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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

12 PYQs at 25% 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, and sin⁻¹(x − 3)/√(9 − x²) twice. The three 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

Intuition

Each part of a formula refuses certain inputs. A square root needs its inside ≥0\ge 0 (>0> 0 in a denominator), a logarithm needs its argument >0> 0, sin⁡−1\sin^{-1} needs its argument in [−1,1][-1, 1], a denominator must be non-zero. The domain is where ALL conditions hold.

Definition

  • f(x)=sin⁡−1(x−3)9−x2f(x) = \dfrac{\sin^{-1}(x - 3)}{\sqrt{9 - x^2}}: −1≤x−3≤1⇒2≤x≤4-1 \le x - 3 \le 1 \Rightarrow 2 \le x \le 4; 9−x2>0⇒−3<x<39 - x^2 > 0 \Rightarrow -3 < x < 3. Intersection [2,3)[2, 3) — closed at 22, OPEN at 33 because the root is in the denominator.
  • f(x)=sin⁡−1 ⁣(log⁡2x2)f(x) = \sin^{-1}\!\left(\log_2\dfrac{x}{2}\right): −1≤log⁡2x2≤1⇒12≤x2≤2⇒1≤x≤4-1 \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⁡2x+3x2+3x+2f(x) = \log_2\dfrac{x + 3}{x^2 + 3x + 2}: need x+3(x+1)(x+2)>0\dfrac{x + 3}{(x + 1)(x + 2)} > 0 — a sign chart on the critical points −3,−2,−1-3, -2, -1 gives (−3,−2)∪(−1,∞)(-3, -2) \cup (-1, \infty). That exact set is not among the options; the official key is (D) (−3,∞)−{−1,−2}(-3, \infty) - \{-1, -2\}, the only option that starts at −3-3 and removes both poles, so pick it on this stem and know that it over-includes (−2,−1)(-2, -1).\n- An implicit equation 2x+2y=22^x + 2^y = 2: 2y=2−2x2^y = 2 - 2^x must be positive, so 2x<22^x < 2, x<1x < 1. Domain (−∞,1)(-\infty, 1) — the four-time repeat.

Standard conditions

u:u≥01u:u>0log⁡u:u>0sin⁡−1u:−1≤u≤11u:u≠0\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
xy−222Domain [−2, 2] (x-shadow) · Range [0, 2] (y-shadow)

Worked example

Find the domain of f(x)=x−1log⁡(5−x)f(x) = \dfrac{\sqrt{x - 1}}{\log(5 - x)}.
Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 1Sets, Relations and FunctionsMODERATE
The domain of the function f(x)=sin⁡−1(x−3)9−x2f(x) = \frac{\sin^{-1}(x-3)}{\sqrt{9-x^2}} is

[Q130 · 16th May Shift 2 · 2023]

Closing the bracket at a root in the denominator

9−x2\sqrt{9 - x^2} underneath means 9−x2>09 - x^2 > 0, strictly. [2,3][2, 3] is option (C) on both sittings; the answer is [2,3)[2, 3).
Drill 7 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|

Intuition

When the sin⁡−1\sin^{-1} argument is a fraction with a positive denominator, '≤1\le 1' clears to a polynomial inequality. With ∣x∣|x| present, solve for x≥0x \ge 0 and reflect.

Definition

  • sin⁡−1∣x∣+5x2+1\sin^{-1}\dfrac{|x| + 5}{x^2 + 1}: the argument is already ≥0>−1\ge 0 > -1, so only ∣x∣+5x2+1≤1\dfrac{|x| + 5}{x^2 + 1} \le 1 matters. Since x2+1>0x^2 + 1 > 0: ∣x∣+5≤x2+1|x| + 5 \le x^2 + 1, i.e. x2−∣x∣−4≥0x^2 - |x| - 4 \ge 0.
  • For x≥0x \ge 0: x2−x−4≥0⇒x≥1+172x^2 - x - 4 \ge 0 \Rightarrow x \ge \dfrac{1 + \sqrt{17}}{2} (the other root is negative). Reflect: domain (−∞,−a]∪[a,∞)(-\infty, -a] \cup [a, \infty) with a=1+172a = \dfrac{1 + \sqrt{17}}{2}.
  • Multiply through by a denominator ONLY when its sign is known; x2+1x^2 + 1 is safe, x−1x - 1 is not.
  • The quadratic-formula root with the '+' is the one that lands in x≥0x \ge 0; the options include the '−' root and half-and-minus variants as distractors.

Clearing a positive denominator

p(x)q(x)≤1, q(x)>0  ⟺  p(x)≤q(x)\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⁡−12∣x∣+3x2+2f(x) = \sin^{-1}\dfrac{2|x| + 3}{x^2 + 2}.
Practice this conceptself-check

From the bank · past-year question

Example 2Sets, Relations and FunctionsHARD
The domain of the function f(x)=sin⁡−1 ⁣(∣x∣+5x2+1)f(x)=\sin^{-1}\!\left(\dfrac{|x|+5}{x^2+1}\right) is (−∞,−a]∪[a,∞)(-\infty,-a]\cup[a,\infty). Then aa is equal to

[Q114 · 10th May Shift 2 · 2024]

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

x2−x−4≥0x^2 - x - 4 \ge 0 holds OUTSIDE the roots. With x≥0x \ge 0 imposed by the ∣x∣|x| split, only x≥1+172x \ge \dfrac{1 + \sqrt{17}}{2} survives; 17−12\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

Intuition

Write y=f(x)y = f(x), cross-multiply to a quadratic in xx, and require its discriminant ≥0\ge 0. The yy that pass are the range — then check the boundary values separately, because clearing the denominator can create or lose an endpoint.

Definition

  • y=x1+x2⇒yx2−x+y=0y = \dfrac{x}{1 + x^2} \Rightarrow yx^2 - x + y = 0; real xx needs 1−4y2≥01 - 4y^2 \ge 0, so ∣y∣≤12|y| \le \dfrac12. Range [−12,12]\left[-\dfrac12, \dfrac12\right].
  • y=x2+x+2x2+x+1=1+1x2+x+1y = \dfrac{x^2 + x + 2}{x^2 + x + 1} = 1 + \dfrac{1}{x^2 + x + 1}: x2+x+1=(x+12)2+34≥34x^2 + x + 1 = \left(x + \tfrac12\right)^2 + \tfrac34 \ge \tfrac34, so the fraction lies in (0,43]\left(0, \tfrac43\right] and y∈(1,73]y \in \left(1, \tfrac73\right]. The maximum IS attained (at x=−12x = -\tfrac12); y=1y = 1 is not.
  • y=x2x2+1=1−1x2+1y = \dfrac{x^2}{x^2 + 1} = 1 - \dfrac{1}{x^2 + 1}: 0≤y<10 \le y < 1, range [0,1)[0, 1).
  • Endpoint check: discriminant =0= 0 gives an attained endpoint; a yy that makes the leading coefficient vanish (y=0y = 0 in yx2−x+y=0yx^2 - x + y = 0) must be tested directly.

Discriminant method

y=f(x) ⇒ a(y)x2+b(y)x+c(y)=0 ⇒ b(y)2−4a(y)c(y)≥0y = 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)=3x2+5x2+1f(x) = \dfrac{3x^2 + 5}{x^2 + 1}.
Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 3Sets, Relations and FunctionsHARD
Range of the function f(x)=x2+x+2x2+x+1,  x∈Rf(x) = \frac{x^2+x+2}{x^2+x+1},\; x \in \mathbb{R} is

[Q126 · 16th May Shift 1 · 2023]

Guessing the bracket from the shape

x2+x+2x2+x+1\dfrac{x^2 + x + 2}{x^2 + x + 1} attains its maximum 73\dfrac73 but never its infimum 11, so the range is (1,73]\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

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