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A function k whose domain is the set of positive integers is defined as k(1) = 4 and k(n) =
k(n - 1) - 2
Function k was evaluated for several numbers. Which of the following are true?
Select each correct answer.

A Function K Whose Domain Is The Set Of Positive Integers Is Defined As K1 4 And Kn Kn 1 2 Function K Was Evaluated For Several Numbers Which Of The Following A class=

Answer :

k(2)=

  • k(1)-2
  • 4-2
  • 2

Option C is true

k(3)

  • k(2)-2
  • 2-2
  • 0

D is true

k(4)

  • k(3)-2=0-2=-2

k(5)

  • -2-2
  • -4

k(6)

  • -4-2
  • -6

E is not true

k(1)

  • k(0)-2

So

  • k(0)-2=4
  • k(0)=6

B is false

Similarly

  • k(-1)-2=k(0)=6
  • k(-1)=8

-1 is not in domain even though

A is false

Only C and D are true

Semsee45go

Answer:

[tex]k(2) = 2[/tex]

[tex]k(3) = 0[/tex]

Step-by-step explanation:

Given:

[tex]\begin{cases}k(1)=4\\k(n)=k(n-1)-2\end{cases}[/tex]

The function [tex]k(n)[/tex] tells us that each term is 2 less than the preceding term.

Therefore:

[tex]\implies k(1)=4[/tex]

[tex]\implies k(2)=k(1)-2=4-2=2[/tex]

[tex]\implies k(3)=k(2)-2=2-2=0[/tex]

[tex]\implies k(4)=k(3)-2=0-2=-2[/tex]

[tex]\implies k(5)=k(4)-2=-2-2=-4[/tex]

[tex]\implies k(6)=k(5)-2=-4-2=-6[/tex]

Rearranging the formula to make the preceding term the subject:

[tex]\begin{aligned} k(n) & = k(n-1) - 2\\\implies k(n)+2 & = k(n-1)\\k(n-1) & =k(n)+2\end{aligned}[/tex]

Therefore:

[tex]\implies k(1)=4[/tex]

[tex]\implies k(0)=k(1)+2=4+2=6[/tex]

[tex]\implies k(-1)=k(0)+2=6+2=8[/tex]

Conclusion

Therefore, the correct answers from the answer options are:

[tex]k(2) = 2[/tex]   and   [tex]k(3) = 0[/tex]

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