A) \[\frac{2}{u}\]
B) \[\frac{3}{u}\]
C) 0
D) \[\frac{1}{u}\]
Correct Answer: A
Solution :
\[2u\frac{\partial u}{\partial x}=2(x-a)\] Þ\[u\,\frac{\partial u}{\partial x}=x-a\] Þ \[u.\frac{{{\partial }^{2}}u}{\partial {{x}^{2}}}+{{\left( \frac{\partial u}{\partial x} \right)}^{2}}=1\] Þ \[u.\frac{{{\partial }^{2}}u}{\partial {{x}^{2}}}=1-{{\left( \frac{x-a}{u} \right)}^{2}}\]Þ \[\frac{{{\partial }^{2}}u}{\partial {{x}^{2}}}=\frac{1}{u}-\frac{{{(x-a)}^{2}}}{{{u}^{3}}}\] Similarly, \[\frac{{{\partial }^{2}}u}{\partial {{y}^{2}}}=\frac{1}{u}-\frac{{{(y-b)}^{2}}}{{{u}^{3}}}\], \[\frac{{{\partial }^{2}}u}{\partial {{z}^{2}}}=\frac{1}{u}-\frac{{{(z-b)}^{2}}}{{{u}^{3}}}\] \[\therefore \] \[\sum \,\frac{{{\partial }^{2}}u}{\partial {{x}^{2}}}=\frac{3}{u}-\frac{1}{{{u}^{3}}}[{{(x-a)}^{2}}+{{(y-b)}^{2}}+{{(z-c)}^{2}}]\] = \[\frac{3}{u}-\frac{1}{{{u}^{3}}}.({{u}^{2}})=\frac{3}{u}-\frac{1}{u}=\frac{2}{u}\].
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