The domain of the function \(\mathbf{f}(\mathbf{x})=\frac{1}{\sqrt{9-\mathrm{x}^{2}}}\) is

This question was previously asked in
VITEEE PYP_125Qs150Min125Marks
View all VITEEE Papers >
  1. −3 ≤ x ≤ 3
  2. −3 < x < 3
  3. −9 ≤ x ≤ 9
  4. −9 < x < 9 

Answer (Detailed Solution Below)

Option 2 : −3 < x < 3
Free
JEE Main 2025 (Session II) All India Live Test
4 K Users
75 Questions 300 Marks 180 Mins

Detailed Solution

Download Solution PDF

Concept Used:

The expression inside the square root must be non-negative.

The denominator cannot be zero.

Calculation:

For f(x) to be defined, the following conditions must be met:

1) 9 - x² > 0 (because it's inside a square root and in the denominator)

⇒ x² < 9

⇒ -3 < x < 3

∴ The domain of the function f(x) is -3 < x < 3.

Hence option 2 is correct

Latest VITEEE Updates

Last updated on Jul 3, 2025

->Vellore Institute of Technology will open its application form for 2026 on November 4, 2025.

->The VITEEE 2026 exam is scheduled to be held from April 20, 2026 to April 27, 2026.

->VITEEE exams are conduted for admission to undergraduate engineering programs at the Vellore Institute of Technology (VIT) and its affiliated campus.

->12th pass candidates can apply for the VITEEE exam.

Get Free Access Now
Hot Links: teen patti real cash 2024 teen patti flush teen patti master game teen patti joy mod apk teen patti gold apk