(a) A function $f : A \to B$ defined as $f(x) = 2x$ is both one-one and onto. If $A = \{1, 2, 3, 4\}$, then find the set $B$. OR (b) Evaluate: $\sin^{-1}\left(\sin\frac{3\pi}{4}\right) + \cos^{-1}\left(\cos\frac{3\pi}{4}\right) + \tan^{-1}(1)$

(a) A function f : A → B defined as f(x) = 2x is both one-one and onto. If A = \1, 2, 3, 4\, then find the set

Question

f : A \to B

(OR)

(b) Evaluate:

\sin^{-1}\left(\sin\frac{3\pi}{4}\right) + \cos^{-1}\left(\cos\frac{3\pi}{4}\right) + \tan^{-1}(1)

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Part (a)Concept understanding — One One Onto

One-One Onto (Bijective) Functions

Picture seating students on chairs so that every student gets a chair, every chair is used, no two students share one and none is left empty. A function that manages this perfect pairing between its domain and codomain is one-one onto, or bijective.

One-one (injective)

$f$ is one-one if different inputs always give different outputs — no two students on one chair. Formally, $f(x_1)=f(x_2)\implies x_1=x_2$ (equivalently $x_1\neq x_2\implies f(x_1)\neq f(x_2)$ ).

$f(x)=2x$ on $\mathbb{R}$ is one-one, since $2a=2b\Rightarrow a=b$ . But $f(x)=x^2$ is not: $f(2)=f(-2)=4$ while $2\neq-2$ .

Onto (surjective)

$f$ is onto if every element of the codomain is actually hit — no chair left empty. Formally, for every $y$ in the codomain there is some $x$ with $f(x)=y$ . Here $f(x)=2x$ is onto (take $x=y/2$ ), whereas $f:\mathbb{R}\to\mathbb{R},\ f(x)=x^2$ is not, since negative values are never outputs.

Both together — bijective

A function that is one-one and onto is bijective: a one-to-one correspondence in which the two sets match up exactly.

Important

One-one and onto are independent properties. $f(x)=e^x$ (from $\mathbb{R}$ to $\mathbb{R}$ ) is one-one but not onto; $f(x)=x^3-x$ is onto but not one-one. You must verify both.

Why it matters

Only a bijection has a genuine inverse function: because each output comes from exactly one input (one-one) and every codomain element is used (onto), the map can be reversed unambiguously. …

Part (b)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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Part (a): for $f(x)=2x$ to be both one-one and onto, the codomain must equal the range, so $B=\{2,4,6,8\}$ . Part (b): applying principal-value ranges gives $\frac\pi4+\frac{3\pi}{4}+\frac\pi4=\frac{5\pi}{4}$ .

Part (a)

A map is one-one when distinct inputs give distinct outputs and onto when every element of the codomain is an image. When both hold, the codomain is exactly the range.

$f(x)=2x$ is one-one: $2x_1=2x_2\Rightarrow x_1=x_2$ . Its images on $A=\{1,2,3,4\}$ are

$f(1)=2,\quad f(2)=4,\quad f(3)=6,\quad f(4)=8.$ …

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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 111 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 markMCQ
$f(x)=2x$
(a) one-one but not onto
(b) one-one and onto
(c) many-one and onto
(d) many-one but not onto
›Reveal solutionSolution

$f(x)=2x$ on $\mathbb{R}$ is a bijection — both one-one and onto — option (b).

One-one: If $f(x_1)=f(x_2)$ then $2x_1=2x_2\Rightarrow x_1=x_2$ . So $f$ is injective.

…

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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
$f: R \to R$
(a) one-one onto
(b) onto only
(c) neither one-one nor onto
(d) none of these
›Reveal solutionSolution

$f(x)=3-4x$ is a linear function with non-zero slope, so it is both one-one and onto.

Given $f:\mathbb{R}\to\mathbb{R}$ , $f(x)=3-4x$ .

One-one: Let $f(x_1)=f(x_2)$ . Then $3-4x_1=3-4x_2 \Rightarrow x_1=x_2$ . So $f$ is injective.

…

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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
$Q. If A = {0, 1, 4, 9, 16, 25, ......} then function defined by f: Z \to A, f(x) = x² is:$
(a) one-one but not onto
(b) onto but not one-one
(c) one-one and onto
(d) neither one-one nor onto
›Reveal solutionSolution

Two different integers with the same absolute value (like $2$ and $-2$ ) give the same square, so $f$ is not one-one; but every element of $A$ is a perfect square that some integer squares to, so $f$ is onto.

$f:\mathbb{Z}\to A$ is defined by $f(x) = x^2$ , where $A=\{0,1,4,9,16,25,\dots\}$ is the set of all perfect squares of non-negative integers.

One-one check: Take $x=2$ and $x=-2$ . Both are in $\mathbb{Z}$ and $f(2) = 4 = f(-2)$ , but $2\ne -2$ . Different inputs give the same output — $f$ is not one-one.

…

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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
$Q. Let f : R \to R be defined by f(x) = 3x, choose the correct answer:$
(a) f is one-one onto
(b) f is many-one onto
(c) f is one-one but not onto
(d) f is neither one-one nor onto
›Reveal solutionSolution

$f(x)=3x$ is a straight-line map with non-zero slope, so it is both injective and surjective on $\mathbb{R}$ — a bijection.

Checking one-one (injective):

Let $f(x_1)=f(x_2)$ . Then $3x_1=3x_2 \Rightarrow x_1=x_2$ . So distinct inputs never share an output — $f$ is one-one.

Checking onto (surjective): …

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Q.(a) A function f:A→Bf : A \to Bf:A→B defined as f(x)=2xf(x) = 2xf(x)=2x is both one-one and onto. If A={1,2,3,4}A = \{1, 2, 3, 4\}A={1,2,3,4}, then find the set BBB.

Part (a)

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Q.(a) A function $f : A \to B$ defined as $f(x) = 2x$ is both one-one and onto. If $A = \{1, 2, 3, 4\}$ , then find the set $B$ .

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