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