Question

Let [𝑡] denote the greatest integer ≤ 𝑡. The number of points of discontinuity of the function 𝑓(𝑥)= [𝑥²−3𝑥+2] for 𝑥∈[0, 4] is _______ (in integer)

Answer 1

Correct Answer: 10 - 8

Given:
\[ f(x)=\lfloor x^2-3x+2\rfloor,\qquad x\in[0,4], \] where $\lfloor t\rfloor$ is the greatest integer $\le t$. $f$ is discontinuous exactly at points where the argument is an integer, i.e. where \[ g(x):=x^2-3x+2 \in \mathbb{Z}. \] 
Step 1 — Range of $g$ on $[0,4]$ 
The quadratic $g(x)=x^2-3x+2$ has vertex at $x=\tfrac{3}{2}$, and \[ g\Big(\tfrac{3}{2}\Big)=\tfrac{1}{4},\qquad g(0)=2,\qquad g(4)=6. \] So on $[0,4]$ we have $g(x)\in\big[\tfrac{1}{4},6\big]$. The integers $n$ that can be attained are $n=1,2,3,4,5,6$. 
Step 2 — Solve $g(x)=n$ for each integer $n$ and count roots in $[0,4]$
Solve $x^2-3x+(2-n)=0$. For each $n$:

• $n=1$: $x^2-3x+1=0$. Roots $x=\dfrac{3\pm\sqrt5}{2}\approx 0.382,\;2.618$  → both in $[0,4]$  → 2 points.
• $n=2$: $x^2-3x=0$. Roots $x=0,\;3$  → both in $[0,4]$  → 2 points.
• $n=3$: $x^2-3x-1=0$. Roots $x\approx -0.303,\;3.303$  → only $3.303\in[0,4]$  → 1 point.
• $n=4$: $x^2-3x-2=0$. Roots $x\approx -0.561,\;3.562$  → only $3.562\in[0,4]$  → 1 point.
• $n=5$: $x^2-3x-3=0$. Roots $x\approx -0.792,\;3.792$  → only $3.792\in[0,4]$  → 1 point.
• $n=6$: $x^2-3x-4=0$. Roots $x=-1,\;4$  → only $x=4\in[0,4]$  → 1 point.

Count them up: $2+2+1+1+1+1 = 8$. 
Conclusion:
The number of points of discontinuity of $f$ on $[0,4]$ is \[ \boxed{8}. \]