functions 700+

[Night reader]

If f(x) = 5 - 2x and f(3k) = f(k + 1), then f(k) =

A. 0.5
B. 1
C. 3
D. 4
E. 6

[anshumishra]

If f(x) = 5 - 2x and f(3k) = f(k + 1), then f(k) =

A. 0.5
B. 1
C. 3
D. 4
E. 6

f(3k) = f(k+1)
=> 5-6k = 5-2(k+1)
=> k = 1/2
f(k) = 5 - 2k = 4

D

[Night reader]

If f(x) = 5 - 2x and f(3k) = f(k + 1), then f(k) =

A. 0.5
B. 1
C. 3
D. 4
E. 6

f(3k) = f(k1)
=> 5-6k = 5-2(k+1)
=> k = 1/2
f(k) = 5 - 2k = 4

D

thanks Anshu, cool solution
i did a bit diff. 3k=k+1, k=1/2 => f(1/2)=5-2*1/2=4

[Anurag@Gurome]

i did a bit diff. 3k=k+1, k=1/2 => f(1/2)=5-2*1/2=4

It is recommended to be careful while using this method to solve these kind of function problem. This is applicable here because it the given function is a linear one. If it is not, for example say f(x) = x², then this method may not give correct results or may miss some possible results.

Anshu's solution is full proof and applicable in general.

[Target2009]

IMO D

[Night reader]

i did a bit diff. 3k=k+1, k=1/2 => f(1/2)=5-2*1/2=4

It is recommended to be careful while using this method to solve these kind of function problem. This is applicable here because it the given function is a linear one. If it is not, for example say f(x) = x², then this method may not give correct results or may miss some possible results.

Anshu's solution is full proof and applicable in general.

Anurag we can always use differentials y=x^n => y`=(x^n)` => y=n*x^(n-1) and so on; I know what is functional relationship and lim-s of y with different functions vary...

[Scott@TargetTestPrep]

If f(x) = 5 - 2x and f(3k) = f(k + 1), then f(k) =

A. 0.5
B. 1
C. 3
D. 4
E. 6

We are given that f(x) = 5 - 2x. So f(3k) = 5 - 2(3k) = 5 - 6k and f(k + 1) = 5 - 2(k + 1) = 3 - 2k. Since f(3k) = f(k + 1), we equate the two expressions, and thus:

5 - 6k = 3 - 2k

2 = 4k

0.5 = k

Therefore, f(k) = f(0.5) = 5 - 2(0.5) = 4.

Answer: D