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

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