Differentiability and continuity of $f(x) =\sin|kx|$ QUESTION(MCQ)  :Let $f(x) =\sin|kx|, \;x$ is real. then 
1.f is continuous nowhere. 
2.f is continuous and differentiable everywhere except at integral values of k
3.f is continuous and differentiable everywhere. 
4.f is continuous but nowhere differentiable. 
MY VIEW :Easily discarding option 1,3and 4.But for option 2, I was trying to solve through drawing the graph. First checking for k=1,first I draw the graph on positive x axis and then making the graph symmetrical about y axis, I draw the graph on negative x axis. Sharpness arises only at x=0.
Also for k=2,we first draw the graph on positive x axis by shrinking the graph 2 times along x axis and then making it symmetrical about y axis, I draw the graph. Sharpness arises only at  x=0 . 
HOW IS OPTION 2 CORRECT?PLEASE GIVE ARGUMENTS. 
 A: If $k=0$ the correct answer is (3). Let now $k\neq0$. 
Since $|kx| = |k|\cdot|x|$, without loss of generality we can suppose $k>0$. Then, the function $f(x)$ is just a rescaled version of $g(x)=\sin|x|$. This function is continuous everywhere, and differentiable infinitely many times everywhere except at $x=0$, where there is a cusp. So, none of the options is correct.
However, if the function was not $f(x)=\sin|kx|$, but rather $f(x)=|\sin(kx)|$, it would have cusps at $kx=n\pi$ with $n\in\mathbb{Z}$, i.e. at integral values of $\pi/k$. This function would be everywhere continuous and infinitely many times differentiable everywhere except at the cusps $x_n = n\pi/k$. This is something similar to (but not quite the same as) answer (2).
A: For $k=0$ your function is identically zero and it is continuous and differentiable everywhere.
For $ k\ne 0 $ your function $$f(x) = |\sin (kx)|$$ is continuous everywhere because it is the composite function of two continuous functions.
Your function is differentiable everywhere except where it takes the value of zero. 
That is where $$\sin (kx) =0 \iff kx=n\pi \iff x=n\pi /k$$
for integer values of $n$ 
