Q112 / 160 · AP EAMCET 2009 · Mathematics

$z=\tan (y+a x)+\sqrt{y-a x} \Rightarrow z_{x x}-a^2 z_{y y} \quad$ is equal to

  • Correct answer
    A
    0
  • B
    1
  • C
    $z_x+z_y$
  • D
    $z_x z_y$

Solution

$\begin{aligned} & \text { Given, } z=\tan (y+a x)+\sqrt{y-a x} \\ & \Rightarrow \quad z_x=\sec ^2(y+a x) a+\frac{1}{2 \sqrt{y-a x}}(-a) \\ & \Rightarrow \quad z_{x x}=2 \sec ^2(y+a x) \tan (y+a x) a^2 \\ & +\frac{1\left(-a^2\right)}{4(y-a x)^{3 / 2}} \\ & \text { and } \quad z_y=\sec ^2(y+a x)+\frac{1}{2 \sqrt{y-a x}} \\ & \Rightarrow \quad z_{y y}=2 \sec ^2(y+a x) \tan (y+a x) \\ & -\frac{1}{4(y-a x)^{3 / 2}} \\ & \therefore \quad z_{x x}-a^2 z_{y y}=0 \\ & \end{aligned}$

Hence, the correct answer is option: 'A'.

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