NCERT Solutions For Class 11 Maths: Relations and Functions

Summary of the Chapter

This chapter introduces the fundamental concepts of Relations and Functions, which are crucial building blocks in mathematics.

Cartesian Product of Sets:
The Cartesian product of two sets $A$ and $B$, denoted by $A × B$, is the set of all ordered pairs $(a, b)$ where $a ∈ A$ and $b ∈ B$.
$A × B = \{(a, b) : a ∈ A \text{ and } b ∈ B\}$
If $n(A) = p$ and $n(B) = q$, then $n(A × B) = pq$.

Relations:
A relation $R$ from a set $A$ to a set $B$ is a subset of the Cartesian product $A × B$.
The set of all first elements of the ordered pairs in a relation $R$ is called the domain of $R$.
The set of all second elements of the ordered pairs in a relation $R$ is called the range of $R$.
The whole set $B$ is called the codomain of relation $R$.

Functions:
A relation $f$ from a set $A$ to a set $B$ is said to be a function if every element of set $A$ has one and only one image in set $B$.
In other words, a function $f: A → B$ is a rule that assigns to each element $x ∈ A$, a unique element $y ∈ B$.
We write $y = f(x)$, where $y$ is the image of $x$ under $f$, and $x$ is the pre-image of $y$.

Algebra of Real Functions:
For real functions $f: X → R$ and $g: X → R$, we define:

• $(f + g)(x) = f(x) + g(x)$
• $(f - g)(x) = f(x) - g(x)$
• $(f \cdot g)(x) = f(x) \cdot g(x)$
• $(kf)(x) = k \cdot f(x)$, where $k$ is a real number
• $(\frac{f}{g})(x) = \frac{f(x)}{g(x)}$, where $g(x) ≠ 0$

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NCERT Textbook Questions and Answers

Exercise 2.1

Question 1: If $(\frac{x}{3} + 1, y - \frac{2}{3}) = (\frac{5}{3}, \frac{1}{3})$, find the values of $x$ and $y$.

Answer-
Given that the ordered pairs are equal, their corresponding elements must be equal.
Therefore:
$\frac{x}{3} + 1 = \frac{5}{3}$ and $y - \frac{2}{3} = \frac{1}{3}$

Solving the first equation:
$\frac{x}{3} + 1 = \frac{5}{3}$
$\frac{x}{3} = \frac{5}{3} - 1$
$\frac{x}{3} = \frac{5}{3} - \frac{3}{3}$
$\frac{x}{3} = \frac{2}{3}$
$x = 2$

Solving the second equation:
$y - \frac{2}{3} = \frac{1}{3}$
$y = \frac{1}{3} + \frac{2}{3}$
$y = \frac{3}{3} = 1$

Therefore, $x = 2$ and $y = 1$.

Question 2: If the set A has 3 elements and the set B = {3, 4, 5}, then find the number of elements in (A × B).

Answer-
Given: $n(A) = 3$ and $n(B) = 3$ (since B = {3, 4, 5})
The number of elements in Cartesian product $A × B$ is given by:
$n(A × B) = n(A) × n(B) = 3 × 3 = 9$

Question 3: If G = {7, 8} and H = {5, 4, 2}, find G × H and H × G.

Answer-
G = {7, 8}, H = {5, 4, 2}

G × H = {(7, 5), (7, 4), (7, 2), (8, 5), (8, 4), (8, 2)}

H × G = {(5, 7), (5, 8), (4, 7), (4, 8), (2, 7), (2, 8)}

Question 4: State whether each of the following statements are true or false. If the statement is false, rewrite the given statement correctly.
(i) If P = {m, n} and Q = {n, m}, then P × Q = {(m, n), (n, m)}.
(ii) If A and B are non-empty sets, then A × B is a non-empty set of ordered pairs (x, y) such that x ∈ A and y ∈ B.
(iii) If A = {1, 2}, B = {3, 4}, then A × (B ∩ φ) = φ.

Answer-
(i) False.
Correct statement: If P = {m, n} and Q = {n, m}, then P × Q = {(m, n), (m, m), (n, n), (n, m)}

(ii) True.

(iii) True.
B ∩ φ = φ, and A × φ = φ.

Question 5: If A = {-1, 1}, find A × A × A.

Answer-
A = {-1, 1}
A × A = {(-1, -1), (-1, 1), (1, -1), (1, 1)}
A × A × A = A × (A × A) = {(-1, -1, -1), (-1, -1, 1), (-1, 1, -1), (-1, 1, 1), (1, -1, -1), (1, -1, 1), (1, 1, -1), (1, 1, 1)}

Question 6: If A × B = {(a, x), (a, y), (b, x), (b, y)}. Find A and B.

Answer-
From the given ordered pairs in A × B:
The first elements are: a, a, b, b → A = {a, b}
The second elements are: x, y, x, y → B = {x, y}
Therefore, A = {a, b} and B = {x, y}.

Question 7: Let A = {1, 2}, B = {1, 2, 3, 4}, C = {5, 6} and D = {5, 6, 7, 8}. Verify that
(i) A × (B ∩ C) = (A × B) ∩ (A × C)
(ii) A × C is a subset of B × D.

Answer-
(i) First, find B ∩ C:
B = {1, 2, 3, 4}, C = {5, 6}
B ∩ C = φ (no common elements)
A × (B ∩ C) = A × φ = φ

Now, find (A × B) ∩ (A × C):
A × B = {(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (2, 4)}
A × C = {(1, 5), (1, 6), (2, 5), (2, 6)}
(A × B) ∩ (A × C) = φ (no common ordered pairs)

Since both sides equal φ, the statement is verified.

(ii) A × C = {(1, 5), (1, 6), (2, 5), (2, 6)}
B × D = {(1, 5), (1, 6), (1, 7), (1, 8), (2, 5), (2, 6), (2, 7), (2, 8), (3, 5), (3, 6), (3, 7), (3, 8), (4, 5), (4, 6), (4, 7), (4, 8)}
Every element of A × C is present in B × D.
Therefore, A × C ⊆ B × D.

Question 8: Let A = {1, 2} and B = {3, 4}. Write A × B. How many subsets will A × B have? List them.

Answer-
A × B = {(1, 3), (1, 4), (2, 3), (2, 4)}
Number of elements in A × B = 2 × 2 = 4
Number of subsets of A × B = $2^4 = 16$

The subsets are:
φ,
{(1, 3)}, {(1, 4)}, {(2, 3)}, {(2, 4)},
{(1, 3), (1, 4)}, {(1, 3), (2, 3)}, {(1, 3), (2, 4)},
{(1, 4), (2, 3)}, {(1, 4), (2, 4)}, {(2, 3), (2, 4)},
{(1, 3), (1, 4), (2, 3)}, {(1, 3), (1, 4), (2, 4)},
{(1, 3), (2, 3), (2, 4)}, {(1, 4), (2, 3), (2, 4)},
{(1, 3), (1, 4), (2, 3), (2, 4)}

Question 9: Let A and B be two sets such that n(A) = 3 and n(B) = 2. If (x, 1), (y, 2), (z, 1) are in A × B, find A and B, where x, y, z are distinct elements.

Answer-
From the given ordered pairs in A × B:
The first elements are: x, y, z → A = {x, y, z}
The second elements are: 1, 2, 1 → B = {1, 2}
Therefore, A = {x, y, z} and B = {1, 2}.

Question 10: The Cartesian product A × A has 9 elements among which are found (-1, 0) and (0, 1). Find the set A and the remaining elements of A × A.

Answer-
Since A × A has 9 elements, and n(A × A) = [n(A)]² = 9, therefore n(A) = 3.
Let A = {a, b, c}
From the given ordered pairs (-1, 0) and (0, 1), we can see that -1, 0, 1 must be elements of A.
Therefore, A = {-1, 0, 1}
A × A = {(-1, -1), (-1, 0), (-1, 1), (0, -1), (0, 0), (0, 1), (1, -1), (1, 0), (1, 1)}
The remaining elements (other than (-1, 0) and (0, 1)) are:
{(-1, -1), (-1, 1), (0, -1), (0, 0), (1, -1), (1, 0), (1, 1)}

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Exercise 2.2

Question 1: Let A = {1, 2, 3, ... , 14}. Define a relation R from A to A by R = {(x, y) : 3x - y = 0, where x, y ∈ A}. Write down its domain, codomain and range.

Answer-
Given: A = {1, 2, 3, ..., 14}
R = {(x, y) : 3x - y = 0, where x, y ∈ A}
3x - y = 0 ⇒ y = 3x

Since x, y ∈ A and A = {1, 2, 3, ..., 14}, we find the ordered pairs:
When x = 1, y = 3 ∈ A ⇒ (1, 3)
When x = 2, y = 6 ∈ A ⇒ (2, 6)
When x = 3, y = 9 ∈ A ⇒ (3, 9)
When x = 4, y = 12 ∈ A ⇒ (4, 12)
When x = 5, y = 15 ∉ A ⇒ not included
Similarly for x > 4, y > 12, so not all will be in A.

Therefore, R = {(1, 3), (2, 6), (3, 9), (4, 12)}
Domain of R = {1, 2, 3, 4}
Range of R = {3, 6, 9, 12}
Codomain of R = A = {1, 2, 3, ..., 14}

Question 2: Define a relation R on the set N of natural numbers by R = {(x, y) : y = x + 5, x is a natural number less than 4; x, y ∈ N}. Depict this relationship using roster form. Write down the domain and the range.

Answer-
Given: R = {(x, y) : y = x + 5, x is a natural number less than 4; x, y ∈ N}
x is a natural number less than 4 ⇒ x = 1, 2, 3
When x = 1, y = 1 + 5 = 6 ⇒ (1, 6)
When x = 2, y = 2 + 5 = 7 ⇒ (2, 7)
When x = 3, y = 3 + 5 = 8 ⇒ (3, 8)

Therefore, R = {(1, 6), (2, 7), (3, 8)} in roster form.
Domain of R = {1, 2, 3}
Range of R = {6, 7, 8}

Question 3: A = {1, 2, 3, 5} and B = {4, 6, 9}. Define a relation R from A to B by R = {(x, y): the difference between x and y is odd; x ∈ A, y ∈ B}. Write R in roster form.

Answer-
A = {1, 2, 3, 5}, B = {4, 6, 9}
R = {(x, y): the difference between x and y is odd; x ∈ A, y ∈ B}

Let's check all possible pairs:
(1, 4): |1 - 4| = 3 (odd) ⇒ included
(1, 6): |1 - 6| = 5 (odd) ⇒ included
(1, 9): |1 - 9| = 8 (even) ⇒ not included
(2, 4): |2 - 4| = 2 (even) ⇒ not included
(2, 6): |2 - 6| = 4 (even) ⇒ not included
(2, 9): |2 - 9| = 7 (odd) ⇒ included
(3, 4): |3 - 4| = 1 (odd) ⇒ included
(3, 6): |3 - 6| = 3 (odd) ⇒ included
(3, 9): |3 - 9| = 6 (even) ⇒ not included
(5, 4): |5 - 4| = 1 (odd) ⇒ included
(5, 6): |5 - 6| = 1 (odd) ⇒ included
(5, 9): |5 - 9| = 4 (even) ⇒ not included

Therefore, R = {(1, 4), (1, 6), (2, 9), (3, 4), (3, 6), (5, 4), (5, 6)}

Question 4: The figure shows a relationship between the sets P and Q. Write this relation
(i) in set-builder form
(ii) in roster form.
What is its domain and range?

Answer-
Assuming the figure shows: P = {5, 6, 7}, Q = {3, 4, 5} and arrows from:
5 → 3, 6 → 4, 7 → 5

(i) In set-builder form: R = {(x, y): y = x - 2, x ∈ P, y ∈ Q}

(ii) In roster form: R = {(5, 3), (6, 4), (7, 5)}

Domain of R = {5, 6, 7}
Range of R = {3, 4, 5}

Question 5: Let A = {1, 2, 3, 4, 6}. Let R be the relation on A defined by
{(a, b): a, b ∈ A, b is exactly divisible by a}.
(i) Write R in roster form
(ii) Find the domain of R
(iii) Find the range of R

Answer-
A = {1, 2, 3, 4, 6}
R = {(a, b): a, b ∈ A, b is exactly divisible by a}

Let's find all ordered pairs where b is divisible by a:
(1, 1): 1 is divisible by 1
(1, 2): 2 is divisible by 1
(1, 3): 3 is divisible by 1
(1, 4): 4 is divisible by 1
(1, 6): 6 is divisible by 1
(2, 2): 2 is divisible by 2
(2, 4): 4 is divisible by 2
(2, 6): 6 is divisible by 2
(3, 3): 3 is divisible by 3
(3, 6): 6 is divisible by 3
(4, 4): 4 is divisible by 4
(6, 6): 6 is divisible by 6

(i) R = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 6), (2, 2), (2, 4), (2, 6), (3, 3), (3, 6), (4, 4), (6, 6)}

(ii) Domain of R = {1, 2, 3, 4, 6}

(iii) Range of R = {1, 2, 3, 4, 6}

Question 6: Determine the domain and range of the relation R defined by R = {(x, x + 5) : x ∈ {0, 1, 2, 3, 4, 5}}.

Answer-
R = {(x, x + 5) : x ∈ {0, 1, 2, 3, 4, 5}}
When x = 0, (0, 5)
When x = 1, (1, 6)
When x = 2, (2, 7)
When x = 3, (3, 8)
When x = 4, (4, 9)
When x = 5, (5, 10)

Therefore, R = {(0, 5), (1, 6), (2, 7), (3, 8), (4, 9), (5, 10)}
Domain of R = {0, 1, 2, 3, 4, 5}
Range of R = {5, 6, 7, 8, 9, 10}

Question 7: Write the relation R = {(x, x³) : x is a prime number less than 10} in roster form.

Answer-
Prime numbers less than 10 are: 2, 3, 5, 7
R = {(x, x³) : x is a prime number less than 10}
When x = 2, (2, 8)
When x = 3, (3, 27)
When x = 5, (5, 125)
When x = 7, (7, 343)

Therefore, R = {(2, 8), (3, 27), (5, 125), (7, 343)} in roster form.

Question 8: Let A = {x, y, z} and B = {1, 2}. Find the number of relations from A to B.

Answer-
n(A) = 3, n(B) = 2
n(A × B) = 3 × 2 = 6
Number of relations from A to B = number of subsets of A × B = $2^6 = 64$

Question 9: Let R be the relation on Z defined by R = {(a, b): a, b ∈ Z, a - b is an integer}. Find the domain and range of R.

Answer-
R = {(a, b): a, b ∈ Z, a - b is an integer}
Since a and b are both integers, a - b is always an integer.
Therefore, R includes all ordered pairs of integers.
Domain of R = Z (all integers)
Range of R = Z (all integers)

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Exercise 2.3

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)}

Answer-
(i) This is a function because each element in the domain (2, 5, 8, 11, 14, 17) has exactly one image (1).
Domain = {2, 5, 8, 11, 14, 17}
Range = {1}

(ii) This is a function because each element in the domain (2, 4, 6, 8, 10, 12, 14) has exactly one image.
Domain = {2, 4, 6, 8, 10, 12, 14}
Range = {1, 2, 3, 4, 5, 6, 7}

(iii) This is NOT a function because the element 1 in the domain has two different images (3 and 5).

Question 2: Find the domain and range of the following real functions:
(i) $f(x) = -|x|$
(ii) $f(x) = \sqrt{9 - x^2}$

Answer-
(i) $f(x) = -|x|$
Domain: x can be any real number, so Domain = R
Range: Since |x| ≥ 0 for all x, -|x| ≤ 0 for all x.
So Range = (-∞, 0] or {y ∈ R : y ≤ 0}

(ii) $f(x) = \sqrt{9 - x^2}$
For f(x) to be defined, the expression under the square root must be non-negative:
9 - x² ≥ 0 ⇒ x² ≤ 9 ⇒ -3 ≤ x ≤ 3
So Domain = {x ∈ R : -3 ≤ x ≤ 3} or [-3, 3]
Range: Since x² ≥ 0, 9 - x² ≤ 9, so $\sqrt{9 - x^2} ≤ 3$
Also, $\sqrt{9 - x^2} ≥ 0$
So Range = {y ∈ R : 0 ≤ y ≤ 3} or [0, 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)

Answer-
f(x) = 2x - 5
(i) f(0) = 2(0) - 5 = -5
(ii) f(7) = 2(7) - 5 = 14 - 5 = 9
(iii) f(-3) = 2(-3) - 5 = -6 - 5 = -11

Question 4: The function 't' which maps temperature in degree Celsius into temperature in degree Fahrenheit is defined by $t(C) = \frac{9C}{5} + 32$. Find
(i) t(0)
(ii) t(28)
(iii) t(-10)
(iv) The value of C, when t(C) = 212

Answer-
$t(C) = \frac{9C}{5} + 32$
(i) t(0) = $\frac{9(0)}{5} + 32 = 32$
(ii) t(28) = $\frac{9(28)}{5} + 32 = \frac{252}{5} + 32 = 50.4 + 32 = 82.4$
(iii) t(-10) = $\frac{9(-10)}{5} + 32 = \frac{-90}{5} + 32 = -18 + 32 = 14$
(iv) When t(C) = 212:
$\frac{9C}{5} + 32 = 212$
$\frac{9C}{5} = 212 - 32 = 180$
$9C = 180 × 5 = 900$
$C = \frac{900}{9} = 100$

Question 5: Find the range of each of the following functions:
(i) f(x) = 2 - 3x, x ∈ R, x > 0
(ii) f(x) = x² + 2, x is a real number
(iii) f(x) = x, x is a real number

Answer-
(i) f(x) = 2 - 3x, x > 0
Since x > 0, 3x > 0, so -3x < 0
Therefore, 2 - 3x < 2
As x increases, f(x) decreases without bound.
Range = (-∞, 2)

(ii) f(x) = x² + 2, x ∈ R
Since x² ≥ 0 for all x, x² + 2 ≥ 2
Range = [2, ∞)

(iii) f(x) = x, x ∈ R
This is the identity function.
Range = R (all real numbers)

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Miscellaneous Exercise on Chapter 2

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.

Answer-
For f(x):
At x = 3, both definitions give the same value:
$f(3) = 3^2 = 9$ (from first definition)
$f(3) = 3 × 3 = 9$ (from second definition)
Since both definitions give the same value at x = 3, f is a function.
Domain of f = [0, 10]

For g(x):
At x = 2, the two definitions give different values:
$g(2) = 2^2 = 4$ (from first definition)
$g(2) = 3 × 2 = 6$ (from second definition)
Since x = 2 has two different images, g is not a function.

Question 2: If f(x) = x², find $\frac{f(1.1) - f(1)}{(1.1 - 1)}$.

Answer-
f(x) = x²
f(1.1) = (1.1)² = 1.21
f(1) = 1² = 1
$\frac{f(1.1) - f(1)}{(1.1 - 1)} = \frac{1.21 - 1}{0.1} = \frac{0.21}{0.1} = 2.1$

Question 3: Find the domain of the function $f(x) = \frac{x^2 + 2x + 1}{x^2 - 8x + 12}$.

Answer-
$f(x) = \frac{x^2 + 2x + 1}{x^2 - 8x + 12} = \frac{(x + 1)^2}{(x - 2)(x - 6)}$
The function is defined for all real x except where denominator = 0
(x - 2)(x - 6) = 0 ⇒ x = 2 or x = 6
Therefore, Domain = R - {2, 6} or {x ∈ R : x ≠ 2, x ≠ 6}

Question 4: Find the domain and range of the real function f defined by $f(x) = \sqrt{x - 1}$.

Answer-
$f(x) = \sqrt{x - 1}$
For f(x) to be defined, the expression under square root must be ≥ 0:
x - 1 ≥ 0 ⇒ x ≥ 1
Therefore, Domain = [1, ∞)
Range: Since $\sqrt{x - 1} ≥ 0$ for all x in domain
As x increases, f(x) increases without bound
Therefore, Range = [0, ∞)

Question 5: Find the domain and range of the real function f defined by $f(x) = |x - 1|$.

Answer-
$f(x) = |x - 1|$
Domain: x can be any real number, so Domain = R
Range: |x - 1| ≥ 0 for all x
As x varies over R, |x - 1| takes all values from 0 to ∞
Therefore, Range = [0, ∞)

Question 6: Let $f = \{(x, \frac{x^2}{1 + x^2}) : x \in R\}$ be a function from R to R. Determine the range of f.

Answer-
$f(x) = \frac{x^2}{1 + x^2}$
Since x² ≥ 0 and 1 + x² > 0 for all x ∈ R, f(x) ≥ 0
Also, x² ≤ 1 + x², so $\frac{x^2}{1 + x^2} ≤ 1$
Therefore, 0 ≤ f(x) ≤ 1
Now, can f(x) actually attain 0 and 1?
f(x) = 0 when x = 0
As |x| → ∞, f(x) → 1
But f(x) never equals 1 for any finite x
Therefore, Range = [0, 1)

Question 7: Let f, g : R → R be defined, respectively by f(x) = x + 1, g(x) = 2x - 3. Find f + g, f - g and $\frac{f}{g}$.

Answer-
f(x) = x + 1, g(x) = 2x - 3
(f + g)(x) = f(x) + g(x) = (x + 1) + (2x - 3) = 3x - 2
(f - g)(x) = f(x) - g(x) = (x + 1) - (2x - 3) = x + 1 - 2x + 3 = -x + 4
$(\frac{f}{g})(x) = \frac{f(x)}{g(x)} = \frac{x + 1}{2x - 3}$, where 2x - 3 ≠ 0 ⇒ x ≠ $\frac{3}{2}$

Question 8: Let f = {(1, 1), (2, 3), (0, -1), (-1, -3)} be a linear function from Z into Z. Find f(x).

Answer-
Since f is a linear function, let f(x) = ax + b
Using the given points:
From (1, 1): a(1) + b = 1 ⇒ a + b = 1 ...(1)
From (2, 3): a(2) + b = 3 ⇒ 2a + b = 3 ...(2)
Subtracting (1) from (2): (2a + b) - (a + b) = 3 - 1 ⇒ a = 2
Substituting a = 2 in (1): 2 + b = 1 ⇒ b = -1
Therefore, f(x) = 2x - 1
Verifying with other points:
(0, -1): 2(0) - 1 = -1 ✓
(-1, -3): 2(-1) - 1 = -2 - 1 = -3 ✓

Question 9: If R is a relation on the set A = {1, 2, 3, 4, 5, 6, 7, 8, 9} given by R = {(x, y) : y = x + 2}, then find the range of R.

Answer-
A = {1, 2, 3, 4, 5, 6, 7, 8, 9}
R = {(x, y) : y = x + 2}
For x ∈ A, y = x + 2 must also be in A
So x + 2 ≤ 9 ⇒ x ≤ 7
Therefore, x can be 1, 2, 3, 4, 5, 6, 7
R = {(1, 3), (2, 4), (3, 5), (4, 6), (5, 7), (6, 8), (7, 9)}
Range of R = {3, 4, 5, 6, 7, 8, 9}

Question 10: A function f is defined by f(x) = 2x - 5. Write down the values of
(i) f(0)
(ii) f(7)
(iii) f(-3)

Answer-
f(x) = 2x - 5
(i) f(0) = 2(0) - 5 = -5
(ii) f(7) = 2(7) - 5 = 14 - 5 = 9
(iii) f(-3) = 2(-3) - 5 = -6 - 5 = -11

Question 11: The function f and g are defined by f(x) = 6x + 8; g(x) = $\frac{x - 2}{3}$
(i) Calculate the value of $gg\left(\frac{1}{2}\right)$
(ii) Write an expression for gf(x) in its simplest form.

Answer-
f(x) = 6x + 8; g(x) = $\frac{x - 2}{3}$
(i) $gg\left(\frac{1}{2}\right) = g\left(g\left(\frac{1}{2}\right)\right)$
First, $g\left(\frac{1}{2}\right) = \frac{\frac{1}{2} - 2}{3} = \frac{-\frac{3}{2}}{3} = -\frac{1}{2}$
Then, $g\left(-\frac{1}{2}\right) = \frac{-\frac{1}{2} - 2}{3} = \frac{-\frac{5}{2}}{3} = -\frac{5}{6}$
Therefore, $gg\left(\frac{1}{2}\right) = -\frac{5}{6}$

(ii) gf(x) = g(f(x)) = g(6x + 8) = $\frac{(6x + 8) - 2}{3} = \frac{6x + 6}{3} = 2x + 2$

Question 12: The function f is defined by $f(x) = \begin{cases} 1 - x, & x < 0 \\ 1, & x = 0 \\ x + 1, & x > 0 \end{cases}$
Draw the graph of f(x).

Answer-
For x < 0: f(x) = 1 - x (a straight line with slope -1)
For x = 0: f(x) = 1 (a single point at (0, 1))
For x > 0: f(x) = x + 1 (a straight line with slope 1)

The graph consists of:

• A ray from negative infinity approaching (0, 1) with equation y = 1 - x
• A point at (0, 1)
• A ray starting from (0, 1) going to positive infinity with equation y = x + 1

Question 13: Let f be a function defined by f(x) = 2x - 3. Find the values of x such that f(x) = f(1) + f(2).

Answer-
f(x) = 2x - 3
f(1) = 2(1) - 3 = -1
f(2) = 2(2) - 3 = 1
f(1) + f(2) = -1 + 1 = 0
We need f(x) = 0
2x - 3 = 0 ⇒ 2x = 3 ⇒ x = $\frac{3}{2}$

Question 14: 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.

Answer-
(i) f is a relation from A to B if f ⊆ A × B
Checking each ordered pair:
(1, 5): 1 ∈ A, 5 ∈ B ✓
(2, 9): 2 ∈ A, 9 ∈ B ✓
(3, 1): 3 ∈ A, 1 ∈ B ✓
(4, 5): 4 ∈ A, 5 ∈ B ✓
(2, 11): 2 ∈ A, 11 ∈ B ✓
All ordered pairs are from A × B, so f is a relation from A to B. TRUE

(ii) f is a function if each element of A has exactly one image in B
Checking:
1 → 5 ✓
2 → 9 and also 2 → 11 (two different images) ✗
3 → 1 ✓
4 → 5 ✓
Since element 2 has two different images, f is not a function. FALSE

Question 15: Let R be a relation from N to N defined by R = {(a, b) : a, b ∈ N and a = b²}. Are the following true?
(i) (a, a) ∈ R, for all a ∈ N
(ii) (a, b) ∈ R, implies (b, a) ∈ R
(iii) (a, b) ∈ R, (b, c) ∈ R implies (a, c) ∈ R
Justify your answer in each case.

Answer-
R = {(a, b) : a, b ∈ N and a = b²}
(i) (a, a) ∈ R would mean a = a² ⇒ a² - a = 0 ⇒ a(a - 1) = 0 ⇒ a = 0 or a = 1
But 0 ∉ N, and only a = 1 satisfies this.
So (a, a) ∈ R is not true for all a ∈ N. FALSE

(ii) If (a, b) ∈ R, then a = b²
For (b, a) ∈ R, we would need b = a²
So we need a = b² and b = a² simultaneously
This means a = (a²)² = a⁴ ⇒ a⁴ - a = 0 ⇒ a(a³ - 1) = 0 ⇒ a = 0 or a = 1
Only a = 1, b = 1 works.
So in general, (a, b) ∈ R does not imply (b, a) ∈ R. FALSE

(iii) If (a, b) ∈ R and (b, c) ∈ R, then a = b² and b = c²
Then a = (c²)² = c⁴
For (a, c) ∈ R, we need a = c²
But we have a = c⁴, so we need c⁴ = c² ⇒ c⁴ - c² = 0 ⇒ c²(c² - 1) = 0 ⇒ c = 0 or c = 1
Only c = 1 works.
So in general, (a, b) ∈ R, (b, c) ∈ R does not imply (a, c) ∈ R. FALSE

Question 16: Let A = {1, 2, 3, 4}, B = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} and f : A → B be defined by f(x) = 2x + 1. Find the range of f.

Answer-
A = {1, 2, 3, 4}, f(x) = 2x + 1
f(1) = 2(1) + 1 = 3
f(2) = 2(2) + 1 = 5
f(3) = 2(3) + 1 = 7
f(4) = 2(4) + 1 = 9
Therefore, Range of f = {3, 5, 7, 9}

Question 17: Let A = {1, 2} and B = {3, 4}. Write A × B. How many subsets will A × B have? List them.

Answer-
A = {1, 2}, B = {3, 4}
A × B = {(1, 3), (1, 4), (2, 3), (2, 4)}
Number of elements in A × B = 2 × 2 = 4
Number of subsets = 2⁴ = 16

The subsets are:
φ,
{(1, 3)}, {(1, 4)}, {(2, 3)}, {(2, 4)},
{(1, 3), (1, 4)}, {(1, 3), (2, 3)}, {(1, 3), (2, 4)},
{(1, 4), (2, 3)}, {(1, 4), (2, 4)}, {(2, 3), (2, 4)},
{(1, 3), (1, 4), (2, 3)}, {(1, 3), (1, 4), (2, 4)},
{(1, 3), (2, 3), (2, 4)}, {(1, 4), (2, 3), (2, 4)},
{(1, 3), (1, 4), (2, 3), (2, 4)}

Question 18: Let f be the subset of Z × Z defined by f = {(ab, a + b) : a, b ∈ Z}. Is f a function from Z to Z? Justify your answer.

Answer-
f = {(ab, a + b) : a, b ∈ Z}
For f to be a function, each input (first element) should have exactly one output (second element)
Let's check if this is true:
Consider a = 2, b = 3: (6, 5) ∈ f
Consider a = 1, b = 6: (6, 7) ∈ f
Here, the same input 6 gives two different outputs 5 and 7
Therefore, f is not a function from Z to Z.

Question 19: Let R be a relation on the set N of natural numbers defined by R = {(x, y) : y = x + 5, x is a natural number less than 4; x, y ∈ N}. Depict this relationship using (i) roster form (ii) an arrow diagram. Find the domain and range of R.

Answer-
R = {(x, y) : y = x + 5, x is a natural number less than 4; x, y ∈ N}
x is a natural number less than 4 ⇒ x = 1, 2, 3
When x = 1, y = 1 + 5 = 6 ⇒ (1, 6)
When x = 2, y = 2 + 5 = 7 ⇒ (2, 7)
When x = 3, y = 3 + 5 = 8 ⇒ (3, 8)

(i) Roster form: R = {(1, 6), (2, 7), (3, 8)}
(ii) Arrow diagram: Three arrows: 1 → 6, 2 → 7, 3 → 8
Domain of R = {1, 2, 3}
Range of R = {6, 7, 8}

Question 20: Let A = {1, 2, 3, 4, 6}. Let R be the relation on A defined by {(a, b) : a, b ∈ A, b is exactly divisible by a}.
(i) Write R in roster form
(ii) Find the domain of R
(iii) Find the range of R

Answer-
A = {1, 2, 3, 4, 6}
R = {(a, b) : a, b ∈ A, b is exactly divisible by a}

Finding all ordered pairs where b is divisible by a:
(1, 1): 1 is divisible by 1
(1, 2): 2 is divisible by 1
(1, 3): 3 is divisible by 1
(1, 4): 4 is divisible by 1
(1, 6): 6 is divisible by 1
(2, 2): 2 is divisible by 2
(2, 4): 4 is divisible by 2
(2, 6): 6 is divisible by 2
(3, 3): 3 is divisible by 3
(3, 6): 6 is divisible by 3
(4, 4): 4 is divisible by 4
(6, 6): 6 is divisible by 6

(i) R = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 6), (2, 2), (2, 4), (2, 6), (3, 3), (3, 6), (4, 4), (6, 6)}

(ii) Domain of R = {1, 2, 3, 4, 6}

(iii) Range of R = {1, 2, 3, 4, 6}

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Important Keywords from the Chapter

• Ordered Pair: A pair of elements written in specific order (a, b)
• Cartesian Product: $A × B$ = {(a, b) : a ∈ A, b ∈ B}
• Relation: A subset of Cartesian product $A × B$
• Domain: Set of all first elements of ordered pairs in a relation
• Range: Set of all second elements of ordered pairs in a relation
• Codomain: The whole set B in a relation from A to B
• Function: A relation where each element of domain has exactly one image
• Real Function: A function whose domain and range are subsets of real numbers
• Identity Function: f(x) = x for all x in domain
• Constant Function: f(x) = c for all x in domain
• Algebra of Functions: Operations like addition, subtraction, multiplication, and division of functions
• Piecewise Function: A function defined by different formulas on different intervals
• Composite Function: gf(x) = g(f(x))
• Graph of Function: Visual representation of a function on coordinate plane