If f:N→Z is defined by fn=2,ifn=3kk∈z10−n,ifn=3k+1k∈z0,ifn=3k+2k∈zthen n∈N,fn>2=

fn=2,n=0,3,6,9−−−−10−n,n=1,4,7,10−−−−0,n=2,5,8−−−− ∴fn>2, if n=1,4,7

If f:N→Z is defined by fn=2,ifn=3kk∈z10−n,ifn=3k+1k∈z0,ifn=3k+2k∈zthen n∈N,fn>2=

fn=2,n=0,3,6,9−−−−10−n,n=1,4,7,10−−−−0,n=2,5,8−−−− ∴fn>2, if n=1,4,7

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If $f : N \to Z$ is defined by $f (n) = \{\begin{cases} \begin{matrix} 2, & i f & n = 3 k & k \in z \end{matrix} \\ \begin{matrix} 10 - n, & i f & n = 3 k + 1 & k \in z \end{matrix} \\ \begin{matrix} 0, & i f & n = 3 k + 2 & k \in z \end{matrix} \end{cases}$ then $\{n \in N, f (n) > 2\} =$

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$\{3, 6, 4\}$

$\{1, 4, 7\}$

$\{4, 7\}$

$\{7\}$

answer is B.

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Detailed Solution

$\begin{array}{l} f (n) = [\begin{matrix} 2, & n = 0, 3, 6, 9 - - - - \\ 10 - n, & n = 1, 4, 7, 10 - - - - \\ 0, & n = 2, 5, 8 - - - - \end{matrix}] \begin{matrix} \end{matrix} \end{array}$

$∴ f (n) > 2,$ if $n = 1, 4, 7$

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If f:N→Z is defined by fn=2,ifn=3kk∈z10−n,ifn=3k+1k∈z0,ifn=3k+2k∈zthen n∈N,fn>2=

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If $f : N \to Z$ is defined by $f (n) = \{\begin{cases} \begin{matrix} 2, & i f & n = 3 k & k \in z \end{matrix} \\ \begin{matrix} 10 - n, & i f & n = 3 k + 1 & k \in z \end{matrix} \\ \begin{matrix} 0, & i f & n = 3 k + 2 & k \in z \end{matrix} \end{cases}$ then $\{n \in N, f (n) > 2\} =$