TS ICET 23 May 2018 Shift 2 Paper

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Question : 114 of 200

Marks: +1, -0

If f (x) =

\x/{\x-1}

then

{\f(a)}/{\f(\a+1)}

=

$\a^2/{\a^2 - 1}$
$\a^2/{\a^2 + 1}$
$\f(\a^2)$
$\f(\a^3)$

Solution:

f(x)=\;{x}/{x-1}

f(a)=\;{a}/{a-1} ; f(a+1)=\;{a+1}/{a}

=\;{f(a)}/{f(a+1)}=\;{a}/{a-1} × \;{a}/{a+1}=\;{a^{2}}/{a^{2}-1}

=f(a^2)

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