Show that the relation $R$ on the set $Z$ of all integers, given by $R = \{(a, b) : 2 \text{ divides } (a - b)\}$, is an equivalence relation. OR If $f(x) = \dfrac{4x+3}{6x-4}$, $x \neq \dfrac{2}{3}$, show that $(f \circ f)(x) = x$ for all $x \neq \dfrac{2}{3}$. Also, find the inverse of $f$.

Show that the relation R on the set Z of all integers, — CBSE Class 12 Mathematics 2019 PYQ

Question

R

(OR)

f(x) = \dfrac{4x+3}{6x-4}

CBSECBSE Class XII Board 2019Subjective· 4mImportance★★★★★

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Part (a)Concept understanding — Equivalence Relation Proof

Proving a Relation is an Equivalence Relation

A relation $R$ on a set $A$ is an equivalence relation when it satisfies exactly three properties: it is reflexive, symmetric, and transitive. To prove a given relation is an equivalence relation, you check these three — in this order — one at a time.

Note

Antisymmetry plays no role here; that property belongs to partial orders. For an equivalence relation you need only reflexive, symmetric, transitive.

The three checks

Reflexive — show $(a, a) \in R$ for every $a \in A$ .
Symmetric — assume $(a, b) \in R$ and deduce $(b, a) \in R$ .
Transitive — assume $(a, b) \in R$ and $(b, c) \in R$ , and deduce $(a, c) \in R$ .

If all three hold, $R$ is an equivalence relation. If even one fails, produce a single counterexample and you are done.

A worked template

Let $R$ be defined on $\mathbb{Z}$ by $a\,R\,b \iff a - b$ is divisible by $5$ .

Reflexive: $a - a = 0$ , and $0$ is divisible by $5$ , so $a\,R\,a$ for every integer $a$ . ✓

Symmetric: if $a\,R\,b$ , then $a - b = 5k$ for some integer $k$ . Then $b - a = -5k = 5(-k)$ , also a multiple of $5$ , so $b\,R\,a$ . ✓

Transitive: if $a\,R\,b$ and $b\,R\,c$ , then $a - b = 5k$ and $b - c = 5m$ . Adding, $a - c = 5(k + m)$ , a multiple of $5$ , so $a\,R\,c$ . ✓

All three hold, so $R$ is an equivalence relation. …

Part (b)Concept understanding — Inverse Function Relationship

Inverse Function Relationship

Two functions are inverses when each undoes the other. If $f$ sends $a$ to $b$ , then $f^{-1}$ sends $b$ back to $a$ . Chain them together and you land exactly where you started.

The defining equations

If $f^{-1}$ is the inverse of $f$ , then

f^{-1}\big(f(x)\big) = x \quad\text{and}\quad f\big(f^{-1}(y)\big) = y.

The first holds for every $x$ in the domain of $f$ ; the second for every $y$ in the range of $f$ . This "round trip returns the input" is what inverse really means.

When does an inverse exist?

Only a one-to-one function (distinct inputs give distinct outputs) can be inverted — otherwise some output would have to map back to two inputs, which no function allows. Graphically, $f$ must pass the horizontal line test.

Important

When a function is not one-to-one over its whole domain (like $\sin x$ or $x^2$ ), we first restrict it to a piece where it is, and the inverse lives on that restricted piece.

The geometry

Because $(a,b)$ lies on $f$ exactly when $(b,a)$ lies on $f^{-1}$ , the graph of $f^{-1}$ is the mirror image of $f$ across the line $y=x$ . Consequently the domain and range swap: the range of $f$ becomes the domain of $f^{-1}$ .

Why the restriction bites — the trig case

For inverse trigonometric functions the relationship is one-sided. The "outer undo" always works:

\sin\big(\sin^{-1}x\big) = x \quad \text{for all } x\in[-1,1].

But the "inner undo" only works on the principal range:

$\sin^{-1}\big(\sin x\big) = x \quad \text{only if } x\in[-\tfrac{\pi}{2},\tfrac{\pi}{2}].$ …

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Part (a): $R$ is reflexive, symmetric and transitive, so it is an equivalence relation on $\mathbb Z$ (its classes are the even and odd integers).

Part (b): $(f\circ f)(x)=x$ , hence $f$ is self-inverse and $f^{-1}(x)=\tfrac{4x+3}{6x-4}$ .

Part (a)

$R=\{(a,b):2\mid(a-b)\}$ on $\mathbb Z$ .

Reflexive. For any $a$ , $a-a=0=2\cdot0$ , so $2\mid(a-a)$ and $(a,a)\in R$ .

Symmetric. If $(a,b)\in R$ then $a-b=2k$ for some integer $k$ , so $b-a=2(-k)$ , giving $(b,a)\in R$ .

Transitive. If $(a,b),(b,c)\in R$ then $a-b=2k$ and $b-c=2m$ ; adding, $a-c=2(k+m)$ , so $(a,c)\in R$ . …

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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 34 on this concept.

CBSE 2026Set ANNUAL1 markMCQ
$Q.The value of \tan^{-1}\left(\frac{1}{x}\right) will be:$
(a) \tan x
(b) \cot x
(c) \cot^{-1}\frac{1}{x}
(d) \cot^{-1} x
›Reveal solutionSolution

$\tan^{-1}(1/x)$ and $\cot^{-1}x$ are the same angle for positive $x$ , because tangent and cotangent are reciprocal functions.

Concept: For $x>0$ , the standard identity is

$\tan^{-1}\left(\frac{1}{x}\right) = \cot^{-1} x$ …

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CBSE 2026Set ANNUAL1 mark
$Q. Write the answer in one word/sentence: Define equivalence relation.$
›Reveal solutionSolution

Equivalence relation = reflexive + symmetric + transitive.

…

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CBSE 2025Set ANNUAL1 markMCQ
$\tan^{-1}\left(\tan \dfrac{7\pi}{6}\right)$
(a) $\dfrac{\pi}{6}$
(b) $\dfrac{7\pi}{6}$
(c) $\dfrac{1}{\sqrt{3}}$
(d) $\dfrac{5\pi}{6}$
›Reveal solutionSolution

Use periodicity of $\tan$ to bring the angle into the principal range $\left(-\tfrac{\pi}{2},\tfrac{\pi}{2}\right)$ of $\tan^{-1}$ .

Since $\tan$ has period $\pi$ : $\tan\dfrac{7\pi}{6} = \tan\left(\dfrac{7\pi}{6}-\pi\right) = \tan\dfrac{\pi}{6}$ .

…

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CBSE 2025Set A1 mark
$-1 \le x \le 1$
›Reveal solutionSolution

$\sin$ and $\sin^{-1}$ are inverse functions of each other on $[-1,1]$ , so applying one after the other returns the original input.

For $-1 \le x \le 1$ , $\sin^{-1}x$ is defined and gives an angle $\theta \in \left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]$ such that $\sin\theta = x$ . Applying $\sin$ to this angle simply recovers $x$ : …

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CBSE 2025Set ANNUAL1 mark
$R = \{(1,-1),(2,-2),(3,-1)\}$
›Reveal solutionSolution
State the three defining properties of an equivalence relation.

A relation $R$ on a non-empty set $A$ is called an equivalence relation if it satisfies all three of the following:

Reflexive: $(a,a)\in R$ for every $a\in A$ .

Symmetric: if $(a,b)\in R$ then $(b,a)\in R$ .

Transitive: if $(a,b)\in R$ and $(b,c)\in R$ then $(a,c)\in R$ . …
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CBSE 2025Set ANNUAL1 markMCQ
$\sin^{-1}\left(\sin\dfrac{2\pi}{3}\right) + \tan^{-1}\left(\tan\dfrac{3\pi}{4}\right)$
(a) $\dfrac{\pi}{3}$
(b) $\dfrac{\pi}{12}$
(c) $\dfrac{5\pi}{12}$
(d) $\dfrac{7\pi}{12}$
›Reveal solutionSolution

Bring each angle into the principal-value range using $\sin^{-1}(\sin\theta)=\pi-\theta$ and $\tan^{-1}(\tan\theta)=\theta-\pi$ for $\theta$ in the second quadrant.

Term 1: $\sin^{-1}\left(\sin\dfrac{2\pi}{3}\right)$

The range of $\sin^{-1}$ is $\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]$ . Since $\dfrac{2\pi}{3}\in\left(\dfrac{\pi}{2},\pi\right)$ lies outside this range, use

$\sin^{-1}(\sin\theta) = \pi-\theta \quad \text{for } \theta\in\left(\tfrac{\pi}{2},\pi\right)$

$\sin^{-1}\left(\sin\dfrac{2\pi}{3}\right) = \pi - \dfrac{2\pi}{3} = \dfrac{\pi}{3}$

Term 2: $\tan^{-1}\left(\tan\dfrac{3\pi}{4}\right)$

…

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CBSE 2024Set D1 markMCQ
$x \in [-1, 1],\ \sin^{-1}(-x) =$
(a) $-\sin^{-1} x$
(b) $\sin^{-1} x$
(c) $-\cos^{-1} x$
(d) $\cos^{-1} x$
›Reveal solutionSolution

$\sin^{-1}$ is odd: $\sin^{-1}(-x)=-\sin^{-1}x$ .

For $x\in[-1,1]$ , the inverse sine is an odd function, so

…

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CBSE 2024Set A1 mark
$R$
›Reveal solutionSolution

A relation that is reflexive, symmetric and transitive is called an equivalence relation.

…

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CBSE 2024Set ANNUAL1 mark
$\sin^{-1}\left(\sin \dfrac{2\pi}{3}\right)$
›Reveal solutionSolution

Since $\dfrac{2\pi}{3}$ is outside the principal range of $\sin^{-1}$ , first rewrite $\sin\dfrac{2\pi}{3}$ as the sine of an angle inside $\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]$ .

$\sin\dfrac{2\pi}{3} = \sin\left(\pi - \dfrac{2\pi}{3}\right) = \sin\dfrac{\pi}{3}$

…

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CBSE 2024Set ANNUAL1 markMCQ
$\tan^{-1}\left(\tan \dfrac{7\pi}{6}\right)$
(a) $\dfrac{\pi}{6}$
(b) $\dfrac{\pi}{4}$
(c) $\dfrac{\pi}{3}$
(d) $\dfrac{\pi}{2}$
›Reveal solutionSolution

Reduce the angle into the principal branch $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$ using the period $\pi$ of $\tan$ .

$\tan^{-1}(\tan\theta)=\theta$ holds only when $\theta$ already lies in the principal value branch $\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)$ .

Here $\theta=\dfrac{7\pi}{6}$ , which does not lie in $\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)$ . Since $\tan$ has period $\pi$ :

$\tan\dfrac{7\pi}{6} = \tan\left(\dfrac{7\pi}{6}-\pi\right) = \tan\dfrac{\pi}{6}$ …

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CBSE 2024Set ANNUAL1 markMCQ
$Q. Let R be the relation in the set Z of all integers defined as, R = {(x, y) : x - y is an integer}, then R is$
(a) Reflexive
(b) Symmetric
(c) Transitive
(d) Equivalence relation
›Reveal solutionSolution

R is reflexive, symmetric, and transitive, hence an equivalence relation.

For any $x \in \mathbb{Z}$ , $x - x = 0$ , which is an integer, so $(x,x) \in R$ for every $x$ . Hence R is reflexive.

If $(x,y) \in R$ , then $x-y$ is an integer. Then $y - x = -(x-y)$ is also an integer, so $(y,x) \in R$ . Hence R is symmetric.

…

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CBSE 2023Set E1 markMCQ
$\sin\left(\sin^{-1}\dfrac{1}{2}\right) =$
(a) $1$
(b) $\dfrac{1}{2}$
(c) $\dfrac{\sqrt{3}}{2}$
(d) $0$
›Reveal solutionSolution

$\sin\left(\sin^{-1}\dfrac{1}{2}\right)=\dfrac{1}{2}$ .

Since $\dfrac{1}{2}\in[-1,1]$ , $\sin(\sin^{-1}x)=x$ holds:

…

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Q.Show that the relation RRR on the set ZZZ of all integers, given by R={(a,b):2 divides (a−b)}R = \{(a, b) : 2 \text{ divides } (a - b)\}R={(a,b):2 divides (a−b)}, is an equivalence relation.

Part (a)

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Q.Show that the relation $R$ on the set $Z$ of all integers, given by $R = \{(a, b) : 2 \text{ divides } (a - b)\}$ , is an equivalence relation.

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