Q132 / 200 · Manipal (MET) 2010 · Mathematics
If $z=\tan (y+a x)+\sqrt{y-a x}$, then $z_{x x}-a^2 z_{y y}$ is equal to
- Correct answerA0
- B2
- C$z_x+z_y$
- D$z_x z_y$
Solution
We have, $z=\tan (y+a x)+\sqrt{y-a x}$ $\Rightarrow z_x=a \sec ^2(y+a x)+\frac{(-a)}{2 \sqrt{y-a x}}$
$\Rightarrow z_{x x}=2 a^2 \sec ^2(y+a x) \tan (y+a x)-\frac{a^2}{4}(y-a x)^{-3 / 2}$
and $z_y=\sec ^2(y+a x)+\frac{1}{2 \sqrt{y-a x}}$
$\Rightarrow z_{y y}=2 \sec ^2(y+a x) \tan (y+a x)-\frac{1}{4}(y-a x)^{-3 / 2}$
$\begin{aligned} & a^2 z_{y y}=2 a^2 \sec ^2(y+a x) \cdot \tan (y+a x) -\frac{a^2}{4}(y-a x)^{-3 / 2}\end{aligned}$
Now, $z_{x x}-a^2 z_{y y}=0$
Hence, the correct answer is option: 'A'.