QJEEM20Ma00bc: JEE Main 2020 Mathematics - Class 11 JEEM20Ma00bc Solution: For a suitably chosen real constant a, let a function, be…

Question

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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

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