"Let a function f : N $\to$ N be defined by $$f(n) = \left[ {\matrix{ {2n,} & {n = 2,4,6,8,......} \cr {n - 1,} & {n = 3,7,11,15,......} \cr {{{n + 1} \over 2},} & {n = 1,5,9,13,......} \cr } } \right.$$ then, f is"

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Accepted Answer

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When n = 1, 5, 9, 13 then ${{n + 1} \over 2}$ will give all odd numbers.

When n = 3, 7, 11, 15 .....

n $-$ 1 will be even but not divisible by 4

When n = 2, 4, 6, 8 .....

Then 2n will give all multiples of 4

So range will be N.

And no two values of n give same y, so function is one-one and onto.

"

Let a function f : N $\to$ N be defined by

$$f(n) = \left[ {\matrix{ {2n,} & {n = 2,4,6,8,......} \cr {n - 1,} & {n = 3,7,11,15,......} \cr {{{n + 1} \over 2},} & {n = 1,5,9,13,......} \cr } } \right.$$

then, f is

Solution

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Let a function f : N $\to$ N be defined by $$f(n) = \left[ {\matrix{ {2n,} & {n = 2,4,6,8,......} \cr {n - 1,} & {n = 3,7,11,15,......} \cr {{{n + 1} \over 2},} & {n = 1,5,9,13,......} \cr } } \right.$$ then, f is

Solution

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More Sets, Relations and Functions questions

Drill 25 more like these. Every day. Free.

Let a function f : N $\to$ N be defined by

$$f(n) = \left[ {\matrix{ {2n,} & {n = 2,4,6,8,......} \cr {n - 1,} & {n = 3,7,11,15,......} \cr {{{n + 1} \over 2},} & {n = 1,5,9,13,......} \cr } } \right.$$

then, f is