Let $$f : [-1, 2] \rightarrow \mathbb{R}$$ be given by $$f(x) = 2x^2 + x + [x^2] - [x]$$, where $$[t]$$ denotes the greatest integer less than or equal to $$t$$. The number of points, where $$f$$ is not continuous, is :

The polynomial part $$2x^{2}+x$$ is continuous everywhere, so any discontinuity of $$f(x)=2x^{2}+x+[x^{2}]-[x]$$ can come only from the step functions $$[x]$$ or $$[x^{2}]$$.

Step 1: Locate the points where $$[x]$$ or $$[x^{2}]$$ can jump.

• $$[x]$$ jumps at every integer lying in $$[-1,2]$$, namely
   $$x=-1,\,0,\,1,\,2$$.

• $$[x^{2}]$$ jumps whenever $$x^{2}=k$$ for an integer $$k$$.
   For $$x\in[-1,2]$$ the values $$k=0,1,2,3,4$$ are possible, giving
   $$x=0,\;x=\pm1,\;x=\pm\sqrt2,\;x=\pm\sqrt3,\;x=\pm2.$$
   Inside $$[-1,2]$$ this list reduces to
   $$x=-1,\,0,\,1,\,\sqrt2,\,\sqrt3,\,2.$$

Thus the only candidates for discontinuity are

$$x=-1,\,0,\,1,\,\sqrt2,\,\sqrt3,\,2.$$

Step 2: Examine each candidate.

Case 1: $$x=0$$

Left limit (take $$x\rightarrow0^{-}$$): $$[x]=-1,\;[x^{2}]=0$$
  $$\displaystyle \lim_{x\to0^{-}}f(x)=2x^{2}+x+0-(-1)=2x^{2}+x+1\longrightarrow1.$$
Right limit (take $$x\rightarrow0^{+}$$): $$[x]=0,\;[x^{2}]=0$$
  $$\displaystyle \lim_{x\to0^{+}}f(x)=2x^{2}+x+0-0\longrightarrow0.$$
Since the two limits differ, $$f$$ is discontinuous at $$x=0$$.

Case 2: $$x=-1$$ (left end-point has only a right‐hand limit)

Value: $$f(-1)=2(1)+(-1)+1-(-1)=3.$$
Right limit (take $$x\rightarrow-1^{+}$$): $$[x]=-1,\;[x^{2}]=0$$
  $$\displaystyle \lim_{x\to-1^{+}}f(x)=2x^{2}+x+0-(-1)=2x^{2}+x+1\longrightarrow2.$$
Right limit $$\neq$$ value, so $$f$$ is discontinuous at $$x=-1$$.

Case 3: $$x=1$$

Left limit ( $$[x]=0,\,[x^{2}]=0$$ ): $$\;2x^{2}+x \longrightarrow3$$.
Value at 1 ( $$[1]=1,\,[1^{2}]=1$$ ): $$2+1+1-1=3$$.
Right limit ( $$[x]=1,\,[x^{2}]=1$$ ): $$\;2x^{2}+x \longrightarrow3$$.
Both limits equal the value, so $$f$$ is continuous at $$x=1$$.

Case 4: $$x=\sqrt2$$

Here $$[x]=1.$$

Left limit ( $$[x^{2}]=1$$ ): $$2x^{2}+x \longrightarrow4+\sqrt2.$$
Right limit ( $$[x^{2}]=2$$ ): $$2x^{2}+x+1 \longrightarrow5+\sqrt2.$$
Limits differ ⇒ discontinuity at $$x=\sqrt2$$.

Case 5: $$x=\sqrt3$$

Again $$[x]=1.$$

Left limit ( $$[x^{2}]=2$$ ): $$2x^{2}+x+1 \longrightarrow7+\sqrt3.$$
Right limit ( $$[x^{2}]=3$$ ): $$2x^{2}+x+2 \longrightarrow8+\sqrt3.$$
Limits differ ⇒ discontinuity at $$x=\sqrt3$$.

Case 6: $$x=2$$ (right end-point has only a left‐hand limit)

Value: $$f(2)=2(4)+2+4-2=12.$$
Left limit ( $$[x]=1,\,[x^{2}]=3$$ ): $$2x^{2}+x+3-1=2x^{2}+x+2\longrightarrow12.$$
Left limit equals the value, so $$f$$ is continuous at $$x=2$$.

Step 3: Count the discontinuities.

The function fails to be continuous exactly at

$$x=-1,\,0,\,\sqrt2,\,\sqrt3.$$

Total number of discontinuity points = $$4$$.

Hence, Option D is correct.