MHT-CET Maths · Sets, Relations and Functions

Domain and Range — Where a Formula Is Defined and What It Produces

Domain: intersect the conditions each piece imposes (log argument > 0, even root ≥ 0, denominator ≠ 0, inverse-sine argument in [−1, 1]). Range: solve y = f(x) for x and ask which y allow a real solution.

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Why this matters

23 PYQs at 22% HARD — the largest page in the chapter and the most repeated: the domain of 2ˣ + 2ʸ = 2 has been set FOUR times with identical options. Three of the HARD ones are a log of a rational function, a sin⁻¹ of a rational function, and a rational-function range whose end-points decide the answer. Every question is answered by the same two routines below; the marks are lost on the bracket at the boundary.

Concept 1 of 3: Domain: Write One Condition Per Piece, Then Intersect

Each part of a formula refuses certain inputs. A square root needs its inside

\ge 0

(

> 0

in a denominator), a logarithm needs its argument

> 0

\sin^{-1}

needs its argument in

[-1, 1]

, a denominator must be non-zero. The domain is where ALL conditions hold.

Definition

$f(x) = \dfrac{\sin^{-1}(x - 3)}{\sqrt{9 - x^2}}$ : $-1 \le x - 3 \le 1 \Rightarrow 2 \le x \le 4$ ; $9 - x^2 > 0 \Rightarrow -3 < x < 3$ . Intersection $[2, 3)$ — closed at $2$ , OPEN at $3$ because the root is in the denominator.
$f(x) = \sin^{-1}\!\left(\log_2\dfrac{x}{2}\right)$ : $-1 \le \log_2\dfrac{x}{2} \le 1 \Rightarrow \dfrac12 \le \dfrac{x}{2} \le 2 \Rightarrow 1 \le x \le 4$ .
$f(x) = \log_2\dfrac{x + 3}{x^2 + 3x + 2}$ : need $\dfrac{x + 3}{(x + 1)(x + 2)} > 0$ — a sign chart on the critical points $-3, -2, -1$ gives $(-3, -2) \cup (-1, \infty)$ . That exact set is not among the options; the official key is (D) $(-3, \infty) - \{-1, -2\}$ , the only option that starts at $-3$ and removes both poles, so pick it on this stem and know that it over-includes $(-2, -1)$ .\n- An implicit equation $2^x + 2^y = 2$ : $2^y = 2 - 2^x$ must be positive, so $2^x < 2$ , $x < 1$ . Domain $(-\infty, 1)$ — the four-time repeat.

Standard conditions

\sqrt{u}: u \ge 0 \quad \frac{1}{\sqrt{u}}: u > 0 \quad \log u: u > 0 \quad \sin^{-1}u: -1 \le u \le 1 \quad \frac{1}{u}: u \ne 0

Worked example

Find the domain of

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

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 16th May Shift 2 · Q130Moderate

Example 1 · Sets, Relations and Functions · Domain and Range — Where a Formula Is Defined and What It Produces

The domain of the function

f(x) = \frac{\sin^{-1}(x-3)}{\sqrt{9-x^2}}

Closing the bracket at a root in the denominator

\sqrt{9 - x^2}

underneath means

9 - x^2 > 0

, strictly.

[2, 3]

is option (C) on both sittings; the answer is

[2, 3)

Drill 6 more on domain: write one condition per piece, then intersect

Concept 2 of 3: Domain Through a Quadratic Inequality: sin⁻¹ of a Rational Function With |x|

When the

\sin^{-1}

argument is a fraction with a positive denominator, '

\le 1

' clears to a polynomial inequality. With

|x|

present, solve for

x \ge 0

and reflect.

Definition

$\sin^{-1}\dfrac{|x| + 5}{x^2 + 1}$ : the argument is already $\ge 0 > -1$ , so only $\dfrac{|x| + 5}{x^2 + 1} \le 1$ matters. Since $x^2 + 1 > 0$ : $|x| + 5 \le x^2 + 1$ , i.e. $x^2 - |x| - 4 \ge 0$ .
For $x \ge 0$ : $x^2 - x - 4 \ge 0 \Rightarrow x \ge \dfrac{1 + \sqrt{17}}{2}$ (the other root is negative). Reflect: domain $(-\infty, -a] \cup [a, \infty)$ with $a = \dfrac{1 + \sqrt{17}}{2}$ .
Multiply through by a denominator ONLY when its sign is known; $x^2 + 1$ is safe, $x - 1$ is not.
The quadratic-formula root with the '+' is the one that lands in $x \ge 0$ ; the options include the '−' root and half-and-minus variants as distractors.

Clearing a positive denominator

\frac{p(x)}{q(x)} \le 1,\ q(x) > 0 \iff p(x) \le q(x)

Worked example

Find the domain of

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

Practise this conceptself-check

The same idea in a real exam question:

MHT-CET · 2024 · 10th May Shift 2 · Q114Hard

Example 2 · Sets, Relations and Functions · Domain and Range — Where a Formula Is Defined and What It Produces

The domain of the function

f(x)=\sin^{-1}\!\left(\dfrac{|x|+5}{x^2+1}\right)

(-\infty,-a]\cup[a,\infty)

. Then

a

is equal to

Solving x² − x − 4 ≥ 0 as x ≥ the smaller root

x^2 - x - 4 \ge 0

holds OUTSIDE the roots. With

x \ge 0

imposed by the

|x|

split, only

x \ge \dfrac{1 + \sqrt{17}}{2}

survives;

\dfrac{\sqrt{17} - 1}{2}

is the wrong sign's root, option (B).

Concept 3 of 3: Range of a Rational Function: Set y = f(x), Clear, and Demand a Real x

Write

y = f(x)

, cross-multiply to a quadratic in

x

, and require its discriminant

\ge 0

. The

y

that pass are the range — then check the boundary values separately, because clearing the denominator can create or lose an endpoint.

Definition

$y = \dfrac{x}{1 + x^2} \Rightarrow yx^2 - x + y = 0$ ; real $x$ needs $1 - 4y^2 \ge 0$ , so $|y| \le \dfrac12$ . Range $\left[-\dfrac12, \dfrac12\right]$ .
$y = \dfrac{x^2 + x + 2}{x^2 + x + 1} = 1 + \dfrac{1}{x^2 + x + 1}$ : $x^2 + x + 1 = \left(x + \tfrac12\right)^2 + \tfrac34 \ge \tfrac34$ , so the fraction lies in $\left(0, \tfrac43\right]$ and $y \in \left(1, \tfrac73\right]$ . The maximum IS attained (at $x = -\tfrac12$ ); $y = 1$ is not.
$y = \dfrac{x^2}{x^2 + 1} = 1 - \dfrac{1}{x^2 + 1}$ : $0 \le y < 1$ , range $[0, 1)$ .
Endpoint check: discriminant $= 0$ gives an attained endpoint; a $y$ that makes the leading coefficient vanish ( $y = 0$ in $yx^2 - x + y = 0$ ) must be tested directly.

Discriminant method

y = f(x) \ \Rightarrow\ a(y)x^2 + b(y)x + c(y) = 0 \ \Rightarrow\ b(y)^2 - 4a(y)c(y) \ge 0

Worked example

Find the range of

f(x) = \dfrac{3x^2 + 5}{x^2 + 1}

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 16th May Shift 1 · Q126Hard

Example 3 · Sets, Relations and Functions · Domain and Range — Where a Formula Is Defined and What It Produces

Range of the function

f(x) = \frac{x^2+x+2}{x^2+x+1},\; x \in \mathbb{R}

Guessing the bracket from the shape

\dfrac{x^2 + x + 2}{x^2 + x + 1}

attains its maximum

\dfrac73

but never its infimum

1

, so the range is

\left(1, \dfrac73\right]

; every other bracket pairing is on the list. Test each endpoint.

Drill 2 more on range of a rational function: set y = f(x), clear, and demand a real x

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

Domain: Write One Condition Per Piece, Then Intersect
Standard conditions
$\sqrt{u}: u \ge 0 \quad \frac{1}{\sqrt{u}}: u > 0 \quad \log u: u > 0 \quad \sin^{-1}u: -1 \le u \le 1 \quad \frac{1}{u}: u \ne 0$
Domain Through a Quadratic Inequality: sin⁻¹ of a Rational Function With |x|
Clearing a positive denominator
$\frac{p(x)}{q(x)} \le 1,\ q(x) > 0 \iff p(x) \le q(x)$
Range of a Rational Function: Set y = f(x), Clear, and Demand a Real x
Discriminant method
$y = f(x) \ \Rightarrow\ a(y)x^2 + b(y)x + c(y) = 0 \ \Rightarrow\ b(y)^2 - 4a(y)c(y) \ge 0$

Watch out for (3)

Closing the bracket at a root in the denominator
→ Domain: Write One Condition Per Piece, Then Intersect
Solving x² − x − 4 ≥ 0 as x ≥ the smaller root
→ Domain Through a Quadratic Inequality: sin⁻¹ of a Rational Function With |x|
Guessing the bracket from the shape
→ Range of a Rational Function: Set y = f(x), Clear, and Demand a Real x

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