We'll solve this problem considering the given constraint
as it follows:
[f(2)]^2 =
(2/3)*f(1)
To calculate [f(2)]^2, w'ell have to calculate
first [f(x)]^2.
[f(x)]^2 = (2/x +
k)^2
We'll expand the
square:
[f(x)]^2 = (2/x)^2 + 4k/x +
k^2
We'll calculate
[f(2)]^2:
[f(2)]^2 = (2/2)^2 + 4k/2 +
k^2
[f(2)]^2 = 1 + 2k + k^2
(1)
(2/3)*f(1) = (2/3)*(2 + k)
(2)
We'll equate (1) and
(2):
1 + 2k + k^2 = (2/3)*(2 +
k)
We'll recognize that 1 + 2k + k^2 is a perfect
square;
1 + 2k + k^2 = (1 +
k)^2
Let (1+k)^2 = t^2 => k + 2 = t +
1
t^2 - 2t/3 - 2/3 = 0
3t^2 -
2t - 2 = 0
t1 = [2 + sqrt(4 +
24)]/6
t1 = (2+2sqrt7)/6
t1 =
(1+sqrt7)/3
t2 =
(1-sqrt7)/3
But k+ 1 = t1 => k = t1 - 1 => k1
= (-2+sqrt7)/3
But k+ 1 = t2 => k = t2 - 1 =>
k2 = (-2-sqrt7)/3
The possible values of k
are: {(-2-sqrt7)/3 ; (-2+sqrt7)/3}.
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