1 markMCQSection A

If $2\cos^{-1}x = y$ , then

Q: If $2\cos^{-1}x = y$, then

**Correct option: (C) $0 \le y \le 2\pi$** The principal value branch of cos⁻¹x is $[0, \pi]$. Since $y = 2\cos^{-1}x$, multiplying the range by 2 gives $0 \le y \le 2\pi$.

Q: What is the principal value branch of $\cos^{-1}x$?

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

Inverse Trigonometric Functions

Range of Inverse Cosine Function

Options

(A)

0 \le y \le \pi

(B)

-\pi \le y \le \pi

(C)

0 \le y \le 2\pi

(D)

-\pi \le y \le 0

Official Answer

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

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

inverse cosineprincipal value branchrangecos inverse xdomain and range0 to 2pi

Marking Scheme

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

Hint

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

Quick Oral Answer

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

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

Why (C) is correct

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

Why the other options are wrong

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

Common Mistakes

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

Interesting Facts

The principal value branches of inverse trigonometric functions were standardized so each function becomes a true bijection — without restricting the domain, $\cos^{-1}x$ 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.

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Frequently Asked Questions

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

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

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

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 $[0, \pi]$ ; for sine, it happens on $[-\pi/2, \pi/2]$ . Using a different branch would make the inverse function ill-defined.