Assertion (A): The maximum value of $(\cos^{-1}x)^2$ is $\pi^2$. Reason (R): The range of the principal value branch of $\cos^{-1}x$ is $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$. (a) Both A and R are true and R is the correct explanation of A. (b) Both A and R are true but R is not the correct explanation of A. (c) A is true but R is false. (d) A is false but R is true.

Assertion (A): The maximum value of (cos⁻¹x)² is π². Reason (R): The range of the principal value branch of co

Question

(\cos^{-1}x)^2

(a) Both A and R are true and R is the correct explanation of A.

(b) Both A and R are true but R is not the correct explanation of A.

(d) A is false but R is true.

CBSECBSE Class XII Board 2023MCQ· 1mImportance★★★★★

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Concept understanding — Principal Value Domain

Principal Value Domain (Principal Branch)

Take $\sin x=\tfrac12$ . It has infinitely many solutions: $x=\tfrac{\pi}{6},\tfrac{5\pi}{6},\tfrac{13\pi}{6},-\tfrac{7\pi}{6},\dots$ — every angle whose sine is $\tfrac12$ . So if we want an inverse that returns a single angle for $\sin^{-1}(0.5)$ , we must first agree on one angle to report. A function is allowed only one output per input, and $\sin x$ over all of $\mathbb{R}$ is many-to-one — it fails the horizontal line test and cannot be inverted as it stands.

The idea: restrict to one clean interval

For each trigonometric ratio we restrict the angle to a single standard interval on which the function is one-to-one while still covering its entire range exactly once. On that interval the inverse becomes well-defined and single-valued. That interval — the set of angles the inverse is allowed to return — is the principal value branch (also called the principal value domain).

The interval is chosen to be strictly monotonic, to hit every output once, and to sit as close to $0$ as possible. For sine that is $\left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right]$ , where $\sin$ increases from $-1$ to $1$ .

The principal value branch of an inverse trig function is the interval of angles it returns — the restricted interval on which the original ratio is one-to-one and onto its range.

Inverse function	Domain (allowed inputs $x$ )	Principal value branch (angles returned)
$\sin^{-1} x$	$[-1,1]$	$\left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right]$
$\cos^{-1} x$	$[-1,1]$	$[0,\pi]$
$\tan^{-1} x$	$\mathbb{R}$	$\left(-\tfrac{\pi}{2},\tfrac{\pi}{2}\right)$
$\cot^{-1} x$	$\mathbb{R}$	$(0,\pi)$
$\sec^{-1} x$	$(-\infty,-1]\cup[1,\infty)$	$[0,\pi]\setminus\{\tfrac{\pi}{2}\}$
$\csc^{-1} x$	$(-\infty,-1]\cup[1,\infty)$	$\left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right]\setminus\{0\}$

Why the intervals differ …

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The assertion is true because $\cos^{-1}x$ reaches its maximum value $\pi$ at $x = -1$ , giving $(\cos^{-1}x)^2 = \pi^2$ . The reason is false because the principal value branch of $\cos^{-1}x$ is $[0, \pi]$ , not $[-\frac{\pi}{2}, \frac{\pi}{2}]$ . The correct option is (c).

The key to this problem is knowing the correct range of the inverse cosine function. Many students confuse the ranges of different inverse trigonometric functions, so let's be precise about what $\cos^{-1}x$ actually means.

The principal value branch of $\cos^{-1}x$ is defined for $x \in [-1, 1]$ and outputs angles in $[0, \pi]$ . This is different from $\sin^{-1}x$ , which outputs angles in $[-\frac{\pi}{2}, \frac{\pi}{2}]$ . The reason for choosing $[0, \pi]$ is that cosine is one-to-one (strictly decreasing) on this interval, making the inverse well-defined.

Watch out

The range $[-\frac{\pi}{2}, \frac{\pi}{2}]$ belongs to $\sin^{-1}x$ and $\tan^{-1}x$ , not to $\cos^{-1}x$ . This is a common source of confusion.

Now let's verify each statement:

Checking Assertion (A):

Since $\cos^{-1}x$ has domain $[-1, 1]$ and range $[0, \pi]$ , the function $y = \cos^{-1}x$ takes values between $0$ and $\pi$ .
The function $f(y) = y^2$ is increasing for $y \geq 0$ . Therefore, $(\cos^{-1}x)^2$ is maximized when $\cos^{-1}x$ itself is maximized. …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 66 on this concept.

CBSE 2026Set 65/1/11 markMCQ
$2\cos^{-1}x = y$
›Reveal solutionSolution
The range of $\cos^{-1}x$ is $[0,\pi]$ , so multiplying by 2 gives $y = 2\cos^{-1}x$ a range of $[0,2\pi]$ . The correct option is (C).

Concept and Intuition

The key to this problem lies entirely in understanding the range of the inverse cosine function. $\cos^{-1}x$ (also written as $\arccos x$ ) is defined as the angle whose cosine is $x$ , and by convention, that angle is always taken from the interval $[0, \pi]$ . This is not arbitrary — it's the standard principal value branch that makes the function one-to-one and therefore invertible.

Once you know that $\cos^{-1}x$ lives between $0$ and $\pi$ (inclusive), finding the range of $y = 2\cos^{-1}x$ is simply a matter of scaling that interval by a factor of 2. No tricky domain restrictions, no sign flips — just multiplication.

Watch out

A common mistake is to confuse the range of $\cos^{-1}x$ with that of $\sin^{-1}x$ (which is $[-\pi/2, \pi/2]$ ). Always recall: $\cos^{-1}x \in [0,\pi]$ , not $[-\pi/2,\pi/2]$ .

Step-by-step solution

Recall the range of $\cos^{-1}x$ The inverse cosine function $\cos^{-1}: [-1,1] \to [0,\pi]$ gives an output angle in radians. This means:

$0 \le \cos^{-1}x \le \pi \quad \text{for all } x \in [-1,1].$

Multiply the inequality by 2 Since 2 is positive, multiplying through preserves the direction of the inequalities:

$2 \cdot 0 \le 2\cos^{-1}x \le 2 \cdot \pi$
which simplifies to:
$0 \le y \le 2\pi.$

Check if every value in $[0,2\pi]$ is actually attained …
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CBSE 2026Set V11 markMCQ
$\tan^{-1} x$
(a) $\left(\frac{-\pi}{2},\frac{\pi}{2}\right)$
(b) $(0, \pi)$
(c) $[-1, 1]$
(d) $(-\infty, \infty)$
›Reveal solutionSolution

The tangent function maps $\left(-\tfrac{\pi}{2},\tfrac{\pi}{2}\right)$ onto all of $\mathbb{R}$ , so $\tan^{-1}x$ accepts every real $x$ ; answer (d).

The principal-branch tangent $\tan:\left(-\tfrac{\pi}{2},\tfrac{\pi}{2}\right)\to\mathbb{R}$ is a bijection onto $\mathbb{R}$ . Its inverse $\tan^{-1}$ therefore has domain equal to the range of $\tan$ , namely all real …
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CBSE 2026Set CX1 mark
$\tan^{-1}\sqrt{3}-\sec^{-1}(-2)$
›Reveal solutionSolution

$\tan^{-1}\sqrt3=\tfrac{\pi}{3}$ , $\sec^{-1}(-2)=\tfrac{2\pi}{3}$ , giving $-\tfrac{\pi}{3}$ .

Concept: Use the principal-value ranges: $\tan^{-1}\in\left(-\tfrac{\pi}{2},\tfrac{\pi}{2}\right)$ and $\sec^{-1}\in[0,\pi]\setminus\{\tfrac{\pi}{2}\}$ .

$\tan^{-1}\sqrt3=\dfrac{\pi}{3}\qquad(\tan\tfrac{\pi}{3}=\sqrt3).$ …

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CBSE 2026Set ANNUAL1 mark
$\sin^{-1}x$
›Reveal solutionSolution

$\sin^{-1}x$ is defined only where $\sin\theta=x$ has a solution, i.e. for $x\in[-1,1]$ .

…

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CBSE 2026Set ANNUAL1 markMCQ
$y = \cos^{-1} x$
(a) $0 \le y \le \pi$
(b) $-\dfrac{\pi}{2} \le y \le \dfrac{\pi}{2}$
(c) $-\pi \le y \le \pi$
(d) None of these
›Reveal solutionSolution

$\cos^{-1}x$ is defined so that its principal value always lies in $[0,\pi]$ .

The function $\cos x$ is one-one and onto from $[0,\pi]$ to $[-1,1]$ , so its inverse $\cos^{-1}x$ is defined on domain $[-1,1]$ with range (principal value …

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CBSE 2026Set ANNUAL1 markMCQ
$\tan^{-1}(-1)$
(a) $\dfrac{\pi}{4}$
(b) $-\dfrac{\pi}{4}$
(c) $\dfrac{3\pi}{4}$
(d) None of these
›Reveal solutionSolution

The principal value of $\tan^{-1}x$ always lies in $\left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$ .

We need $y$ such that $\tan y = -1$ and $y \in \left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)$ .

…

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CBSE 2026Set ANNUAL1 markMCQ
$\cos^{-1} x$
(a) $[0, \pi]$
(b) $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$
(c) $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$
(d) None of these
›Reveal solutionSolution

The principal value branch of $\cos^{-1}x$ is $[0,\pi]$ by definition.

The function $\cos:[0,\pi]\to[-1,1]$ is a bijection, so its inverse $\cos^{-1}:[-1,1]\to[0,\pi]$ is defined w …

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CBSE 2026Set ANNUAL1 markMCQ
$Q. Principal value of cos⁻¹(1/2) is:$
(a) π/2
(b) π/3
(c) π/4
(d) π/6
›Reveal solutionSolution

The principal value of $\cos^{-1}x$ lies in $[0,\pi]$ , and $\cos\left(\frac{\pi}{3}\right) = \frac12$ .

We need $\theta \in [0,\pi]$ such that $\cos\theta = \dfrac12$ .

…

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CBSE 2026Set ANNUAL1 markMCQ
$\cos^{-1}\left(\dfrac{\sqrt{3}}{2}\right)$
(a) $\dfrac{\pi}{6}$
(b) $\dfrac{\pi}{3}$
(c) $\dfrac{\pi}{4}$
(d) $\dfrac{\pi}{2}$
›Reveal solutionSolution

Find the angle in the principal-value range $[0,\pi]$ of $\cos^{-1}$ whose cosine equals $\dfrac{\sqrt3}{2}$ .

We need $\theta\in[0,\pi]$ (the principal-value branch of $\cos^{-1}$ ) such that

$\cos\theta = \frac{\sqrt3}{2}$

…

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CBSE 2026Set ANNUAL1 markMCQ
$Q. The principal value of cot⁻¹(-1/\sqrt3) is ................. .$
(a) π/3
(b) π/4
(c) 2π/3
(d) 4π/3
›Reveal solutionSolution

The principal range of $\cot^{-1}$ is $(0,\pi)$ ; find the angle in that range with cotangent $-1/\sqrt3$ .

We know $\cot(\pi/3)=1/\sqrt3$ . Since the given value is negative and the principal range of $\cot^{-1}$ is $(0,\pi)$ , the required angle lies in the second quadrant, where cotangent is negative.

…

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CBSE 2026Set ANNUAL1 markMCQ
$Q.The principal value of tan^{-1}(-\sqrt{3}) is:$
(a)
(i) \pi/3
(b)
(ii) -\pi/3
(c)
(iii) \pi/6
(d)
(iv) -\pi/6
›Reveal solutionSolution
$\tan^{-1}(-\sqrt{3}) = -\dfrac{\pi}{3}$ — option (ii).

Concept. The principal value of $\tan^{-1} x$ is the unique angle $\theta$ lying in the open interval $\left(-\frac{\pi}{2},\ \frac{\pi}{2}\right)$ such that $\tan\theta = x$ . This is the standard NCERT/CBSE principal-value branch that the UBSE Class-12 syllabus also follows.

Why this branch. Tangent is one-to-one on $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$ , so exactly one angle there gives each real value.

Steps.

We need $\theta$ with $\tan\theta = -\sqrt{3}$ and $-\frac{\pi}{2}<\theta<\frac{\pi}{2}$ . …
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CBSE 2025Set X11 markMCQ

\sin^{-1} x

List - I	List - II
A) Domain of $\sin^{-1} x$	i) $\left(\dfrac{-\pi}{2}, \dfrac{\pi}{2}\right)$
B) Range of $\tan^{-1} x$	ii) $[0, \pi]$
C) Range of $\cos^{-1} x$	iii) $[-1, 1]$

(a) A-i, B-ii, C-iii

(b) A-iii, B-ii, C-i

(d) A-iii, B-i, C-ii

›Reveal solution

Matching inverse-trig domains/ranges — correct option is (d).

Recall the standard facts: the domain of $\sin^{-1}x$ is $[-1,1]$ (matches iii), the principal range of $\tan^{-1}x$ is the open interval $\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)$ (matches i), and the principal range …

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Q.Assertion (A): The maximum value of (cos⁡−1x)2(\cos^{-1}x)^2(cos−1x)2 is π2\pi^2π2. Reason (R): The range of the principal value branch of cos⁡−1x\cos^{-1}xcos−1x is [−π2,π2]\left[-\frac{\pi}{2}, \frac{\pi}{2}\right][−2π​,2π​].

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Q.Assertion (A): The maximum value of $(\cos^{-1}x)^2$ is $\pi^2$ . Reason (R): The range of the principal value branch of $\cos^{-1}x$ is $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ .

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