Domain of $\sqrt{a^2 - x^2}$ $(a > 0)$ is (A) $(-a,\ a)$ (B) $[-a,\ a]$ (C) $[0,\ a]$ (D) $(-a,\ 0]$

Domain of √(a² - x²) (a > 0) is (A) (-a,a) (B) [-a – Class 11 Mathematics

NCERT Exemplar · Q28

\sqrt{a^2 - x^2}

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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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For a square root to be defined in the reals, its argument must be non-negative; solving $a^2 - x^2 \geq 0$ gives $-a \leq x \leq a$ , so the domain is $[-a, a]$ .

Why the square root restricts the domain

When we write $\sqrt{a^2 - x^2}$ , we're asking: for which values of $x$ does this expression produce a real number? The square root function in the real number system is only defined when its argument is non-negative. If $a^2 - x^2$ becomes negative, $\sqrt{a^2 - x^2}$ has no real value.

This is fundamentally different from, say, $\frac{1}{x}$ , where we exclude points that make the denominator zero. Here we need the entire expression under the root to be zero or positive.

Finding where $a^2 - x^2 \geq 0$

The condition for the domain is:

a^2 - x^2 \geq 0

Notice the inequality is non-strict ( $\geq$ , not $>$ ) because $\sqrt{0} = 0$ is perfectly well-defined.

Step-by-step solution:

Rearrange the inequality:

a^2 \geq x^2

Interpret what $x^2 \leq a^2$ means geometrically:

The square of $x$ cannot exceed the square of $a$ . Since both $a^2$ and $x^2$ are non-negative, this means $|x| \leq a$ (the absolute value of $x$ is at most $a$ ).
Translate the absolute value inequality:

|x| \leq a \quad \iff \quad -a \leq x \leq a

This captures all real numbers whose distance from zero is at most $a$ .

Check the boundary points:
- At $x = a$ : $\sqrt{a^2 - a^2} = \sqrt{0} = 0$ ✓
- At $x = -a$ : $\sqrt{a^2 - (-a)^2} = \sqrt{0} = 0$ ✓ …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2026Set 1A1 markMCQ
$\frac{1}{\sqrt{x^2 - 25}}$
(1) $(-\infty, -5) \cup (5, \infty)$
(2) $(-\infty, -5] \cup [5, \infty)$
(3) $(-\infty, -5] \cup (5, \infty)$
(4) $(-\infty, -5) \cup [5, \infty)$
›Reveal solutionSolution

$\dfrac{1}{\sqrt{x^2-25}}$ is defined when $x^2-25>0$ , giving $(-\infty,-5)\cup(5,\infty)$ .

The expression $\dfrac{1}{\sqrt{x^2-25}}$ needs the quantity under the root to be positive (it is in the denominator, so it cannot be zero and cannot be negative).

…

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CBSE 2025Set ANNUAL1 markMCQ
$f(x) = \dfrac{x^2 + 1}{x^2 - 1}$
(a) $R - \{1\}$
(b) $R - \{-1, 1\}$
(c) $R - \{-1\}$
(d) $R$
›Reveal solutionSolution

The domain excludes values of $x$ that make the denominator $x^2-1$ zero, i.e. $x = \pm 1$ .

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

$f(x)$ is defined for all real $x$ except where $x^2 - 1 = 0$ :

$x^2 = 1 \implies x = 1 \text{ or } x = -1$

…

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CBSE 2023Set ANNUAL1 mark
$f(x) = x^2 + 2$
›Reveal solutionSolution

Since $x^2 \ge 0$ always, $f(x)=x^2+2$ never goes below $2$ , so its range is $[2,\infty)$ .

For any real $x$ , $x^2 \ge 0$ , with equality only at $x = 0$ .

So $f(x) = x^2 + 2 \ge 0 + 2 = 2$ , with the minimum value $2$ attained at $x=0$ .

…

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CBSE 2023Set ANNUAL1 markMCQ
$f(x) = |x|$
(a) $N$ (Set of all natural number)
(b) $R$ (Set of all real number)
(c) $I$ (set of integers)
(d) None of these
›Reveal solutionSolution

Domain $=\mathbb{R}$ ; option (b).

$|x|$ can be computed for any real $x$ (it equals $x$ if $x\ge0$ and $-x$ if $x<0$ ). Hence the domain of the modulus function is the set of all real numbers — a standard NCE …

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CBSE 2023Set ANNUAL1 markMCQ
$\frac{1}{\sqrt{x^2 - a^2}}$
(a) $(-\infty, a) \cup (a, \infty)$
(b) $(a, -\infty) \cup (\infty, a)$
(c) $(-\infty, a) \cap (a, \infty)$
(d) None of these
›Reveal solutionSolution

Domain $=(-\infty,-a)\cup(a,\infty)$ ; option (a).

For $\dfrac{1}{\sqrt{x^2-a^2}}$ to be real, we need $x^2-a^2>0$ (strict, since it is in the denominator), i.e. $x^2>a^2\Rightarrow |x|>|a|$ .

Assuming $a>0$ , this means $x<-a$ or $x>a$ , i.e. $(-\infty,-a)\cup(a,\infty)$ . …

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CBSE 2022Set TERM11 markMCQ
$f(x)=\dfrac{4-x}{x-4}$
(a) Domain $= R$ , Range $= \{-1, 1\}$
(b) Domain $= R-\{1\}$ , Range $= R$
(c) Domain $= R-\{4\}$ , Range $= \{-1\}$
(d) Domain $= R-\{-4\}$ , Range $= [-1, 1]$
›Reveal solutionSolution

Simplify the function first; it collapses to a single constant value away from $x=4$ .

…

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CBSE 2022Set ANNUAL1 markMCQ
$f(x) = |x|$
(a) N (set of all natural number)
(b) R (Set of all real number)
(c) I (set of integers)
(d) None of these
›Reveal solutionSolution

Domain of $f(x)=|x|$ is $R$ .

$|x|$ equals $x$ for $x\ge 0$ and $-x$ for $x<0$ ; it is well-defined for every real number. Hence the domain is all of $R$ (the range is $[0,\infty)$ ).

…

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CBSE 2020Set ANNUAL1 mark
$f = \{(2,1), (3,1), (4,1), (5,1)\}$
›Reveal solutionSolution

The range collects the distinct second elements of every ordered pair; here every pair outputs $1$ , so the range is $\{1\}$ .

The function is given as a set of ordered pairs: $f = \{(2,1), (3,1), (4,1), (5,1)\}$ .

Domain (first elements) $= \{2, 3, 4, 5\}$ .

…

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CBSE 2019Set ANNUAL1 mark
$f(x) = \sqrt{9-x^2}$
›Reveal solutionSolution

The domain of $f(x)=\sqrt{9-x^2}$ is $[-3,3]$ .

For the square root $\sqrt{9-x^2}$ to give a real value, the expression under the root must be non-negative:

$9-x^2 \ge 0$

$x^2 \le 9$

$-3 \le x \le 3$

…

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Q.Domain of a2−x2\sqrt{a^2 - x^2}a2−x2​ (a>0)(a > 0)(a>0) is (A) (−a, a)(-a,\ a)(−a, a) (B) [−a, a][-a,\ a][−a, a] (C) [0, a][0,\ a][0, a] (D) (−a, 0](-a,\ 0](−a, 0]

Why the square root restricts the domain

Finding where a2−x2≥0a^2 - x^2 \geq 0a2−x2≥0

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Q.Domain of $\sqrt{a^2 - x^2}$ $(a > 0)$ is
(A) $(-a,\ a)$
(B) $[-a,\ a]$
(C) $[0,\ a]$
(D) $(-a,\ 0]$

Finding where $a^2 - x^2 \geq 0$

Unlock everything free for 14 days