NCERT Class 11 Mathematics — Relations and Functions, Miscellaneous Exercise on Chapter 2. All 12 questions solved.

This exercise pulls together everything from the chapter and adds one new tool, the algebra of functions. For real functions $f$ and $g$ with a common domain:

$$(f+g)(x) = f(x) + g(x), \qquad (f-g)(x) = f(x) – g(x), \qquad \left(\frac{f}{g}\right)(x) = \frac{f(x)}{g(x)},\ g(x) \ne 0$$

The recurring question is still the one from Exercise 2.3 — is this a function? — but now the failures are subtler: a piecewise definition whose two branches disagree where they meet (question 1), and a rule whose input can be produced in two different ways (question 11).

Key insight. A piecewise definition on overlapping intervals is legitimate only if the branches agree at the overlap. In question 1 both $f$ and $g$ are defined twice at their joining point; $f$ survives because $x^2$ and $3x$ both give $9$ at $x = 3$, and $g$ fails because at $x = 2$ they give $4$ and $6$. The definition looks identical in shape — only the number where the pieces meet differs.

Question 1

The relation $f$ is defined by

$$f(x) = \begin{cases} x^2, & 0 \le x \le 3 \\ 3x, & 3 \le x \le 10 \end{cases}$$

The relation $g$ is defined by

$$g(x) = \begin{cases} x^2, & 0 \le x \le 2 \\ 3x, & 2 \le x \le 10 \end{cases}$$

Show that $f$ is a function and $g$ is not a function.

Solution. Away from the joining point there is nothing to check — on $0 \le x < 3$ only the first branch applies, and on $3 < x \le 10$ only the second. The whole question is what happens where the two intervals overlap, and both definitions overlap at exactly one point because their intervals are written with $\le$ at both ends.

For $f$, the overlap is at $x = 3$. The two branches give

$$x^2 = 3^2 = 9 \qquad\text{and}\qquad 3x = 3(3) = 9$$

They agree, so $x = 3$ has the single image $9$. Every other point of $[0, 10]$ is covered by exactly one branch, so every input has exactly one output and $f$ is a function.

For $g$, the overlap is at $x = 2$. Now

$$x^2 = 2^2 = 4 \qquad\text{but}\qquad 3x = 3(2) = 6$$

The input $2$ is assigned two different images, $4$ and $6$. That violates the definition of a function outright, so $g$ is a relation but not a function. $\blacksquare$

Question 2

If $f(x) = x^2$, find $\dfrac{f(1.1) – f(1)}{(1.1 – 1)}$.

Solution. Evaluate the two function values, then divide:

$$f(1.1) = (1.1)^2 = 1.21, \qquad f(1) = 1^2 = 1$$

$$\frac{f(1.1) – f(1)}{1.1 – 1} = \frac{1.21 – 1}{0.1} = \frac{0.21}{0.1} = 2.1$$

This quantity is the average rate of change of $f$ between $x = 1$ and $x = 1.1$ — the slope of the chord joining those two points on the parabola. As the gap shrinks towards zero, it approaches $f'(1) = 2$, which is why this expression reappears at the start of the Limits and Derivatives chapter.

Question 3

Find the domain of the function $f(x) = \dfrac{x^2 + 2x + 1}{x^2 – 8x + 12}$.

Solution. A rational function is defined wherever its denominator is non-zero, so factorise the denominator and exclude its roots:

$$x^2 – 8x + 12 = (x – 2)(x – 6)$$

This vanishes at $x = 2$ and $x = 6$. Everywhere else the quotient is a perfectly good real number, so

$$\text{Domain} = \mathbf{R} – \{2, 6\}$$

The numerator factorises as $(x+1)^2$, but that is irrelevant to the domain — a zero numerator causes no difficulty, only a zero denominator does.

Question 4

Find the domain and the range of the real function $f$ defined by $f(x) = \sqrt{(x – 1)}$.

Solution. The square root of a negative number is not real, so the domain is fixed by

$$x – 1 \ge 0 \quad\Longrightarrow\quad x \ge 1$$

$$\text{Domain} = [1,\ \infty)$$

For the range: as $x$ runs over $[1, \infty)$, the expression $x – 1$ runs over $[0, \infty)$, and the non-negative square root of a non-negative number is again non-negative, taking every such value exactly once. So

$$\text{Range} = [0,\ \infty)$$

Both endpoints are attained at $x = 1$, where $f(1) = 0$.

Question 5

Find the domain and the range of the real function $f$ defined by $f(x) = |x – 1|$.

Solution. The modulus is defined for every real number and involves no division or root, so nothing is excluded:

$$\text{Domain} = \mathbf{R}$$

For the range, $|x – 1|$ is a distance and therefore never negative. It equals $0$ at $x = 1$, and for any $k > 0$ it equals $k$ at $x = 1 + k$. Every non-negative value is achieved:

$$\text{Range} = [0,\ \infty)$$

Compare with question 4: the same range, reached quite differently — there the domain had to be restricted to make the formula defined, here it did not.

Question 6

Let $f = \left\{ \left(x,\ \dfrac{x^2}{1 + x^2}\right) : x \in \mathbf{R} \right\}$ be a function from $\mathbf{R}$ into $\mathbf{R}$. Determine the range of $f$.

Solution. Write $y$ for the output and solve for $x$ — the range is the set of $y$ for which a real $x$ exists.

$$y = \frac{x^2}{1 + x^2}$$

Note first that $1 + x^2 > 0$ always, so the function is defined for every real $x$ and no value is lost. Clearing the denominator:

$$y(1 + x^2) = x^2 \quad\Longrightarrow\quad y = x^2 – yx^2 = x^2(1 – y) \quad\Longrightarrow\quad x^2 = \frac{y}{1 – y}$$

A real $x$ exists precisely when the right-hand side is non-negative (and the denominator is non-zero):

$$\frac{y}{1 – y} \ge 0 \quad\Longrightarrow\quad 0 \le y < 1$$

since numerator and denominator must share a sign: $y \ge 0$ with $1 – y > 0$. (The alternative, $y \le 0$ with $1 – y < 0$, requires $y \le 0$ and $y > 1$ simultaneously, which is impossible.)

$$\text{Range} = [0,\ 1)$$

A quick sanity check without algebra: dividing numerator and denominator by $x^2$ for $x \ne 0$ gives $y = \dfrac{1}{\frac{1}{x^2} + 1}$, which is always less than $1$ and approaches $1$ as $|x|$ grows. And $y = 0$ occurs at $x = 0$.

Question 7

Let $f, g : \mathbf{R} \to \mathbf{R}$ be defined respectively by $f(x) = x + 1$, $g(x) = 2x – 3$. Find $f + g$, $f – g$ and $\dfrac{f}{g}$.

Solution. Combine the formulae pointwise.

$$(f + g)(x) = (x + 1) + (2x – 3) = 3x – 2$$

$$(f – g)(x) = (x + 1) – (2x – 3) = x + 1 – 2x + 3 = -x + 4$$

For the quotient, the denominator must not vanish. $2x – 3 = 0$ at $x = \tfrac{3}{2}$, so that point is removed from the domain:

$$\left(\frac{f}{g}\right)(x) = \frac{x + 1}{2x – 3}, \qquad x \ne \frac{3}{2}$$

The subtraction is where marks are lost: the minus sign must be distributed across both terms of $g(x)$, turning $-3$ into $+3$.

Question 8

Let $f = \{(1,1), (2,3), (0,-1), (-1,-3)\}$ be a function from $\mathbf{Z}$ to $\mathbf{Z}$ defined by $f(x) = ax + b$, for some integers $a$, $b$. Determine $a$, $b$.

Solution. Each ordered pair gives one equation. Choose the pair that isolates a single unknown — here $(0, -1)$, because $x = 0$ kills the $a$ term:

$$f(0) = a(0) + b = b = -1$$

Now use $(1, 1)$:

$$f(1) = a(1) + b = a – 1 = 1 \quad\Longrightarrow\quad a = 2$$

So $f(x) = 2x – 1$. Verify against the two pairs not yet used:

$$f(2) = 4 – 1 = 3 \quad ✓ \qquad f(-1) = -2 – 1 = -3 \quad ✓$$

All four pairs are consistent with the same $a$ and $b$, which confirms both the answer and the question’s premise that $f$ really is linear.

Question 9

Let $R$ be a relation from $\mathbf{N}$ to $\mathbf{N}$ defined by $R = \{(a, b) : a, b \in \mathbf{N} \text{ and } a = b^2\}$. Are the following true?

(i) $(a, a) \in R$, for all $a \in \mathbf{N}$ (ii) $(a, b) \in R$ implies $(b, a) \in R$ (iii) $(a, b) \in R$, $(b, c) \in R$ implies $(a, c) \in R$

Justify your answer in each case.

Solution. Membership in $R$ means the first entry is the square of the second. Each part is a general claim, so one counterexample refutes it.

(i) False. $(a, a) \in R$ would need $a = a^2$, which holds only for $a = 1$. Taking $a = 2$: $(2, 2) \in R$ would require $2 = 4$, which is false. So the statement fails for every natural number except $1$.

(ii) False. Take $a = 4$, $b = 2$. Then $(4, 2) \in R$ because $4 = 2^2$. But $(2, 4) \in R$ would require $2 = 4^2 = 16$, which is false.

(iii) False. Take $c = 2$, $b = 4$, $a = 16$. Then

$$(16, 4) \in R \text{ since } 16 = 4^2, \qquad (4, 2) \in R \text{ since } 4 = 2^2$$

But $(16, 2) \in R$ would require $16 = 2^2 = 4$, which is false. In general the relation composes to $a = c^4$, not $a = c^2$.

Question 10

Let $A = \{1, 2, 3, 4\}$, $B = \{1, 5, 9, 11, 15, 16\}$ and $f = \{(1,5), (2,9), (3,1), (4,5), (2,11)\}$. Are the following true?

(i) $f$ is a relation from $A$ to $B$    (ii) $f$ is a function from $A$ to $B$

Justify your answer in each case.

Solution.

(i) True. A relation from $A$ to $B$ is any subset of $A \times B$. Check each pair: the first entries $1, 2, 3, 4, 2$ all lie in $A$, and the second entries $5, 9, 1, 5, 11$ all lie in $B$. So every pair belongs to $A \times B$, and $f$ is a subset of it.

(ii) False. The element $2$ appears as a first entry twice, in $(2, 9)$ and $(2, 11)$, so it has two distinct images. A function requires exactly one.

Note that $(1,5)$ and $(4,5)$ sharing the image $5$ is not a problem — repeated outputs are allowed, repeated inputs are not.

Question 11

Let $f$ be the subset of $\mathbf{Z} \times \mathbf{Z}$ defined by $f = \{(ab,\ a + b) : a, b \in \mathbf{Z}\}$. Is $f$ a function from $\mathbf{Z}$ to $\mathbf{Z}$? Justify your answer.

Solution. The first entry is a product and the second a sum, and a single integer can be written as a product in many ways — each factorisation giving a different sum. That is exactly what breaks the function condition.

Take the first entry $6$:

• $a = 2$, $b = 3$ gives the pair $(6, 5)$;
• $a = 1$, $b = 6$ gives the pair $(6, 7)$.

Both belong to $f$, so the input $6$ has two different images, $5$ and $7$. Hence $f$ is not a function.

(The failure is thorough rather than incidental: $0$ has infinitely many images, since $a = 0$ with any $b$ gives the pair $(0, b)$.)

Question 12

Let $A = \{9, 10, 11, 12, 13\}$ and let $f : A \to \mathbf{N}$ be defined by $f(n) = $ the highest prime factor of $n$. Find the range of $f$.

Solution. Factorise each element of $A$ and take the largest prime that appears.

$n$	Prime factorisation	Highest prime factor
$9$	$3 \times 3$	$3$
$10$	$2 \times 5$	$5$
$11$	$11$ (prime)	$11$
$12$	$2 \times 2 \times 3$	$3$
$13$	$13$ (prime)	$13$

The images are $3, 5, 11, 3, 13$. The range is the set of these values, so the repeated $3$ is written once:

$$\text{Range} = \{3, 5, 11, 13\}$$

A prime number is its own highest prime factor, which is what makes $11$ and $13$ map to themselves.

Common mistakes

• Question 1, checking the wrong point or no point at all. Both $f$ and $g$ look identical in structure; the only thing separating them is whether the branches agree where the intervals overlap. Test $x = 3$ for $f$ and $x = 2$ for $g$ — and nothing else is needed.
• Question 3, excluding the roots of the numerator. $(x+1)^2 = 0$ at $x = -1$, but $f(-1) = 0$ is a perfectly good value. Only the denominator restricts a domain.
• Question 4, giving the range as $\mathbf{R}$. The radical symbol means the non-negative root, so no output is negative.
• Question 6, forgetting to check the sign condition. Reaching $x^2 = \dfrac{y}{1-y}$ is half the work; the range comes from requiring that expression to be non-negative, which is what excludes $y \ge 1$.
• Question 7, mishandling the minus sign in $f – g$. $(x+1) – (2x-3)$ is $-x + 4$, not $-x – 2$. Distribute the minus across both terms.
• Question 8, using two awkward pairs. Starting from $(0, -1)$ gives $b$ immediately, because the $a$ term vanishes. Choosing $(2,3)$ and $(-1,-3)$ instead means solving simultaneous equations for no reason.
• Question 9, answering (i) “true because $1 = 1^2$”. The claim is “for all $a \in \mathbf{N}$”, so a single value that works does not establish it — and $a = 2$ already fails.
• Question 10(ii), objecting to $(1,5)$ and $(4,5)$. Two inputs sharing an output is ordinary. The disqualifying pair is $(2,9)$ with $(2,11)$.
• Question 11, testing only one factorisation. Any composite first entry can be produced two ways. Writing down $6 = 2 \times 3 = 1 \times 6$ is the entire proof.

Practise next

• Exercise 3.1 — the next chapter, Trigonometric Functions, where domain and range questions return for $\sin$, $\cos$ and $\tan$, and the range work of questions 4 to 6 is directly reused.
• Exercise 2.3 — worth revising if question 1 or 10 felt uncertain, since both turn on the single definition tested there.