CBSE Sample Paper 2026-27 · Section D · 5 marks

Q32: Is $(a, b)R(c, d) \Leftrightarrow a^2 + d^2 = b^2 + c^2$ an equivalence relation; or is $y = x^2 + 5$ injective, surjective

Sample 2026-27Ch 1 · Relations and Functions

The question

(a) Let $A = N \times N$ ⁠, where $N$ is the set of natural numbers and a relation $R$ is defined on $A$ such that $(a, b)R(c, d)$ if and only if $a^2 + d^2 = b^2 + c^2$ ⁠. Check whether $R$ is an equivalence relation or not?

(b) A relation $R: (-\infty, \infty) \to [0, \infty)$ is defined as $R = \{(x, y) : y = x^2 + 5\}$

Draw its graph and check whether it is a function or not.

If the given relation is a function, then examine whether the function is injective, and whether it is surjective or not.

For Visually Impaired:

A relation $R: (-\infty, \infty) \to [0, \infty)$ is defined as $R = \{(x, y) : y = x^2 + 7\}$ ⁠.

Check whether it is a function or not. If the given relation is a function, then examine whether the function is injective, surjective, or both, and conclude whether it is bijective or not.

Exam answer

(a)

Reflexive: Let $(p, q) \in A = N \times N$ ⁠.

Then $p^2 + q^2 = q^2 + p^2$ ⁠, so $(p, q)R(p, q)$ for all $(p, q) \in N \times N$ ⁠.

Why?

Reflexive means every element is related to itself. Read the rule with $(a, b) = (p, q)$ and $(c, d) = (p, q)$ ⁠: we need

p^2 + q^2 = q^2 + p^2

That is true for all $p, q$ ⁠, because addition can be done in any order.

$\Rightarrow R$ is a reflexive relation.

Symmetric: Let $(p, q), (r, s) \in A$ with $(p, q)R(r, s)$ ⁠.

\begin{aligned} &\Rightarrow p^2 + s^2 = q^2 + r^2 \\ &\Rightarrow r^2 + q^2 = s^2 + p^2 \\ &\Rightarrow (r, s)R(p, q) \end{aligned}

Why?

Symmetric means: if $(p, q)R(r, s)$ ⁠, then $(r, s)R(p, q)$ ⁠.

The rule for $(a, b)R(c, d)$ is $a^2 + d^2 = b^2 + c^2$ ⁠. For $(r, s)R(p, q)$ ⁠, put $a = r$ ⁠, $b = s$ ⁠, $c = p$ ⁠, $d = q$ ⁠:

r^2 + q^2 = s^2 + p^2

This is the given equation with its two sides swapped and each side reordered.

\begin{aligned} p^2 + s^2 &= q^2 + r^2 \\ q^2 + r^2 &= p^2 + s^2 \quad \text{(swap sides)} \\ r^2 + q^2 &= s^2 + p^2 \quad \text{(reorder)} \end{aligned}

Thus, $R$ is a symmetric relation.

Transitive: Let $(p, q), (r, s), (u, v) \in A$ with $(p, q)R(r, s)$ and $(r, s)R(u, v)$ ⁠.

p^2 + s^2 = q^2 + r^2 \quad\ldots(\text{i})

r^2 + v^2 = s^2 + u^2 \quad\ldots(\text{ii})

Why?

Transitive means: if the first pair is related to the second, and the second to the third, then the first is related to the third.

Write out both given relations from the rule. The target is $(p, q)R(u, v)$ ⁠, that is $p^2 + v^2 = q^2 + u^2$ ⁠.

Adding (i) and (ii),

\begin{aligned} &p^2 + s^2 + r^2 + v^2 \\ &\quad = q^2 + r^2 + s^2 + u^2 \\ &\Rightarrow p^2 + v^2 = q^2 + u^2 \\ &\Rightarrow (p, q)R(u, v) \end{aligned}

Why?

The middle pair's terms $r^2$ and $s^2$ appear on both sides after adding. Subtract $r^2 + s^2$ from both sides, and only the target equation is left.

\begin{aligned} &\text{Left: } p^2 + s^2 + r^2 + v^2 \\ &\quad - (r^2 + s^2) = p^2 + v^2 \\[4pt] &\text{Right: } q^2 + r^2 + s^2 + u^2 \\ &\quad - (r^2 + s^2) = q^2 + u^2 \end{aligned}

Thus, $R$ is a transitive relation.

So, $R$ is reflexive, symmetric and transitive. Hence $R$ is an equivalence relation.

Why?

An equivalence relation is one that is reflexive, symmetric and transitive, all three. Always write the conclusion line in words.

Another way to see it: the rule is the same as $a^2 - b^2 = c^2 - d^2$ ⁠. So R relates $(a, b)$ and $(c, d)$ exactly when $a^2 - b^2$ and $c^2 - d^2$ are equal.

(b)

Graph of $y = x^2 + 5$ ⁠: a parabola opening upwards with vertex $(0, 5)$ ⁠.

Why?

$y = x^2 + 5$ is the parabola $y = x^2$ moved up by 5. To draw it, plot a few points:

\begin{aligned} x = 0 &\Rightarrow y = 0 + 5 = 5 \\ x = \pm 1 &\Rightarrow y = 1 + 5 = 6 \\ x = \pm 2 &\Rightarrow y = 4 + 5 = 9 \\ x = \pm 3 &\Rightarrow y = 9 + 5 = 14 \end{aligned}

( $x$ and $-x$ have the same square, for example $(-2)^2 = 4 = 2^2$ ⁠.)

The graph of

y = x^2 + 5

⁠, drawn through

(\pm 1, 6)

⁠,

(\pm 2, 9)

⁠,

(\pm 3, 14)

and

(\pm 4, 21)

⁠.

Yes, $R$ is a function, as the graph shows that every element of $(-\infty, \infty)$ has a unique image in $[0, \infty)$ ⁠.

Why?

A relation is a function when every element of the domain has exactly one image in the co-domain.

On the graph: every vertical line meets the curve exactly once. Each image $x^2 + 5$ is at least 5, so it lies in $[0, \infty)$ ⁠.

Let $f: (-\infty, \infty) \to [0, \infty)$ ⁠, $f(x) = x^2 + 5$ ⁠.

Injective: $f(-1) = (-1)^2 + 5 = 6$ and $f(1) = 1^2 + 5 = 6$ ⁠.

So $f(-1) = f(1)$ with $-1 \ne 1$ ⁠. Thus $f$ is many-one, so not injective.

Why?

Injective (one-one) means different inputs always give different outputs. To show a function is not injective, one counterexample is enough: two different inputs with the same output.

On the graph, the horizontal line $y = 6$ meets the curve twice.

Surjective: Range of $f = [5, \infty) \ne$ co-domain $[0, \infty)$ ⁠.

Why?

Surjective (onto) means the range is the whole co-domain.

Since $x^2 \ge 0$ ⁠, $x^2 + 5 \ge 5$ ⁠. So a number such as 2 in the co-domain is never an output: $x^2 + 5 = 2$ needs $x^2 = -3$ ⁠, which has no real solution.

\begin{aligned} x^2 + 5 &= 2 \\ x^2 &= 2 - 5 = -3 \end{aligned}

A square is never negative, so there is no such real $x$ ⁠.

So, $f$ is not surjective.

For Visually Impaired

For every element $a$ of $(-\infty, \infty)$ ⁠, $a^2 \in [0, \infty)$ ⁠, so $a^2 + 7 \in [7, \infty)$ ⁠. Hence every element of $(-\infty, \infty)$ has a unique image in $[0, \infty)$ ⁠.

Why?

Each real $a$ gives one value $a^2 + 7$ ⁠, and that value is at least 7, so it is in the co-domain $[0, \infty)$ ⁠. That is what "function" requires.

So $f: (-\infty, \infty) \to [0, \infty)$ ⁠, $f(x) = x^2 + 7$ is a function.

Injective: $f(-1) = (-1)^2 + 7 = 8$ and $f(1) = 1^2 + 7 = 8$ ⁠.

So $f(-1) = f(1)$ with $-1 \ne 1$ ⁠. Thus $f$ is many-one, so not injective.

Why?

As before, $x$ and $-x$ have the same square, so one pair such as $-1$ and $1$ shows two inputs with the same output.

Surjective: Range of $f = [7, \infty) \ne$ co-domain $[0, \infty)$ ⁠. So, it is not surjective.

Why?

Since $x^2 + 7 \ge 7$ ⁠, the numbers from 0 up to (but not including) 7 are never outputs.

For example, take 0 in the co-domain:

\begin{aligned} x^2 + 7 &= 0 \\ x^2 &= 0 - 7 = -7 \end{aligned}

No real square is negative, so 0 has no pre-image.

Bijective means injective and surjective both. Failing either one is enough to be not bijective.

$\therefore\ f(x)$ is neither injective nor surjective, and hence not bijective.

Where this comes from

Chapter 1, Relations and Functions: Equivalence relation; injective and surjective.

Question type: (a) Show that a relation is an equivalence relation; (b) Not one-one or not onto

How to spot it: click to see

(a) Type 3. Show that a relation is an equivalence relation

Spot it: the words "is an equivalence relation", often with "2 divides $a - b$ " or " $|a - b|$ is even".

Here: The rule is $a^2 + d^2 = b^2 + c^2$ on pairs. For transitive, add the two given equations: $r^2$ and $s^2$ cancel, leaving $p^2 + v^2 = q^2 + u^2$ ⁠.

(b) Type 6. Not one-one or not onto: one example is enough

Spot it: functions such as $x^2$ ⁠, $|x|$ or $[x]$ on $\R$ ⁠, where two inputs share an output, or some numbers are never reached.

Here: $x^2 + 5$ on all reals: $f(-1) = f(1) = 6$ ⁠, so it is not one-one. And $x^2 + 5 \ge 5$ ⁠, so 2 is never reached, and it is not onto.

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Relations and Functions in past papers: question 8 of 18 · see them all

Q32: Is (a,b)R(c,d)⇔a2+d2=b2+c2(a, b)R(c, d) \Leftrightarrow a^2 + d^2 = b^2 + c^2(a,b)R(c,d)⇔a2+d2=b2+c2 an equivalence relation; or is y=x2+5y = x^2 + 5y=x2+5 injective, surjective

Q32: Is $(a, b)R(c, d) \Leftrightarrow a^2 + d^2 = b^2 + c^2$ an equivalence relation; or is $y = x^2 + 5$ injective, surjective