Find the domain of each of the following functions – Class 11 Mathematics

NCERT Exemplar · Q17

Q. Find the domain of each of the following functions given by

(i)

f(x) = \dfrac{1}{\sqrt{1 - \cos x}}

(ii)

f(x) = \dfrac{1}{\sqrt{x + |x|}}

(iii)

f(x) = x\,|x|

(iv)

f(x) = \dfrac{x^3 - x + 3}{x^2 - 1}

(v)

f(x) = \dfrac{3x}{2x - 8}

Puducherry CbseLong· 3mImportance★★★★★est

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Concept understanding — Rational Function Domain

What is a Rational Function Domain?

Imagine you're baking a cake and the recipe says "add flour until the mixture is smooth." If you add too much flour, the mixture becomes a dry lump — it stops being a proper batter. A rational function is like that mixture: it's a fraction made of two polynomials, and it only "works" when the denominator isn't zero.

A rational function looks like this:

f(x) = \frac{P(x)}{Q(x)}

where $P(x)$ and $Q(x)$ are polynomials, and $Q(x) \neq 0$ .

The domain of a rational function is simply the set of all real numbers $x$ for which the function is defined — meaning, all $x$ except those that make the denominator zero.

The Intuition First

Think of division in everyday life. You can divide 10 apples among 5 people — that's fine. You can divide 10 apples among 2 people — also fine. But can you divide 10 apples among 0 people? That doesn't make sense. You can't split something among nobody.

In the same way, a rational function is a division. The denominator tells you "how many groups" you're splitting into. If the denominator is zero, the division is impossible — the function has no value there.

So the domain is: all real numbers, except the ones that make the bottom zero.

The Precise Statement

\text{Domain of } f(x) = \frac{P(x)}{Q(x)} \text{ is } \{ x \in \mathbb{R} \mid Q(x) \neq 0 \}

In plain words: find every $x$ that makes $Q(x) = 0$ , and remove those from the set of all real numbers.

How to Find the Domain — Step by Step

Step 1: Write down the denominator $Q(x)$ .

Step 2: Set $Q(x) = 0$ and solve for $x$ .

Step 3: The domain is all real numbers except those solutions.

Tip

You only care about the denominator. The numerator $P(x)$ can be anything — even zero — and the function is still defined (it just equals zero). Only the denominator matters for domain.

Examples

Example 1: $f(x) = \frac{1}{x-3}$

Denominator: $x - 3 = 0 \implies x = 3$

Domain: all real numbers except $3$ . In interval notation: $(-\infty, 3) \cup (3, \infty)$

Example 2: $f(x) = \frac{x^2 + 1}{x^2 - 4}$

Denominator: $x^2 - 4 = 0 \implies (x-2)(x+2) = 0 \implies x = 2$ or $x = -2$

Domain: all real numbers except $2$ and $-2$ . In interval notation: $(-\infty, -2) \cup (-2, 2) \cup (2, \infty)$

Example 3: $f(x) = \frac{2x+5}{x^2 + 1}$

Denominator: $x^2 + 1 = 0 \implies x^2 = -1$ — no real solution.

Domain: all real numbers, i.e., $(-\infty, \infty)$ …

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Domain means all real $x$ for which the function is defined. For (i) $x \neq 2n\pi$ ;

(ii) $x > 0$ ;

(iii) all real numbers;

(iv) $x \neq \pm 1$ ;

(v) $x \neq 4$ .

The Core Idea: What is a Domain?

The domain of a function is the set of all real numbers $x$ for which the function gives a real, defined output. When finding a domain, we look for the "trouble spots" — places where the rule breaks. The main culprits are:

Division by zero (denominator cannot be zero)
Even roots of negatives (square root, fourth root, etc., must have non-negative radicand)
Logarithms of non-positive numbers (not in these problems, but good to remember)

We check each function for these issues.

(i) $f(x) = \dfrac{1}{\sqrt{1 - \cos x}}$

Step 1: Identify the restrictions.

We have a square root in the denominator. Two conditions must hold:

The expression inside the square root must be positive (not just non-negative, because it's in the denominator — zero would make the denominator zero).
The denominator itself cannot be zero.

So we need: $1 - \cos x > 0$ .

Step 2: Solve the inequality.

$1 - \cos x > 0 \implies \cos x < 1$ .

When does $\cos x = 1$ ? At $x = 2n\pi$ , where $n$ is any integer ( $n \in \mathbb{Z}$ ). For all other $x$ , $\cos x < 1$ .

Step 3: Check if any other restrictions exist.

The square root of a positive number is defined and positive. No other issues.

Watch out

A common mistake is to write $1 - \cos x \geq 0$ and then include $x = 2n\pi$ in the domain. But at those points, the denominator becomes $\sqrt{0} = 0$ , and division by zero is undefined. So those points must be excluded.

Domain: All real numbers except $x = 2n\pi$ , where $n \in \mathbb{Z}$ .

(ii) $f(x) = \dfrac{1}{\sqrt{x + |x|}}$

Step 1: Identify the restrictions.

Again, a square root in the denominator. We need $x + |x| > 0$ .

Step 2: Understand $|x|$ .

Recall: $|x| = x$ when $x \geq 0$ , and $|x| = -x$ when $x < 0$ .

Step 3: Consider the two cases.

Case 1: $x \geq 0$ .

Then $|x| = x$ , so $x + |x| = x + x = 2x$ .

Condition: $2x > 0 \implies x > 0$ .

Since we are in the case $x \geq 0$ , this gives $x > 0$ .
Case 2: $x < 0$ .

Then $|x| = -x$ , so $x + |x| = x + (-x) = 0$ .

Condition: $0 > 0$ is false. So no negative $x$ works.

Step 4: Combine.

Only $x > 0$ satisfies the condition.

Tip

Notice that $x + |x|$ is always $0$ for $x \leq 0$ and $2x$ for $x > 0$ . So the expression inside the root is zero for all non-positive $x$ , and positive only for $x > 0$ .

Domain: All positive real numbers, i.e., $(0, \infty)$ .

(iii) $f(x) = x\,|x|$ …

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Q.Find the domain of each of the following functions given by

The Core Idea: What is a Domain?

(i) f(x)=11−cos⁡xf(x) = \dfrac{1}{\sqrt{1 - \cos x}}f(x)=1−cosx​1​

(ii) f(x)=1x+∣x∣f(x) = \dfrac{1}{\sqrt{x + |x|}}f(x)=x+∣x∣​1​

(iii) f(x)=x ∣x∣f(x) = x\,|x|f(x)=x∣x∣ …

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(i) $f(x) = \dfrac{1}{\sqrt{1 - \cos x}}$

(ii) $f(x) = \dfrac{1}{\sqrt{x + |x|}}$

(iii) $f(x) = x\,|x|$ …

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