The function $f:\mathbb {R}^2 \to \mathbb{R}$ is defined by $$f(x,y) = \begin{cases} 0 , & \text{ y$\lt$o} \\ \sin x+2\cos y, & \text{ 0 $\le{y}\lt{\frac{\pi}{2}}$}\\ \cos x+2\sin y, &\text{$\frac{\pi}{2} \le{y}\lt\pi$}\\ \cos x ,&\text{$y\ge\pi$} \end{cases}$$
The axes where $f$ is discontinuous are
i) $y=0, y={\frac{\pi}{2}}$ & $y={\pi}$
ii) $y=0, y={\frac{\pi}{2}}$
iii) $y={\frac{\pi}{2}}$ & $y={\pi}$
iv) nowhere
Fix $x$
$\lim_{y\to 0^{-}}f(x,y)=0$ and $\lim_{y\to 0^{+}}f(x,y)=\lim_{y\to 0^{+}}\sin x+2\cos y=\sin x+2$
Left and right hand side limits are not same . Hence not continuous at $y=0$.
$\lim_{y\to {\frac{\pi}{2}}^{-}}f(x,y)=\lim_{y\to {\frac{\pi}{2}}^{-}}\sin x+2\cos y=\sin x$ and $\lim_{y\to {\frac{\pi}{2}}^{+}}f(x,y)=\lim_{y\to 0^{+}}\cos x+2\sin y=\cos x+2$
Hence not continuous at $\frac{\pi}{2}$
similarly I can show that function is continuous at $\pi$
Hence option (ii) is correct.
Am I correct?
Yes, option $(ii)$ is the right answer. To prove discontinuity at a point, we just have to show that two path that leads to different value at that point.