The domain of the function f(x) = 1 x ^2 - 3 x - 10, where x denotes the greatest integer less than or equal to x, is:

Given: f(x) = 1 x ^2 - 3 x - 10 Find: The domain of f(x). Since the square root is in the denominator, the quantity inside the square root must be strictly positive: x ^2 - 3 x - 10 0 Factorizing, x ^2 - 3 x - 10 = ( x + 2)( x - 5) So we solve ( x + 2)( x - 5) 0 This gives x < -2 or x 5 Because x is an integer, this refines to x -3 or x 6 Hence the corresponding values of x are x (-, -2) [6,) Therefore, the correct option is D. Given: x is the greatest integer less than or equal to x. Find: How the inequality in x translates into intervals of x. From x -3 we get all real numbers whose floor is at most -3, that is x < -2 Similarly, from x 6 we get x 6 Thus, x (-, -2) [6,) This matches option D. Note that the solution says "The Correct Option is A", but the actual worked result and final interval clearly correspond to option D.

The domain of the function f(x) = 1 x ^2 - 3 x - 10, where x deno… — Solution | JEE Mathematics

MCQMediumJEE 2023Functions

The domain of the function $f(x) = \frac{1}{\sqrt{\lfloor x \rfloor^2 - 3\lfloor x \rfloor - 10}}$ , where $\lfloor x \rfloor$ denotes the greatest integer less than or equal to $x$ , is:

A
$(-\infty, -3] \cup [6, \infty)$ ](streamdown:incomplete-link)
B
$(-\infty, -2) \cup (5, \infty)$
C
$(-\infty, -3] \cup (5, \infty)$
D
$(-\infty, -2) \cup [6, \infty)$ ](streamdown:incomplete-link)

Answer

Correct answer:D

Step-by-step solution

Standard Method

Given: $f(x) = \frac{1}{\sqrt{\lfloor x \rfloor^2 - 3\lfloor x \rfloor - 10}}$

Find: The domain of $f(x)$ .

Since the square root is in the denominator, the quantity inside the square root must be strictly positive:

\lfloor x \rfloor^2 - 3\lfloor x \rfloor - 10 > 0

Factorizing,

\lfloor x \rfloor^2 - 3\lfloor x \rfloor - 10 = (\lfloor x \rfloor + 2)(\lfloor x \rfloor - 5)

So we solve

(\lfloor x \rfloor + 2)(\lfloor x \rfloor - 5) > 0

This gives

\lfloor x \rfloor < -2 \quad \text{or} \quad \lfloor x \rfloor > 5

Because $\lfloor x \rfloor$ is an integer, this refines to

\lfloor x \rfloor \le -3 \quad \text{or} \quad \lfloor x \rfloor \ge 6

Hence the corresponding values of $x$ are

x \in (-\infty, -2) \cup [6, \infty)

Therefore, the correct option is D.](streamdown:incomplete-link)

Floor Function Interpretation

Given: $\lfloor x \rfloor$ is the greatest integer less than or equal to $x$ .

Find: How the inequality in $\lfloor x \rfloor$ translates into intervals of $x$ .

From

\lfloor x \rfloor \le -3

we get all real numbers whose floor is at most $-3$ , that is

x < -2

Similarly, from

\lfloor x \rfloor \ge 6

we get

x \ge 6

Thus,

x \in (-\infty, -2) \cup [6, \infty)

This matches option D. Note that the solution says "The Correct Option is A", but the actual worked result and final interval clearly correspond to option D.](streamdown:incomplete-link)

Common mistakes

Requiring only the denominator to be nonzero is incorrect. Since the denominator is $\sqrt{\lfloor x \rfloor^2 - 3\lfloor x \rfloor - 10}$ , the expression inside the square root must be strictly positive, not merely nonzero. Use $\lfloor x \rfloor^2 - 3\lfloor x \rfloor - 10 > 0$ .
Stopping at $\lfloor x \rfloor < -2$ or $\lfloor x \rfloor > 5$ without using the integer nature of the floor function is incomplete. Because $\lfloor x \rfloor$ is an integer, convert this to $\lfloor x \rfloor \le -3$ or $\lfloor x \rfloor \ge 6$ before finding the interval for $x$ .
Writing the left interval as $(-\infty, -3]$ is wrong for $x$ . The condition $\lfloor x \rfloor \le -3$ means all real $x < -2$ , not only values up to $-3$ . Always translate floor conditions back to real-number intervals carefully.

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