Find the domain and Range of the function $f(x) = \dfrac{1}{\sqrt{x - 5}}$.

Find the domain and Range of the function f(x) = d – Class 11 Mathematics

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f(x) = \dfrac{1}{\sqrt{x - 5}}

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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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The function $f(x) = \frac{1}{\sqrt{x - 5}}$ requires $x - 5 > 0$ (positive under the square root, non-zero in denominator), giving domain $(5, \infty)$ ; as $x$ ranges over this interval, $f(x)$ takes all positive values, so range is $(0, \infty)$ .

Understanding the constraints

A rational function with a square root in the denominator imposes two simultaneous conditions. The expression under the square root must be non-negative (since we're working with real numbers), and the denominator itself cannot be zero. These aren't separate checks that we combine—they work together to carve out the valid input values.

For $f(x) = \frac{1}{\sqrt{x - 5}}$ , the square root $\sqrt{x - 5}$ appears in the denominator. This immediately tells us that $x - 5$ must be strictly positive, not merely non-negative. If $x - 5 = 0$ , the square root would be zero and we'd be dividing by zero. If $x - 5 < 0$ , the square root wouldn't be real.

Finding the domain

Set up the inequality. We need $x - 5 > 0$ (strict inequality, not $\geq$ ).
Solve for $x$ . Adding $5$ to both sides gives $x > 5$ .
Express in interval notation. The domain is all real numbers strictly greater than $5$ , which we write as $(5, \infty)$ .

Watch out

A common mistake is writing the domain as $[5, \infty)$ , including the endpoint. But $x = 5$ makes the denominator zero, so it must be excluded. The parenthesis matters.

Finding the range

Now we ask: as $x$ varies over $(5, \infty)$ , what values can $f(x)$ take?

Analyze the behavior near the boundary. As $x \to 5^+$ (approaching $5$ from the right), the quantity $x - 5 \to 0^+$ , so $\sqrt{x - 5} \to 0^+$ , and therefore $f(x) = \frac{1}{\sqrt{x - 5}} \to +\infty$ .
Analyze the behavior at infinity. As $x \to \infty$ , we have $x - 5 \to \infty$ , so $\sqrt{x - 5} \to \infty$ , and thus $f(x) = \frac{1}{\sqrt{x - 5}} \to 0^+$ . …

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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.Find the domain and Range of the function f(x)=1x−5f(x) = \dfrac{1}{\sqrt{x - 5}}f(x)=x−5​1​.

Understanding the constraints

Finding the domain

Finding the range

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Q.Find the domain and Range of the function $f(x) = \dfrac{1}{\sqrt{x - 5}}$ .

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