JEE MAIN - Mathematics (2020 - 6th September Evening Slot - No. 1)
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 :
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 :
$$ {1 \over 3}$$
–3
$$ - {1 \over 3}$$
3
Explanation
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
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
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