Question

Let \[ f(x) = [x]^2 - [x+3] - 3,\quad x \in \mathbb{R}, \] where \([\,\cdot\,]\) denotes the greatest integer function. Then

• A. \(f(x)>0\) only for \(x \in [4,\infty)\)
• B. \(f(x)<0\) only for \(x \in [-1,3)\)
• C. \(\displaystyle \int_{0}^{2} f(x)\,dx = -6\)
• D. \(f(x) = 0\) for finitely many values of \(x\)

Hint:

For expressions involving greatest integer functions, always analyze the function over intervals \([n, n+1)\) where the value remains constant.

Answer 1

The Correct Option is B

Let \(n = [x]\), where \(n \in \mathbb{Z}\). Then \(x \in [n, n+1)\).

Step 1: Express the function in terms of \(n\).
\[ f(x) = n^2 - (n+3) - 3 = n^2 - n - 6. \]

Step 2: Determine when \(f(x)<0\).
Solve the inequality: \[ n^2 - n - 6<0. \] Factorizing, \[ (n-3)(n+2)<0. \] This gives \[ -2<n<3. \] Thus, \[ n = -1,\,0,\,1,\,2. \]

Step 3: Translate back to intervals of \(x\).
For \(n = -1,0,1,2\), \[ x \in [-1,3). \] Hence, \[ f(x)<0 \text{ only for } x \in [-1,3). \]

Final Answer: \[ \boxed{f(x)<0 \text{ only for } x \in [-1,3)} \]