A) \[0\]
B) \[2\]
C) \[{{z}_{x}}+{{z}_{y}}\]
D) \[{{z}_{x}}{{z}_{y}}\]
Correct Answer: A
Solution :
We have,\[z=\tan (y+ax)+\sqrt{y-ax}\] \[\Rightarrow \]\[{{z}_{x}}=a{{\sec }^{2}}(y+ax)+\frac{(-a)}{2\sqrt{y-ax}}\] \[\Rightarrow \]\[{{z}_{xx}}=2{{a}^{2}}{{\sec }^{2}}(y+ax)\tan (y+ax)\] \[-\frac{{{a}^{2}}}{4}{{(y-ax)}^{-3/2}}\] and \[{{z}_{y}}={{\sec }^{2}}(y+ax)+\frac{1}{2\sqrt{y-ax}}\] \[\Rightarrow \]\[{{z}_{yy}}=2{{\sec }^{2}}(y+ax)\tan (y+ax)\] \[-\frac{{{a}^{2}}}{4}{{(y-ax)}^{-3/2}}\] \[{{a}^{2}}{{z}_{yy}}\,=2{{a}^{2}}{{\sec }^{2}}(y+ax).\tan (y+ax)\] \[-\frac{{{a}^{2}}}{4}{{(y-ax)}^{-3/2}}\] Now, \[{{z}_{xx}}-{{a}^{2}}{{z}_{yy}}=0\]
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