Question 11 markMCQSection A

If 2cos⁡−1x=y2\cos^{-1}x = y, then

Inverse Trigonometric Functions
Range of Inverse Cosine Function

Options

(A)0≤y≤π0 \le y \le \pi
(B)−π≤y≤π-\pi \le y \le \pi
(C)0≤y≤2π0 \le y \le 2\pi
(D)−π≤y≤0-\pi \le y \le 0

Official answer

Correct option: (C) 0≤y≤2π0 \le y \le 2\pi


The principal value branch of cos⁻¹x is [0,π][0, \pi]. Since y=2cos⁡−1xy = 2\cos^{-1}x, multiplying the range by 2 gives 0≤y≤2π0 \le y \le 2\pi.

Answer keywords:inverse cosineprincipal value branchrangecos inverse xdomain and range0 to 2pi

Marking scheme

  1. 1.1 mark: correct option (C) selected — no partial credit for MCQs, full mark only for the exact correct choice.

Hint

Recall that cos⁡−1x\cos^{-1}x has principal value range [0,π][0, \pi]; then simply double every part of that inequality.

Quick oral answer

Since cos⁡−1x\cos^{-1}x lies in [0,π][0, \pi], doubling it makes y lie in [0,2π][0, 2\pi], so the answer is option (C).

Common mistakes

  1. 1.Forgetting to multiply the range [0,π][0, \pi] by 2 and selecting option (A) instead.
  2. 2.Confusing the principal value branch of cos⁡−1x\cos^{-1}x with that of sin⁡−1x\sin^{-1}x [−π/2,π/2][-\pi/2, \pi/2] or tan⁡−1x\tan^{-1}x (−π/2,π/2)(-\pi/2, \pi/2).

Analysis and explanation

This tests whether a student can correctly scale the range of an inverse trigonometric function under a simple algebraic operation.


Concept

  • cos⁻¹x is defined for x∈[−1,1]x \in [-1, 1] with principal value branch [0,π][0, \pi] (this is the range chosen to make cosine a bijection).
  • If y=2cos⁡−1xy = 2\cos^{-1}x, then since cos⁻¹x ∈ [0,π][0, \pi], multiplying throughout by 2 gives y ∈ [0,2π][0, 2\pi].

Why (C) is correct

  • Direct scaling of [0,π][0, \pi] by 2 gives [0,2π][0, 2\pi], matching option (C) exactly.

Why the other options are wrong

  • (A) 0≤y≤π0 \le y \le \pi is simply the range of cos⁻¹x itself, ignoring the factor of 2 — a common oversight.
  • (B) −π≤y≤π-\pi \le y \le \pi confuses this with the range of tan⁻¹x or sin⁻¹x doubled, not cos⁻¹x.
  • (D) −π≤y≤0-\pi \le y \le 0 is the range doubled but with an incorrect sign, as if the branch were [−π/2,0][-\pi/2, 0].

Interesting facts

The principal value branches of inverse trigonometric functions were standardized so each function becomes a true bijection — without restricting the domain, cos⁡−1x\cos^{-1}x would have infinitely many valid outputs for a single input.

Inverse trigonometric functions are used extensively in robotics and navigation to compute angles from ratios of known distances, where staying within the correct principal branch is critical for accurate results.

Question details

Q1 of 44 questions

Board
CBSE
Class
Class 12
Subject
Mathematics
Year
2026
Section
Section A
Marks
1
Type
MCQ
Chapter
Inverse Trigonometric Functions
Topic
Range of Inverse Cosine Function
Bloom's level
understand
Syllabus topic
Unit I: Relations and Functions — Inverse Trigonometric Functions: Principal Value Branch
Author
CBSE
Reviewed by
Padhantu Academic Team

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Frequently asked questions

What is the principal value branch of cos⁡−1x\cos^{-1}x?

The principal value branch of cos⁡−1x\cos^{-1}x is [0,π][0, \pi], meaning for any x in the domain [−1,1][-1, 1], cos⁡−1x\cos^{-1}x always returns a value between 0 and π inclusive.

Why can't cos⁡−1x\cos^{-1}x have the same range as sin⁡−1x\sin^{-1}x?

Each inverse trigonometric function needs its own restricted domain (branch) on which the original function is one-one and onto. For cosine, this happens on [0,π][0, \pi]; for sine, it happens on [−π/2,π/2][-\pi/2, \pi/2]. Using a different branch would make the inverse function ill-defined.