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The number of roots of the eqn., $x-\large\frac{2}{x-1}$$=1-\large\frac{2}{x-1}$ is

$\begin{array}{1 1}(A) \;1 \\(B)\;2\;imaginary \;solutions \\(C)\;0\;or\;no\;solution \\(D)\;Infinite\;solution \end{array}$

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The given equation $x-\large\frac{2}{x-1}$$=1-\large\frac{2}{x-1}$ is defined only if $x\neq 1$
Because for $x=1$ the denominator becomes $0$
$\therefore$ for $x\neq 1$ the equation becomes $x=1$
which is contradiction to $x\neq 1$
answered Jul 25, 2013 by rvidyagovindarajan_1

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