For a suitably chosen real constant a, let a function, f: R − {−a} → R be defined by f(x) = (a-x)/(a+x). Further suppose that for any real number x ≠ −a and f(x) ≠ −a, (f ? f)(x) = x. Then f(-1/3) is equal to:
For a suitably chosen real constant a, let a function, f: R − {−a} → R be defined by f(x) = (a-x)/(a+x). Further suppose that for any real number x ≠ −a and f(x) ≠ −a, (f ? f)(x) = x. Then f(-1/3) is equal to:
Option 1 -
1/3
Option 2 -
-1/3
Option 3 -
-3
Option 4 -
3
Asked by Shiksha User
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1 Answer
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Correct Option - 4
Detailed Solution:f (f (x) = (a-f (x)/ (a+f (x) = x
Let f (x) = y. (a-y)/ (a+y) = x ⇒ a-y = ax + xy ⇒ a (1-x) = y (1+x) ⇒ y = a (1-x)/ (1+x)
⇒ f (x) = a (1-x)/ (1+x)
From the given options, we infer that comparing the derived f (x) leads to a=1.
⇒ a = 1
So f (x) = (1-x)/ (1+x)
f (-1/2) = (1 - (-1/2)/ (1 + (-1/2) = (3/2)/ (1/2) = 3
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