NCERT Class 11 Mathematics — Relations and Functions, Exercise 2.3. All 5 questions solved.

A function is a relation with one extra requirement:

Every element of the domain has exactly one image.

So a relation fails to be a function when some first entry is paired with two different second entries. Nothing forbids the reverse — many inputs may share the same output, and question 1(i) is exactly that case.

For a real function given by a formula, the domain is the largest set of real numbers for which the formula makes sense (no negative number under a square root, no zero in a denominator), and the range is the set of values it then actually produces.

Key insight. Domain and range are decided by different questions. For domain, ask what can go in — which is a matter of what the formula forbids. For range, ask what comes out — which usually means tracking the formula from the inside out. In question 2(ii), $9 – x^2$ must not be negative (that fixes the domain), and then it runs from $0$ to $9$, so its square root runs from $0$ to $3$ (that fixes the range).

Question 1

Which of the following relations are functions? Give reasons. If it is a function, determine its domain and range.

(i) $\{(2,1), (5,1), (8,1), (11,1), (14,1), (17,1)\}$ (ii) $\{(2,1), (4,2), (6,3), (8,4), (10,5), (12,6), (14,7)\}$ (iii) $\{(1,3), (1,5), (2,5)\}$

Solution. The test is mechanical: list the first entries and check whether any appears twice.

(i) A function. The first entries $2, 5, 8, 11, 14, 17$ are all different, so each has a single image. That every image happens to be $1$ is irrelevant — a function may send many inputs to the same output. This is the constant function on that domain.

$$\text{Domain} = \{2, 5, 8, 11, 14, 17\}, \qquad \text{Range} = \{1\}$$

(ii) A function. The first entries $2, 4, 6, 8, 10, 12, 14$ are distinct. The rule at work is $y = \tfrac{x}{2}$.

$$\text{Domain} = \{2, 4, 6, 8, 10, 12, 14\}, \qquad \text{Range} = \{1, 2, 3, 4, 5, 6, 7\}$$

(iii) Not a function. The element $1$ appears as a first entry twice, in $(1,3)$ and $(1,5)$, so it has two different images. A function must assign exactly one.

Question 2

Find the domain and range of the following real functions:

(i) $f(x) = -|x|$      (ii) $f(x) = \sqrt{9 – x^2}$

Solution.

(i) The modulus $|x|$ is defined for every real $x$ and negating it creates no obstruction, so the domain is all of $\mathbf{R}$.

For the range, note $|x| \ge 0$ for every $x$, with equality only at $x = 0$. Multiplying by $-1$ reverses the inequality:

$$-|x| \le 0$$

Every value from $0$ downwards is attained — $-|x| = -k$ at $x = k$ for any $k \ge 0$. Hence

$$\text{Domain} = \mathbf{R}, \qquad \text{Range} = (-\infty,\ 0]$$

The bracket at $0$ is square because $f(0) = 0$ is attained.

(ii) A square root of a real number requires a non-negative radicand:

$$9 – x^2 \ge 0 \quad\Longrightarrow\quad x^2 \le 9 \quad\Longrightarrow\quad -3 \le x \le 3$$

So the domain is $[-3, 3]$.

For the range, work outwards. On $[-3, 3]$, $x^2$ runs over $[0, 9]$, so $9 – x^2$ runs over $[0, 9]$ as well (largest at $x = 0$, smallest at $x = \pm 3$). Taking the non-negative square root of a quantity in $[0, 9]$ gives a value in $[0, 3]$, and every such value occurs. Hence

$$\text{Domain} = [-3, 3] = \{x : -3 \le x \le 3\}, \qquad \text{Range} = [0, 3] = \{y : 0 \le y \le 3\}$$

Geometrically, $y = \sqrt{9 – x^2}$ is the upper half of the circle $x^2 + y^2 = 9$ — which is why the range stops at $0$ rather than continuing to $-3$.

Question 3

A function $f$ is defined by $f(x) = 2x – 5$. Write down the values of

(i) $f(0)$    (ii) $f(7)$    (iii) $f(-3)$

Solution. Substitute each value in turn, keeping the brackets around a negative argument:

$$f(0) = 2(0) – 5 = -5$$ $$f(7) = 2(7) – 5 = 14 – 5 = 9$$ $$f(-3) = 2(-3) – 5 = -6 – 5 = -11$$

Question 4

The function $t$ which maps temperature in degrees Celsius into temperature in degrees Fahrenheit is defined by $t(\mathrm{C}) = \dfrac{9\mathrm{C}}{5} + 32$. Find

(i) $t(0)$    (ii) $t(28)$    (iii) $t(-10)$    (iv) the value of $\mathrm{C}$ when $t(\mathrm{C}) = 212$

Solution.

(i) $t(0) = \dfrac{9(0)}{5} + 32 = 32$ — the freezing point of water, $0^\circ$C being $32^\circ$F.

(ii)

$$t(28) = \frac{9 \times 28}{5} + 32 = \frac{252}{5} + 32 = \frac{252 + 160}{5} = \frac{412}{5} = 82.4$$

(iii)

$$t(-10) = \frac{9(-10)}{5} + 32 = -18 + 32 = 14$$

(iv) Here the function is run backwards: solve $t(\mathrm{C}) = 212$ for $\mathrm{C}$.

$$\frac{9\mathrm{C}}{5} + 32 = 212 \quad\Longrightarrow\quad \frac{9\mathrm{C}}{5} = 180 \quad\Longrightarrow\quad 9\mathrm{C} = 900 \quad\Longrightarrow\quad \mathrm{C} = 100$$

$100^\circ$C is the boiling point of water, and $212^\circ$F is its Fahrenheit equivalent — a useful check that the arithmetic has landed somewhere real.

Question 5

Find the range of each of the following functions:

(i) $f(x) = 2 – 3x$, $x \in \mathbf{R}$, $x > 0$ (ii) $f(x) = x^2 + 2$, $x$ is a real number (iii) $f(x) = x$, $x$ is a real number

Solution.

(i) The domain is restricted to $x > 0$, and that restriction is the whole question. Multiply through by $-3$, remembering that multiplying an inequality by a negative number reverses it:

$$x > 0 \quad\Longrightarrow\quad -3x < 0 \quad\Longrightarrow\quad 2 – 3x < 2$$

So every value produced is strictly below $2$, and as $x$ grows the output falls without bound. The value $2$ itself would require $x = 0$, which is excluded.

$$\text{Range} = (-\infty,\ 2)$$

(ii) Here $x$ ranges over all of $\mathbf{R}$. Since $x^2 \ge 0$ with equality only at $x = 0$,

$$x^2 + 2 \ge 2$$

The minimum $2$ is attained, at $x = 0$, and every larger value occurs (for $y \ge 2$, take $x = \sqrt{y – 2}$). Hence

$$\text{Range} = [2,\ \infty)$$

Note the contrast with (i): a square bracket here because the endpoint is reached, a round bracket there because it is not.

(iii) The identity function returns whatever it is given, and it is given every real number:

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

Common mistakes

• Question 1(i), rejecting it because every image is $1$. A function may repeat outputs freely; what it may not do is repeat an input with different outputs. Only (iii) breaks that rule.
• Question 2(i), giving the range as $\mathbf{R}$. The minus sign in front of the modulus caps the output at $0$. Sketching $y = |x|$ and reflecting it in the $x$-axis settles it in a second.
• Question 2(ii), giving the range as $[-3, 3]$. The symbol $\sqrt{\ }$ denotes the non-negative square root, so no output is negative. The graph is a semicircle, not a circle.
• Question 2(ii), solving $x^2 \le 9$ as $x \le 3$. The solution is $-3 \le x \le 3$; dropping the lower half loses half the domain.
• Question 4(ii), rounding the fraction away. $\tfrac{412}{5}$ is exactly $82.4$; both forms are correct, but writing $82$ is not.
• Question 5(i), including $2$ in the range. $f(x) = 2$ needs $x = 0$, and the domain excludes it. The endpoint of a range is included only when some point of the domain actually reaches it.
• Question 5(i), forgetting to flip the inequality. Multiplying $x > 0$ by $-3$ gives $-3x < 0$, not $-3x > 0$. That single sign decides whether the answer is $(-\infty, 2)$ or $(2, \infty)$.

Practise next

• Miscellaneous Exercise on Chapter 2 — domains of more awkward formulae, the algebra of functions ($f + g$, $f – g$, $f/g$), and further tests of whether a given relation is a function.
• Exercise 2.2 — worth revising if question 1 felt uncertain, since a function is defined as a particular kind of relation.