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If ${x \over {(x - 1)\,{{({x^2} + 1)}^2}}} = {1 \over 4}\left[ {{1 \over {(x - 1)}} - {{x + 1} \over {{x^2} + 1}}} \right] + y$ then $y =$
${{(1 - x)} \over {2{{({x^2} + 1)}^2}}}$
${{(1 - x)} \over {3({x^2} + 1)}}$
${{1 + x} \over {2\,{{({x^2} - 1)}^2}}}$
None of these
Solution
(a) ${x \over {(x – 1){{({x^2} + 1)}^2}}} = {1 \over 4}\left[ {{1 \over {(x – 1)}} – {{x + 1} \over {{x^2} + 1}}} \right] + y$
==>${x \over {(x – 1)\,{{({x^2} + 1)}^2}}} = {1 \over 4}\left[ {{1 \over {(x – 1)}} – {{x + 1} \over {{x^2} + 1}}} \right] + {{Ax + B} \over {{{({x^2} + 1)}^2}}}$
==> $4x = {({x^2} + 1)^2} – (x + 1)\,(x – 1)\,({x^2} + 1) + 4(Ax + B)\,(x – 1)$
==> $4A + 2 = 0$,$4B – 4A = 4$ ==> $A = {{ – 1} \over 2}$, $B = {1 \over 2}$
$\therefore y = {{Ax + B} \over {{{({x^2} + 1)}^2}}} = {1 \over 2}{{(1 – x)} \over {{{({x^2} + 1)}^2}}}$
