1. Analyze $f(x)=[x^2-3]$ on $\left[-\tfrac12,2\right]$

For the greatest integer function $[u]$, discontinuity occurs when $u$ crosses an integer.

So $f(x)=[x^2-3]$ can be discontinuous at those $x$ where $x^2-3\in \mathbb Z.$ But we only need points where the value actually jumps in the given interval.

Now for $x\in \left[-\tfrac12,2\right]$, $x^2\in [0,4],$ so $x^2-3\in [-3,1].$

Thus possible integer values crossed are $-3,-2,-1,0,1.$

Solve:

• $x^2-3=-3 \implies x^2=0 \implies x=0$
• $x^2-3=-2 \implies x^2=1 \implies x=\pm 1$
• $x^2-3=-1 \implies x^2=2 \implies x=\pm \sqrt2$
• $x^2-3=0 \implies x^2=3 \implies x=\pm \sqrt3$
• $x^2-3=1 \implies x^2=4 \implies x=\pm 2$

Restricting to $\left[-\tfrac12,2\right]$, we get candidates: $0,1,\sqrt2,\sqrt3,2.$

Now check whether these are actual discontinuities.

Since $x^2-3$ has a minimum at $x=0$, near $x=0$ we have $x^2-3\ge -3,$ and for $x\ne 0$ close to $0$, $-3<x^2-3<-2.$ Hence $f(0)=[-3]=-3,$ but for nearby $x\ne 0$, $f(x)=-3.$ So there is no jump at $x=0$.

At $x=1,\sqrt2,\sqrt3,2$, the expression $x^2-3$ crosses an integer level and $f$ jumps. Therefore, $f$ is discontinuous exactly at $x=1,\sqrt2,\sqrt3,2,$ which are 4 points.

So:

• A is false
• B is true

2. Find explicit form of $f(x)$ on subintervals

Using thresholds:

• For $x^2-3\in[-3,-2)$ i.e. $0\le x^2<1$, we get $f(x)=-3$.
• For $x^2-3\in[-2,-1)$ i.e. $1\le x^2<2$, we get $f(x)=-2$.
• For $x^2-3\in[-1,0)$ i.e. $2\le x^2<3$, we get $f(x)=-1$.
• For $x^2-3\in[0,1)$ i.e. $3\le x^2<4$, we get $f(x)=0$.
• At $x=2$, $f(2)=1$, but $2$ is not in the open interval for differentiability of $g$.

Now on $\left(-\tfrac12,2\right)$, only nonnegative values except the small negative part occur:

• For $x\in\left(-\tfrac12,1\right)$, $x^2<1 \implies f(x)=-3$
• For $x\in[1,\sqrt2)$, $f(x)=-2$
• For $x\in[\sqrt2,\sqrt3)$, $f(x)=-1$
• For $x\in[\sqrt3,2)$, $f(x)=0$

3. Simplify $g(x)$

Given $g(x)=|x|f(x)+|4x-7|f(x)=f(x)\big(|x|+|4x-7|\big).$

In the interval $\left(-\tfrac12,2\right)$:

• $|x|$ is non-differentiable at $x=0$
• $|4x-7|$ is non-differentiable at $x=\tfrac74$

Also $f(x)$ has jump discontinuities at $x=1,\sqrt2,\sqrt3.$ (We exclude $x=2$ since interval is open.)

So possible non-differentiability points of $g$ are among $0,1,\sqrt2,\sqrt3,\tfrac74.$

4. Check each candidate point

Let $h(x)=|x|+|4x-7|.$ Then $g(x)=f(x)h(x).$

(i) At $x=0$

Here $f(x)=-3$ in a neighborhood of $0$, so $g(x)=-3(|x|+|4x-7|).$ Since $|x|$ is not differentiable at $0$, $g$ is not differentiable at $0$.

(ii) At $x=\tfrac74$

Since $\tfrac74\in(\sqrt3,2)$, we have $f(x)=0$ in a whole neighborhood of $\tfrac74$. Thus near $x=\tfrac74$, $g(x)=0.$ Hence $g$ is actually differentiable there. So $x=\tfrac74$ is not a non-differentiability point.

(iii) At $x=1$

Here $h(x)$ is continuous and $h(1)=|1|+|4-7|=1+3=4\ne 0.$ Since $f$ jumps from $-3$ to $-2$ at $x=1$, $g=f\,h$ also jumps at $x=1$:

• left value near $1$: $g=-3h$
• right value near $1$: $g=-2h$ Hence $g$ is discontinuous, so not differentiable at $x=1$.

(iv) At $x=\sqrt2$

Again $h(\sqrt2)>0$, and $f$ jumps from $-2$ to $-1$. So $g$ is discontinuous, hence not differentiable at $x=\sqrt2$.

(v) At $x=\sqrt3$

Again $h(\sqrt3)>0$, and $f$ jumps from $-1$ to $0$. So $g$ is discontinuous, hence not differentiable at $x=\sqrt3$.

5. Count non-differentiability points of $g$ in $\left(-\tfrac12,2\right)$

They are exactly $0,1,\sqrt2,\sqrt3,$ so total = $4$ points.

Therefore:

• C is true
• D is false

6. Final option check

• A: False
• B: True
• C: True
• D: False

So the correct answers are $\boxed{B,\ C}.$