The function f is defined by f „(x) =… ƒ -1/x for all non-zero numbers x. If f „(a) …= ƒ-1/2 and f „(ab)… = ƒ1/6, then b ƒ=
(A) 3
(B)1/3
(C) -1/3
(D) -3
(E) -12
OA D
Pl. help me out with this question.
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The function f is defined by
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Jay@ManhattanReview
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f „(x)… ƒ = -1/xjack0997 wrote:The function f is defined by f „(x) =… ƒ -1/x for all non-zero numbers x. If f „(a) …= ƒ-1/2 and f „(ab)… = ƒ1/6, then b ƒ=
(A) 3
(B)1/3
(C) -1/3
(D) -3
(E) -12
OA D
Pl. help me out with this question.
=ƒ> f „(a)… ƒ = -1/a = -1/2
ƒ=> a ƒ= 2
Also, we have:
f „(ab)… ƒ = -1/ab = 1/6
=> ab = -6
=> b = -6/a = -6/2
=> [spoiler]b = -3[/spoiler]
The correct answer: D
Hope this helps!
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The function f is defined by f „(x) =… ƒ -1/x for all non-zero numbers x. If f „(a) …= ƒ-1/2 and f „(ab)… = ƒ1/6, then b ƒ=
f(a) = -1/2
and, by definition,
f(a) = -1/a
so a = 2
f(ab) = f(2b) = 1/6
and, by definition
f(2b) = -1/2b
so b = -3
f(a) = -1/2
and, by definition,
f(a) = -1/a
so a = 2
f(ab) = f(2b) = 1/6
and, by definition
f(2b) = -1/2b
so b = -3
