Q.Let $f(x) = |\sin x|$. Then
(A) $f$ is everywhere differentiable
(B) $f$ is everywhere continuous but not differentiable at $x = n\pi$, $n \in \mathbb{Z}$
(C) $f$ is everywhere continuous but not differentiable at $x = (2n + 1)\dfrac{\pi}{2}$, $n \in \mathbb{Z}$
(D) none of these

The key idea is that $|\sin x|$ inherits the continuity of $\sin x$ but loses differentiability where $\sin x$ changes sign — that is, at its zeros.

• $\sin x$ is continuous everywhere, so $f(x) = |\sin x|$ is also continuous everywhere.
• $\sin x$ is zero at $x = n\pi$, $n \in \mathbb{Z}$. At these points, the graph of $|\sin x|$ has a sharp corner (the derivative from left and right differ in sign). …

The key to this problem is understanding how the absolute value operation affects differentiability. When you wrap a function inside $|\cdot|$, the result inherits the continuity of the original function, but differentiability can break at points where the inside function equals zero — because that's where the graph gets a sharp "V" or inverted "V" corner.

Let's think about $\sin x$ first. It's smooth and differentiable everywhere. Its graph crosses the x-axis at every integer multiple of $\pi$: $x = 0, \pm\pi, \pm 2\pi, \dots$. At each of these points, $\sin x$ changes sign — from positive to negative or vice versa.

Now apply the absolute value. Where $\sin x$ is positive, $|\sin x| = \sin x$ — the graph is unchanged. Where $\sin x$ is negative, $|\sin x| = -\sin x$ — the graph gets reflected upward across the x-axis. So at each zero crossing, instead of a smooth transition through the axis, you get a sharp point: the left-hand slope and right-hand slope are opposites. That's a classic non-differentiable corner.

Let's verify this step by step.

1. Continuity first. The function $\sin x$ is continuous everywhere. The absolute value function $| \cdot |$ is continuous everywhere. The composition of continuous functions is continuous. So $f(x) = |\sin x|$ is continuous for all real $x$. No issues here.

2. Where could differentiability fail? Differentiability can only fail where the inside function $\sin x$ is zero, because that's where the absolute value might create a corner. At any point where $\sin x \neq 0$, there is a small neighbourhood where $\sin x$ keeps the same sign, so $f(x)$ equals either $\sin x$ or $-\sin x$ locally — both are differentiable. So the only candidates for non-differentiability are $x = n\pi$, $n \in \mathbb{Z}$.

3. Check at a typical zero, say $x = 0$. Compute the left-hand derivative and right-hand derivative separately.

   • For $x$ just to the right of $0$, $\sin x > 0$, so $f(x) = \sin x$. The derivative from the right is $\cos 0 = 1$.
   • For $x$ just to the left of $0$, $\sin x < 0$, so $f(x) = -\sin x$. The derivative from the left is $-\cos 0 = -1$. Since $1 \neq -1$, the derivative does not exist at $x=0$. The same reasoning applies at every $x = n\pi$ because $\sin x$ changes sign across each of these points. …

Method: Differentiability of $|g(x)|$ Where $g$ Is Periodic/Oscillating

Use this whenever you must find every point where the absolute value of a smooth, sign-changing function (like $\sin x$) fails to be differentiable.

Steps

Step 1: Confirm continuity everywhere first.

$g$ continuous $\Rightarrow |g|$ continuous, by composition of continuous functions. This is rarely the issue for such questions — the real question is differentiability.

Step 2: Locate every zero of $g(x)$ — these are the only candidate points where $|g(x)|$ can have a corner, since away from a zero, $g$ keeps one constant sign in a small neighbourhood and $|g(x)|$ equals $\pm g(x)$ smoothly there.

Step 3: At each zero, determine whether $g$ actually changes sign across it (a genuine zero-crossing) as opposed to just touching zero. …

Common Mistakes

Mistake 1: Choosing option (C) — flagging the points $x=(2n+1)\frac{\pi}{2}$ (where $\sin x = \pm 1$) as non-differentiable, confusing extrema with zero-crossings.

Why it's wrong: at these points $\sin x$ reaches its maximum or minimum and does not change sign nearby, so $|\sin x|$ equals a single smooth branch ($\sin x$ or $-\sin x$) in a neighbourhood of that point — there is no corner there.

Correct approach: only the actual zero-crossings of $\sin x$ (where the sign flips) create corners in $|\sin x|$ — locate those specifically ($x=n\pi$), not the extrema.

Mistake 2: Assuming $|\sin x|$ is differentiable everywhere simply because $\sin x$ is. …