1. Given function

We have

We need the set $K$ of points where $f$ is not differentiable.

2. Compare the two expressions inside max

Let

Since $f(x)=\max\{g_1(x),g_2(x)\}$, the active branch changes where

Multiplying by $-1$,

Now square both sides:

So the two curves meet at

3. Determine which branch is larger

We compare $|x|$ and $\sqrt{1-x^2}$.

• If $|x|<\frac1{\sqrt2}$, then

  $$x^2<\frac12 \implies 1-x^2>x^2 \implies \sqrt{1-x^2}>|x|.$$

  Hence

  $$-|x|>-\sqrt{1-x^2},$$

  so

  $$f(x)=-|x|.$$

• If $|x|>\frac1{\sqrt2}$, then

  $$\sqrt{1-x^2}<|x|,$$

  hence

  $$-\sqrt{1-x^2}>-|x|,$$

  so

  $$f(x)=-\sqrt{1-x^2}.$$

Therefore,

4. Check differentiability of each branch

(i) For $- |x|$

The function $|x|$ is not differentiable at $x=0$, so $-|x|$ is also not differentiable at

Everywhere else it is differentiable.

(ii) For $-\sqrt{1-x^2}$

For $x\in(-1,1)$,

This is differentiable for all $x\in(-1,1)$.

So possible non-differentiable points are:

• $x=0$,
• and the switching points $x=\pm \dfrac1{\sqrt2}$.

5. Check the switching points carefully

A max-function can fail to be differentiable where the two branches meet.

At $x=\dfrac1{\sqrt2}$

Left side uses $f(x)=-|x|=-x$ (since $x>0$), so

Right side uses $f(x)=-\sqrt{1-x^2}$, so

At $x=\dfrac1{\sqrt2}$,

Thus,

not equal. So $f$ is not differentiable at

At $x=-\dfrac1{\sqrt2}$

For $x<0$, $-|x|=x$, so on the inner interval the derivative is

On the outer interval,

At $x=-\dfrac1{\sqrt2}$,

Thus,

(up to side assignment, the two one-sided derivatives are unequal), so $f$ is not differentiable at

6. Collect all non-differentiable points

Therefore,

So $K$ has exactly

elements.

7. Option check

• A: one element ❌
• B: three elements ✅
• C: five elements ❌
• D: two elements ❌

Hence the correct option is