Q.The range of the principal value branch of the function $y = \sec^{-1} x$ is ____________ .

Showing the 12 most recent of 77 on this concept.

1. CBSE 2026Set 65/1/11 markMCQ
   Q.If $2\cos^{-1}x = y$, then (A) $0 \le y \le \pi$ (B) $-\pi \le y \le \pi$ (C) $0 \le y \le 2\pi$ (D) $-\pi \le y \le 0$

   ›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

   1. 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:

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

   2. 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:

   $0 \le y \le 2\pi.$

   3. Check if every value in $[0,2\pi]$ is actually attained …

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2. CBSE 2026Set 65/3/11 markMCQ
   Q.The domain of $f(x) = \cos^{-1}(2x-5)$ is: (A) $[-1, 1]$ (B) $[4, 6]$ (C) $[-7, -3]$ (D) $[2, 3]$

   ›Reveal solutionSolution

   The inverse cosine function requires its argument to lie in $[-1, 1]$. Solving $-1 \le 2x - 5 \le 1$ gives the domain $[2, 3]$.

   The inverse cosine function $\cos^{-1}(u)$ is defined only when its input $u$ satisfies $-1 \le u \le 1$. This restriction comes from the fact that the cosine of any real angle always produces a value between $-1$ and $1$, so we can only "invert" the process for inputs in that range.

   For $f(x) = \cos^{-1}(2x - 5)$ to be defined, the expression inside—namely $2x - 5$—must satisfy this fundamental constraint.

   Finding the domain

   We need to solve the compound inequality:

   $-1 \le 2x - 5 \le 1$

   1. Add 5 to all parts:

   $-1 + 5 \le 2x - 5 + 5 \le 1 + 5$

   $4 \le 2x \le 6$

   2. Divide all parts by 2:

   $\frac{4}{2} \le \frac{2x}{2} \le \frac{6}{2}$

   $2 \le x \le 3$

   So the domain is the closed interval $[2, 3]$. …

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3. CBSE 2026Set V11 markMCQ
   Q.The domain of $\tan^{-1} x$ is

   (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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4. CBSE 2026Set CX1 mark
   Q.Find the value of $\tan^{-1}\sqrt{3}-\sec^{-1}(-2)$.

   ›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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5. CBSE 2026Set ANNUAL1 mark
   Q.$\sin^{-1}x$ is a function whose domain is __________.

   ›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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6. CBSE 2026Set ANNUAL1 markMCQ
   Q.If $y = \cos^{-1} x$ then

   (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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7. CBSE 2026Set ANNUAL1 markMCQ
   Q.Principal value of $\tan^{-1}(-1)$ is

   (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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8. CBSE 2026Set ANNUAL1 markMCQ
   Q.The principal value of $\cos^{-1} x$ is:

   (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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9. CBSE 2026Set ANNUAL1 markMCQ
   Q.Domain of function cosec⁻¹ is:

   (a) [-1, 1]

   (b) R - (-1, 1)

   (c) R

   (d) (-1, 1)

   ›Reveal solutionSolution

   Since $|\csc\theta|\ge 1$ always, its inverse function is defined only for inputs with absolute value at least $1$.

   The cosecant function $\csc\theta = \dfrac{1}{\sin\theta}$ satisfies $|\csc\theta| \ge 1$ for all $\theta$ (except where $\sin\theta=0$), since $|\sin\theta|\le1$.

   …

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10. 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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11. CBSE 2026Set ANNUAL1 markMCQ
    Q.sec⁻¹(−x) is equal to

    (a) sec⁻¹ x

    (b) −sec⁻¹ x

    (c) π − sec⁻¹ x

    (d) π + sec⁻¹ x

    ›Reveal solutionSolution

    Unlike an odd function, $\sec^{-1}$ is NOT odd on its restricted range; the correct identity is $\sec^{-1}(-x)=\pi-\sec^{-1}x$.

    The principal value branch of $\sec^{-1}$ is $[0,\pi]-\{\pi/2\}$. Let $\sec^{-1}x=\theta$, so $\sec\theta=x$ with $\theta \in [0,\pi]$.

    …

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12. CBSE 2026Set ANNUAL1 markMCQ
    Q.The principal value of $\cos^{-1}\left(\dfrac{\sqrt{3}}{2}\right)$ is

    (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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