1 \)
So, \( f(1)=f(2) \) but \( 1 \neq 2 \). Therefore, \( f(x) \) is not one-one.
Now, \( f(x)=\frac{n+1}{2} \) if \( \mathrm{n} \) is odd
if \( y=\frac{n+1}{2} \), then \( n=2 y-1, \forall y \)
Also, \( f(x)=\frac{n}{2} \) if \( n \) is even. That is,
\( y=\frac{n}{2} \) or \( n=2 y, \forall y \)
Therefore, \( f(x) \) is onto.
\( f(n)=\left\{\begin{array}{cc}\frac{n+1}{2} & n \text { is odd } \\ \frac{n}{2} & n \text { is even }\end{array}\right. \)
For \( n=1 \), we have \( f(1)=\frac{1+1}{2}=1 \)
and, if \( n=2 \), we have \( f(2)=\frac{2}{2}=1 \)
So, \( f(1)=f(2) \) but \( 1 \neq 2 \). Therefore, \( f(x) \) is not one-one.
Now, \( f(x)=\frac{n+1}{2} \) if \( \mathrm{n} \) is odd
if \( y=\frac{n+1}{2} \), then \( n=2 y-1, \forall y \)
Also, \( f(x)=\frac{n}{2} \) if \( n \) is even. That is,
\( y=\frac{n}{2} \) or \( n=2 y, \forall y \)
Therefore, \( f(x) \) is onto.
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