Question

Let [x] denote the greatest integer ≤ x, where x ∈ R. If the domain of the real valued function $f(x) = \sqrt{\frac{|[x]|-2}{|[x]|-3}}$ is $(-\infty, a) \cup [b, c) \cup [4, \infty), a < b < c$, then the value of a + b + c is :

• A. a = -3, b = -2, c = 3
• B. a = -4, b = -2, c = 3
• C. a = -3, b = -1, c = 3
• D. a = -3, b = -2, c = 4

Answer 1

Answer: (A)

a = -3, b = -2, c = 3

Answer 2

The function given is $f(x) = \sqrt{\frac{|[x]|-2}{|[x]|-3}}$. For the function to be defined, two conditions must be met:

1. The expression inside the square root must be non-negative: $\frac{|[x]|-2}{|[x]|-3} \ge 0$
2. The denominator must not be zero: $|[x]|-3 \ne 0 \implies |[x]| \ne 3$

Let $y = |[x]|$. Since $[x]$ is an integer, $y$ must be a non-negative integer. The inequality becomes: $\frac{y-2}{y-3} \ge 0$ This inequality holds if both the numerator and denominator are positive, or both are negative.

Case 1: $y-2 \ge 0$ and $y-3 > 0$ $y \ge 2$ and $y > 3$. This implies $y > 3$.

Case 2: $y-2 \le 0$ and $y-3 < 0$ $y \le 2$ and $y < 3$. This implies $y \le 2$.

So, the condition on $y$ is $y \le 2$ or $y > 3$. Substituting back $y = |[x]|$, we get: $|[x]| \le 2 \quad \text{or} \quad |[x]| > 3$

Let's solve these conditions for $[x]$:

Condition 1: $|[x]| \le 2$ This means $-2 \le [x] \le 2$. Since $[x]$ must be an integer, the possible values for $[x]$ are -2, -1, 0, 1, 2.

• If $[x] = -2$, then $-2 \le x < -1$.
• If $[x] = -1$, then $-1 \le x < 0$.
• If $[x] = 0$, then $0 \le x < 1$.
• If $[x] = 1$, then $1 \le x < 2$.
• If $[x] = 2$, then $2 \le x < 3$. The union of these intervals is $[-2, 3)$.

Condition 2: $|[x]| > 3$ This means $[x] > 3$ or $[x] < -3$.

• If $[x] > 3$: Since $[x]$ is an integer, this means $[x] \ge 4$. If $[x] = 4$, then $4 \le x < 5$. If $[x] = 5$, then $5 \le x < 6$, and so on. This corresponds to the interval $[4, \infty)$.
• If $[x] < -3$: Since $[x]$ is an integer, this means $[x] \le -4$. If $[x] = -4$, then $-4 \le x < -3$. If $[x] = -5$, then $-5 \le x < -4$, and so on. This corresponds to the interval $(-\infty, -3)$.

Combining the results from both conditions, the domain of $f(x)$ is the union of these intervals: Domain = $(-\infty, -3) \cup [-2, 3) \cup [4, \infty)$.

The problem states that the domain is $(-\infty, a) \cup [b, c) \cup [4, \infty)$, with $a < b < c$. Comparing our derived domain with the given format:

• $(-\infty, a)$ corresponds to $(-\infty, -3)$, so $a = -3$.
• $[b, c)$ corresponds to $[-2, 3)$, so $b = -2$ and $c = 3$.
• $[4, \infty)$ matches the given interval.

We check the condition $a < b < c$: $-3 < -2 < 3$. This condition is satisfied.

The question asks for the value of $a + b + c$. $a + b + c = (-3) + (-2) + 3$ $a + b + c = -5 + 3$ $a + b + c = -2$.