Let f = \(2, 4), (5, 6), (8, -1), (10, -3)\, g = \(2, 5), (7, 1), (8, 4), (10, 13), (11, 5)\ be two real funct

NCERT Exemplar · Q42

f = \{(2, 4), (5, 6), (8, -1), (10, -3)\}

(a)

f - g

;

(b)

f + g

;

(c)

f \cdot g

;

(d)

\dfrac{f}{g}

. Column II —

(i)

\left\{\left(2,\ \dfrac{4}{5}\right),\ \left(8,\ \dfrac{-1}{4}\right),\ \left(10,\ \dfrac{-3}{13}\right)\right\}

;

(ii)

\{(2, 20),\ (8, -4),\ (10, -39)\}

;

(iii)

\{(2, -1),\ (8, -5),\ (10, -16)\}

;

(iv)

\{(2, 9),\ (8, 3),\ (10, 10)\}

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Concept understanding — Function Operations

Function Operations: Combining Machines

Think of a function as a machine. You feed it an input (say, a number $x$ ), it does something, and out comes an output $f(x)$ . Now imagine you have two such machines, $f$ and $g$ . Function operations are simply ways to hook these machines together — to add, subtract, multiply, or divide their outputs, or to feed one machine's output into the other.

The core idea is simple: if you can do arithmetic with numbers, you can do arithmetic with functions. The only catch is that both functions must be "ready to work" on the same input at the same time.

The Four Arithmetic Operations

Let $f$ and $g$ be two functions. For any input $x$ that belongs to both their domains (the set of numbers each can accept), we define:

Operation	Notation	What it means
Sum	$(f+g)(x)$	$f(x) + g(x)$
Difference	$(f-g)(x)$	$f(x) - g(x)$
Product	$(f \cdot g)(x)$	$f(x) \cdot g(x)$
Quotient	$\left(\frac{f}{g}\right)(x)$	$\frac{f(x)}{g(x)}$ , provided $g(x) \neq 0$

Note

The domain of the new function is the intersection of the domains of $f$ and $g$ — the numbers both machines can handle. For the quotient, you must also exclude any $x$ where $g(x) = 0$ , because division by zero is undefined.

Example. Let $f(x) = \sqrt{x}$ (domain: $x \geq 0$ ) and $g(x) = x - 1$ (domain: all real numbers). Then:

$(f+g)(x) = \sqrt{x} + x - 1$ , domain: $x \geq 0$ .
$\left(\frac{f}{g}\right)(x) = \frac{\sqrt{x}}{x-1}$ , domain: $x \geq 0$ and $x \neq 1$ .

Composition: Feeding One Machine into Another

This is the most powerful operation. Instead of adding outputs side by side, you take the output of one function and feed it as the input to the other.

(f \circ g)(x) = f\big(g(x)\big)

Read " $f$ composed with $g$ ". You do $g$ first, then $f$ on the result.

Intuition. Suppose $g$ is a machine that converts Celsius to Fahrenheit, and $f$ is a machine that converts Fahrenheit to Kelvin. Then $f \circ g$ converts Celsius directly to Kelvin — one combined machine.

Domain trap. For $f(g(x))$ to make sense, two conditions must hold:

$x$ must be in the domain of $g$ (so $g(x)$ exists).
$g(x)$ must be in the domain of $f$ (so $f$ can accept it).

So the domain of $f \circ g$ is: all $x$ in the domain of $g$ such that $g(x)$ is in the domain of $f$ .

Watch out

Composition is not commutative. $f \circ g$ is almost never the same as $g \circ f$ . For example, if $f(x) = x^2$ and $g(x) = x+1$ , then:

$(f \circ g)(x) = (x+1)^2 = x^2 + 2x + 1$
$(g \circ f)(x) = x^2 + 1$ …

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The key idea here is Function Operations, specifically arithmetic operations on functions defined as sets of ordered pairs. For any arithmetic operation ( $f+g$ , $f-g$ , $f \cdot g$ , $f/g$ ), the domain of the resulting function is the intersection of the domains of $f$ and $g$ . For division, an additional condition is that the denominator function $g(x)$ must not be zero.

First, identify the domains of $f$ and $g$ :

$D_f = \{2, 5, 8, 10\}$

$D_g = \{2, 7, 8, 10, 11\}$

The common domain for all operations is $D_f \cap D_g = \{2, 8, 10\}$ .
Perform each operation for the elements in the common domain:
- (a) $f - g$ :
  
  $(f - g)(2) = f(2) - g(2) = 4 - 5 = -1$
  
  $(f - g)(8) = f(8) - g(8) = -1 - 4 = -5$
  
  $(f - g)(10) = f(10) - g(10) = -3 - 13 = -16$
  
  So, $f - g = \{(2, -1), (8, -5), (10, -16)\}$ , which matches (iii).
- (b) $f + g$ :
  
  $(f + g)(2) = f(2) + g(2) = 4 + 5 = 9$
  
  $(f + g)(8) = f(8) + g(8) = -1 + 4 = 3$
  
  $(f + g)(10) = f(10) + g(10) = -3 + 13 = 10$
  
  So, $f + g = \{(2, 9), (8, 3), (10, 10)\}$ , which matches (iv).
- (c) $f \cdot g$ :
  
  $(f \cdot g)(2) = f(2) \cdot g(2) = 4 \cdot 5 = 20$
  
  $(f \cdot g)(8) = f(8) \cdot g(8) = -1 \cdot 4 = -4$
  
  $(f \cdot g)(10) = f(10) \cdot g(10) = -3 \cdot 13 = -39$
  
  So, $f \cdot g = \{(2, 20), (8, -4), (10, -39)\}$ , which matches (ii). …

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To perform operations like addition, subtraction, multiplication, or division on two functions $f$ and $g$ defined by ordered pairs, we first identify the common domain where both functions are defined. Then, we apply the operation to the corresponding function values for each element in this common domain. For division, we additionally exclude any points where the denominator function $g(x)$ is zero. The final matching is (a)-(iii), (b)-(iv), (c)-(ii), (d)-(i).

When we talk about functions, especially in the context of operations like addition or multiplication, it's crucial to understand their domains. A function given as a set of ordered pairs, like $f = \{(2, 4), (5, 6)\}$ , simply means that $f(2)=4$ and $f(5)=6$ . The domain of $f$ is the set of all first elements in these pairs, so $D_f = \{2, 5\}$ .

For any binary operation (like $+$ , $-$ , $\cdot$ , $/$ ) between two functions $f$ and $g$ , the resulting function is only defined for those input values $x$ that are present in both the domain of $f$ and the domain of $g$ . This is because to calculate, say, $f(x) + g(x)$ , we need both $f(x)$ and $g(x)$ to exist.

f

Let's apply this understanding to the given functions.

Identify the domains of $f$ and $g$ .

The function $f = \{(2, 4), (5, 6), (8, -1), (10, -3)\}$ has its domain $D_f$ as the set of all first components of its ordered pairs.

$D_f = \{2, 5, 8, 10\}$ .

Similarly, for $g = \{(2, 5), (7, 1), (8, 4), (10, 13), (11, 5)\}$ , its domain $D_g$ is:

$D_g = \{2, 7, 8, 10, 11\}$ .
Determine the common domain for $f+g$ , $f-g$ , and $f \cdot g$ .

The common domain is the intersection of $D_f$ and $D_g$ .

$D_f \cap D_g = \{2, 5, 8, 10\} \cap \{2, 7, 8, 10, 11\} = \{2, 8, 10\}$ .

All operations (a),

(b),

(c) will be defined only for $x \in \{2, 8, 10\}$ .
Calculate $f-g$ .

For each $x$ in the common domain $\{2, 8, 10\}$ , we find $(f-g)(x) = f(x) - g(x)$ .
- For $x=2$ : $f(2) - g(2) = 4 - 5 = -1$ . So, $(2, -1)$ is an ordered pair in $f-g$ .
- For $x=8$ : $f(8) - g(8) = -1 - 4 = -5$ . So, $(8, -5)$ is an ordered pair in $f-g$ .
- For $x=10$ : $f(10) - g(10) = -3 - 13 = -16$ . So, $(10, -16)$ is an ordered pair in $f-g$ . Thus, $f-g = \{(2, -1), (8, -5), (10, -16)\}$ . This matches Column II (iii). So, (a) $\rightarrow$ (iii).
Calculate $f+g$ .

For each $x$ in the common domain $\{2, 8, 10\}$ , we find $(f+g)(x) = f(x) + g(x)$ .
- For $x=2$ : $f(2) + g(2) = 4 + 5 = 9$ . So, $(2, 9)$ is an ordered pair in $f+g$ .
- For $x=8$ : $f(8) + g(8) = -1 + 4 = 3$ . So, $(8, 3)$ is an ordered pair in $f+g$ .
- For $x=10$ : $f(10) + g(10) = -3 + 13 = 10$ . So, $(10, 10)$ is an ordered pair in $f+g$ . Thus, $f+g = \{(2, 9), (8, 3), (10, 10)\}$ . This matches Column II (iv). So, (b) $\rightarrow$ (iv).
Calculate $f \cdot g$ .

For each $x$ in the common domain $\{2, 8, 10\}$ , we find $(f \cdot g)(x) = f(x) \cdot g(x)$ .
- For $x=2$ : $f(2) \cdot g(2) = 4 \cdot 5 = 20$ . So, $(2, 20)$ is an ordered pair in $f \cdot g$ .
- For $x=8$ : $f(8) \cdot g(8) = -1 \cdot 4 = -4$ . So, $(8, -4)$ is an ordered pair in $f \cdot g$ .
- For $x=10$ : $f(10) \cdot g(10) = -3 \cdot 13 = -39$ . So, $(10, -39)$ is an ordered pair in $f \cdot g$ . Thus, $f \cdot g = \{(2, 20), (8, -4), (10, -39)\}$ . This matches Column II (ii). So, (c) $\rightarrow$ (ii).
Calculate $\dfrac{f}{g}$ . …

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