Domain of a Function Year-Wise PYQs: JEE Main & WBJEE Solved Problems

Year-Wise PYQs: Domain of a Function (JEE Main & WBJEE)

Mastering the Domain of a Function is one of the highest-yield strategies for cracking the calculus section in both JEE Main and WBJEE. It tests your fundamental grasp of inequalities, logarithms, trigonometric constraints, and special functions like the Greatest Integer Function (\([x]\)) or Fractional Part Function (\(\{x\}\)).

Below is a beautifully formatted, year-wise compilation of actual exam problems with their core logic and detailed step-by-step solutions perfect for your prep roadmap.

Part 1: JEE Main Problems

JEE Main 2024

Question

Find the domain of the function:

$$f(x) = \frac{1}{\sqrt{[x]^2 - [x] - 6}}$$

(where \([x]\) denotes the greatest integer function).

Core Logic: For the square root in the denominator to be well-defined, the quadratic expression inside must be strictly positive.

Step-by-Step Solution:

1. Set up the mandatory radical inequality: \([x]^2 - [x] - 6 > 0\)
2. Factorize the quadratic expression: \(([x] - 3)([x] + 2) > 0\)
3. By the wavy curve method, this splits into two distinct boundary zones:
   • \([x] > 3 \implies \text{since } [x] \text{ is an integer, } [x] \ge 4 \implies x \in [4, \infty)\)
   • \([x] < -2 \implies \text{similarly, } [x] \le -3 \implies x \in (-\infty, -2)\)

Final Answer: \(x \in (-\infty, -2) \cup [4, \infty)\)

JEE Main 2023

Question

Determine the domain of the function:

$$f(x) = \log_{10} \left( \frac{x^2 - 5x + 6}{x^2 + 1} \right) + \sqrt{\log_{10} \left( \frac{x-1}{x} \right)}$$

Core Logic: Both the independent logarithmic argument condition and the square root boundary condition must be satisfied simultaneously (Intersection method).

Step-by-Step Solution:

1. For the first log term: We need \(\frac{x^2 - 5x + 6}{x^2 + 1} > 0\). Since the denominator \(x^2 + 1\) is unconditionally positive for all real numbers, we simply solve:
   \(x^2 - 5x + 6 > 0 \implies (x-2)(x-3) > 0 \implies\)
   Condition 1: \(x \in (-\infty, 2) \cup (3, \infty)\)
2. For the second radical term: The inner log function must stay non-negative:
   \(\log_{10} \left( \frac{x-1}{x} \right) \ge 0 \implies \frac{x-1}{x} \ge 10^0 \implies \frac{x-1}{x} \ge 1\)
3. Simplify the inequality: \(\frac{x-1}{x} - 1 \ge 0 \implies \frac{-1}{x} \ge 0 \implies x < 0 \implies\)
   Condition 2: \(x \in (-\infty, 0)\)
4. Taking the intersection of Condition 1 and Condition 2 gives our final set.

Final Answer: \(x \in (-\infty, 0)\)

JEE Main 2022

Question

Find the domain of the definition of the function:

$$f(x) = \sqrt{4 - x^2} + \frac{1}{\sqrt{\log_{10}(x - [x])}}$$

Core Logic: Identify properties of the fractional part function \(\{x\} = x - [x]\) and check validity constraints.

Step-by-Step Solution:

1. For the first part \(\sqrt{4 - x^2}\) to be defined: \(4 - x^2 \ge 0 \implies x^2 \le 4 \implies x \in [-2, 2]\).
2. For the log term sitting in the denominator: \(\log_{10}(x - [x]) > 0 \implies \log_{10}(\{x\}) > 0\).
3. This simplifies directly to: \(\{x\} > 10^0 \implies \{x\} > 1\).
4. By standard definitions, the fractional part function can only map to values in the range \([0, 1)\). It can never be greater than or equal to 1.

Final Answer: \(\phi\) (Empty Set / No real solution exists)

JEE Main 2021

Question

Find the domain of the function:

$$f(x) = \sin^{-1} \left( \frac{x^2 - 3x + 2}{x^2 + 2x + 7} \right)$$

Core Logic: The valid input domain of an inverse sine function \(\sin^{-1}(t)\) is strictly bounded: \(-1 \le t \le 1\).

Step-by-Step Solution:

1. Set up the constraint: \(-1 \le \frac{x^2 - 3x + 2}{x^2 + 2x + 7} \le 1\)
2. Notice that the denominator \(x^2 + 2x + 7\) has a negative discriminant (\(D = 4 - 28 = -24 < 0\)), meaning it is strictly positive for all real numbers. We can safely cross-multiply.
3. Left Side: \(-x^2 - 2x - 7 \le x^2 - 3x + 2 \implies 2x^2 - x + 9 \le 0\). This quadratic also has \(D < 0\), meaning it's never \(\le 0\). No real solution here.
4. Right Side: \(x^2 - 3x + 2 \le x^2 + 2x + 7 \implies -5x \le 5 \implies x \ge -1\).

Final Answer: \(x \in \left[ -1, \infty \right)\)

JEE Main 2019

Question

The domain of the definition of \(f(x) = \sqrt{\log_x(\cos 2\pi x)}\) is:

Core Logic: Carefully track base restrictions for the logarithm alongside periodic trigonometric cycles.

Step-by-Step Solution:

1. Base conditions: \(x > 0\) and \(x \neq 1\).
2. Log argument constraint: \(\cos 2\pi x > 0\).
3. Radical constraint: \(\log_x(\cos 2\pi x) \ge 0\).
4. Assuming base \(x > 1\), removing the log preserves inequality: \(\cos 2\pi x \ge x^0 \implies \cos 2\pi x \ge 1\).
5. Since cosine can never output a value greater than 1, it must exactly equal 1:
   \(\cos 2\pi x = 1 \implies 2\pi x = 2n\pi \implies x = n\) (integers). Given \(x > 1\), \(x\) can be any integer greater than 1.

Final Answer: \(x \in \{2, 3, 4, 5, \dots\}\)

Part 2: WBJEE Problems

WBJEE 2024

Question

The domain of definition of the real-valued function \(f(x) = \sqrt{\log_{0.5} (x^2 - 5x + 6)}\) is:

Core Logic: Critical Rule! When the base of a logarithm is less than 1 (\(0 < \text{base} < 1\)), the inequality sign flips when dropping the log.

Step-by-Step Solution:

1. Log argument constraint: \(x^2 - 5x + 6 > 0 \implies (x-2)(x-3) > 0 \implies x \in (-\infty, 2) \cup (3, \infty)\).
2. Square root constraint: \(\log_{0.5} (x^2 - 5x + 6) \ge 0\).
3. Flip inequality direction because base is \(0.5\): \(x^2 - 5x + 6 \le (0.5)^0 \implies x^2 - 5x + 6 \le 1\).
4. Solve the resulting quadratic: \(x^2 - 5x + 5 \le 0\). Roots are \(x = \frac{5 \pm \sqrt{5}}{2}\), giving the interval \(\left[ \frac{5-\sqrt{5}}{2}, \frac{5+\sqrt{5}}{2} \right]\).
5. Intersect this with step 1. Note that \(\frac{5-\sqrt{5}}{2} \approx 1.38\) and \(\frac{5+\sqrt{5}}{2} \approx 3.62\).

Final Answer: \(x \in \left[ \frac{5-\sqrt{5}}{2}, 2 \right) \cup \left( 3, \frac{5+\sqrt{5}}{2} \right]\)

WBJEE 2022

Question

Find the domain of the function:

$$f(x) = \sqrt{\frac{1 - |x|}{2 - |x|}}$$

Core Logic: Substitute complex modulus terms temporarily to keep algebraic wavy-curve calculations neat and clean.

Step-by-Step Solution:

1. Set radicand to non-negative: \(\frac{1 - |x|}{2 - |x|} \ge 0\) along with denominator safety constraint \(|x| \neq 2\).
2. Let \(|x| = t\). The working equation turns into \(\frac{1-t}{2-t} \ge 0 \implies \frac{t-1}{t-2} \ge 0\).
3. By standard wavy curve method: \(t \in [0, 1] \cup (2, \infty)\) (remembering \(t = |x| \ge 0\)).
4. Map back to original \(x\) variables:
   • \(0 \le |x| \le 1 \implies x \in [-1, 1]\)
   • \(|x| > 2 \implies x \in (-\infty, -2) \cup (2, \infty)\)

Final Answer: \(x \in (-\infty, -2) \cup [-1, 1] \cup (2, \infty)\)

WBJEE 2020

Question

The domain of the function \(f(x) = \sqrt{\exp \left(\sin^{-1}\left(\log_2 x\right)\right) - 1}\) is:

Core Logic: Peel layers systematically from the innermost functions out to the parent functions.

Step-by-Step Solution:

1. For inner log term \(\log_2 x\), we explicitly require \(x > 0\).
2. For the inverse sine boundary constraints: \(-1 \le \log_2 x \le 1 \implies 2^{-1} \le x \le 2^1 \implies x \in \left[\frac{1}{2}, 2\right]\).
3. For the structural square root condition: \(e^{\sin^{-1}(\log_2 x)} - 1 \ge 0 \implies e^{\sin^{-1}(\log_2 x)} \ge e^0\).
4. This strips down to: \(\sin^{-1}(\log_2 x) \ge 0 \implies \log_2 x \ge 0 \implies x \ge 2^0 \implies x \ge 1\).
5. Intersecting our bounds \(x \in \left[\frac{1}{2}, 2\right]\) and \(x \in [1, \infty)\) gives the valid core.

Final Answer: \(x \in [1, 2]\)

WBJEE 2018

Question

The domain of the function \(f(x) = \ln \left(x - [x]\right)\) is:

Core Logic: Determine precisely where the natural fractional part function vanishes or stays invalid.

Step-by-Step Solution:

1. For natural logs, arguments must stay strictly positive: \(x - [x] > 0\).
2. Recall that \(x - [x] = \{x\}\), which maps out the fractional part of a number.
3. The standard mathematical range of \(\{x\}\) is \([0, 1)\). It is positive almost everywhere except exactly when \(x\) hits an integer value (where fractional parts clear out to zero, e.g., \(\{4\} = 0\)).
4. Hence, we simply throw out integers from the valid real number lines.

Final Answer: \(\mathbb{R} - \mathbb{Z}\) (All Real Numbers except Integers)

💡 Pro-Tip for Domain Problems

When facing multi-layered functions, do not get overwhelmed! Break them down into isolated, simple math expressions. Find the valid solution boundaries for each component individually, and then use a layout sketch to extract their final intersection. Watch out for inequalities flipping with fractional bases!