For a suitably chosen real constant a, let a function, [f:R - - a R] be defined by [f(x) =…

Question

For a suitably chosen real constant a, let a

function, $$f:R - \left\{ { - a} \right\} \to R$$  be defined by

$$f(x) = {{a - x} \over {a + x}}$$. Further suppose that for any real
number $$x \ne  - a$$ and $$f(x) \ne  - a$$,

(fof)(x) = x. Then $$f\left( { - {1 \over 2}} \right)$$ is equal to :

Accepted Answer

Given, $$f(x) = {{a - x} \over {a + x}}$$

and (fof)(x) = x

$$ \Rightarrow $$ f(f(x)) = $${{a - f\left( x \right)} \over {a + f\left( x \right)}}$$ = x

$$ \Rightarrow $$ $${{a - \left( {{{a - x} \over {a + x}}} \right)} \over {a + \left( {{{a - x} \over {a + x}}} \right)}}$$ = x

$$ \Rightarrow $$ $${{{a^2} + ax - a + x} \over {{a^2} + ax + a + x}}$$ = x

$$ \Rightarrow $$ (a2
 -- a) + x(a + 1) = (a2
 + a)x + x2(a -- 1)

$$ \Rightarrow $$ a(a -- 1) + x(1 -- a2) -- x2(a -- 1) = 0

$$ \Rightarrow $$ a = 1

$$ \therefore $$ f(x) = $${{1 - x} \over {1 + x}}$$

So, $$f\left( { - {1 \over 2}} \right)$$ = $${{1 + {1 \over 2}} \over {1 - {1 \over 2}}}$$ = 3

Solutions · 1