Q.The set of points where the function $f$ given by $f(x) = |2x - 1|\sin x$ is differentiable is
(A) $\mathbb{R}$
(B) $\mathbb{R} - \left\{\dfrac{1}{2}\right\}$
(C) $(0, \infty)$
(D) none of these

The key idea is that $|2x-1|$ is not differentiable at $x = \frac{1}{2}$, and the product rule for differentiability fails at that point unless the other factor is zero.

1. $|2x-1|$ is differentiable everywhere except at $x = \frac{1}{2}$, where it has a sharp corner.
2. $\sin x$ is differentiable everywhere.
3. The product $f(x) = |2x-1|\sin x$ is differentiable wherever both factors are differentiable, except possibly at $x = \frac{1}{2}$. At $x = \frac{1}{2}$, check differentiability directly: …

The key to this problem lies in understanding how differentiability interacts with absolute value functions. An absolute value function $|g(x)|$ is differentiable at points where $g(x) \neq 0$, but at points where $g(x) = 0$, it typically has a sharp corner — unless the other factor in the product "saves" it by being zero there too, smoothing out the kink.

Here, $f(x) = |2x-1| \sin x$. The absolute value part $|2x-1|$ has a potential problem at $x = \frac{1}{2}$, where $2x-1 = 0$. At that point, $\sin\left(\frac{1}{2}\right)$ is not zero — it's about $0.479$. So the product inherits the corner from the absolute value. Everywhere else, $|2x-1|$ is just a nice linear function (either $2x-1$ or $-(2x-1)$), and $\sin x$ is smooth, so their product is differentiable.

Let's check this carefully.

1. Where is $|2x-1|$ differentiable?

   The function $|2x-1|$ is differentiable for all $x \neq \frac{1}{2}$. At $x = \frac{1}{2}$, it has a sharp V-shaped corner — the left-hand derivative is $-2$ and the right-hand derivative is $+2$, so they don't match.

2. What about the product with $\sin x$?

   Since $\sin x$ is differentiable everywhere, the product $f(x) = |2x-1| \sin x$ will be differentiable wherever both factors are differentiable — that is, for all $x \neq \frac{1}{2}$. The only question is whether the corner at $x = \frac{1}{2}$ gets "fixed" by $\sin x$ being zero there. But $\sin\left(\frac{1}{2}\right) \neq 0$, so no smoothing occurs.

3. Check the derivative from first principles at $x = \frac{1}{2}$

   Let's compute the left-hand and right-hand derivatives at $x = \frac{1}{2}$ to confirm.

   For $x < \frac{1}{2}$, $2x-1 < 0$, so $|2x-1| = -(2x-1) = 1 - 2x$.

   For $x > \frac{1}{2}$, $2x-1 > 0$, so $|2x-1| = 2x-1$.

   The left-hand derivative:

For $h < 0$, $\frac{1}{2}+h < \frac{1}{2}$, so $|2(\frac{1}{2}+h)-1| = |2h| = -2h$ (since $h<0$). Also $f\left(\frac{1}{2}\right) = |0|\sin\frac{1}{2} = 0$. So:

The right-hand derivative: for $h > 0$, $|2(\frac{1}{2}+h)-1| = |2h| = 2h$, so:

$f'_+\left(\frac{1}{2}\right) = \lim_{h \to 0^+} \frac{(2h) \sin\left(\frac{1}{2}+h\right)}{h} = \lim_{h \to 0^+} 2 \sin\left(\frac{1}{2}+h\right) = 2 \sin\frac{1}{2}$ …

Method: Differentiability of $|\text{linear}(x)| \times (\text{smooth function})$

Use this whenever a function is a product of an absolute value of a linear expression and another differentiable function.

Steps

Step 1: Find the corner point — where the expression inside the modulus equals zero.

For $|ax+b|$, solve $ax+b=0$. Away from this single point, $|ax+b|$ is just $\pm(ax+b)$, a smooth linear function, so the product with any differentiable function is automatically differentiable there.

Step 2: Check whether the OTHER factor vanishes at that same corner point.

If $h(\text{corner}) = 0$, the kink can be smoothed out and the product may still be differentiable there — this needs a direct check. If $h(\text{corner}) \neq 0$, the corner survives into the product.

Step 3: When the corner survives, confirm it rigorously using one-sided derivatives. …

Common Mistakes

Mistake 1: Concluding that $f(x)=|2x-1|\sin x$ must be non-differentiable at $x=\frac{1}{2}$ purely because $|2x-1|$ itself is non-differentiable there, without checking the other factor.

Why it's wrong: a corner in one factor of a product can be 'smoothed away' if the other factor happens to be zero at exactly that point — this is not automatic, and must be checked.

Correct approach: evaluate the other factor (here $\sin x$) at the suspected corner; since $\sin\frac{1}{2} \neq 0$, the corner genuinely survives into the product, and only then is it safe to conclude non-differentiability there.

Mistake 2: Skipping the explicit left/right derivative computation and just asserting non-differentiability from intuition. …