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

To solve the problem, we need to evaluate the given function \( f(x) = [x]^2 - [x+3] - 3 \) and match it with the given conditions.

The function is expressed in terms of the greatest integer function, denoted by \([x]\), which gives the greatest integer less than or equal to \(x\).

1. Consider the function \( f(x) = [x]^2 - [x+3] - 3 \).
2. Let's break down \([x+3]\):
   • \([x+3] = [x] + 3\) when \(x\) is an integer.
   • \([x+3] = [x] + 2\) when \(x\) is not an integer (since \([x]\) rounds down \(x\) and \([x+3]\) will add three to the greatest integer less than \(x\)).
3. Consider \(x\) as an integer:
   • In this situation, \([x]^2 - ([x] + 3) - 3 = [x]^2 - [x] - 6\).
   • Solving this for \(f(x) = 0\):
     • The equation becomes \([x]^2 = [x] + 6\).
     • This yields no integer solutions since it requires \([x]([x] - 1) = 6\), which does not hold for any integer.
4. Consider \(x\) as a non-integer:
   • In this situation, \([x]^2 - ([x] + 3 - 1) - 3 = [x]^2 - ([x] + 2) - 3 = [x]^2 - [x] - 5\).
   • The equation becomes \([x]^2 - [x] - 5 = 0\). Solving for integer parts \([x] = n\):
     • \(n^2 - n - 5 = 0\), which holds for \(x \in [-1, 0)\) or \(x \in [2, 3)\).
5. Evaluate \( f(x) < 0 \) for the potential intervals:
   • In \(x \in [-1, 0) \cup [2, 3)\),
     • Check the function \(f(x)\): For each segment, calculate \([x]\) and \([x+3]\), then substitute back into \(f(x)\).

Therefore, the correct solution is \(f(x) < 0\) only for \(x \in [-1, 3)\).