NCERT Solutions for Class 11 Maths Chapter 2 - Relations and Functions

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Exercise 2.1

$$\textbf{1. If}\space\bigg(\frac{\textbf{x}}{\textbf{3}}+\textbf{1},\textbf{y}-\frac{\textbf{2}}{\textbf{3}}\bigg)=\bigg(\frac{\textbf{5}}{\textbf{3}},\frac{\textbf{1}}{\textbf{3}}\bigg),$$

find the values of x and y

Sol. Given

$$\bigg(\frac{x}{3}+1,y-\frac{2}{3}\bigg)=\frac{5}{3},\frac{1}{3}\\\Rarr\space\frac{x}{3}+1=\frac{5}{3}\space\text{and}\space y-\frac{2}{3}=\frac{1}{3}\\\Rarr\space\frac{x}{3}=\frac{2}{3}\space\text{and} \space y=\frac{3}{3}$$

⇒ x = 2 and y = 1.

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

Sol. Given, number of elements in set A = 3

Number of elements in set B = 3 (Set B = 3, 4, 5)

According to question, A × B = 3 × 3 = 9

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

Sol. Given, G = (7, 8) and H = (5, 4, 2)

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

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

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 ∩ f) = Φ.

Sol. (i) False, if P = {m, n} and Q = {n, m}

then P × Q = {(m, m), (m, n), (n, n), (n, m)}

(ii) True

(iii) True

5. If A = {– 1, 1}, find A × A × A.

Sol. Given A = (– 1, 1)

So A × A = (– 1, 1) × (– 1, 1)

= {(– 1, 1), (– 1, 1), (1, – 1), (1, 1)}

and 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, – 1) ,(1, – 1, 1), (1, 1, 1), (1, 1, – 1)}

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

Sol. Given, A × B = {(a, x), (a, y), (b, x), (b, y)}

A = {a, b} and B = {x, y}.

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.

Sol. Given, A = {1, 2}, B = {1, 2, 3, 4}, C = {5, 6} and D = {5, 6, 7, 8}

(i) B ∩ C = Φ

⇒ A × (B ∩ C) = A × Φ = Φ …(1)

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) and (A × B) ∩ (A × C) = Φ …(2)

By equation (1) and equation (2)

A × (B ∩ C) = (A × B) ∩ (A × C) = Φ

(ii) A × C = {(1, 5), (1, 6), (2, 5), (2, 6)}

and 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)}

Hence, A × C is a subset of B × D.

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

Sol. Given, A = {(1, 2)} and B = {3, 4}

So, A × B = {(1, 3), (1, 4), (2, 3), (2, 4)}

Number of elements in A × B = 4

Therefore, the number of subsets = 2⁴ = 16

Subsets of A × B = f, {(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)}.

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 and z are distinct elements.

Sol. If (x, 1), (y, 2), (z, 1) lies in (A × B), where n(A) = 3 and n(B) = 2, then

A = (x, y, z) and B = {1, 2}.

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.

Sol. (– 1, 0) and (0, 1) are in the cartesian product of A × A having 9 elements.

So, A = {– 1, 0, 1}

A × A = {– 1, 0, 1} × {– 1, 0, 1}

= {(– 1, – 1), (– 1, 0), (– 1, 1), (0, – 1), (0, 0), (0, 1), (1, – 1), (1, 0), (1, 1)}

Exercise 2.2

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.

Sol. The relation R from A to A is given as R = {(x, y) : 3x – y = 0}, where x, y ∈ A} that is R = {(x, y) : 3x = y, where x, y ∈ A}

The roster form is given by R = {(1, 3), (2, 6), (3, 9), (4, 12)}

The domain of R is the set of all first elements of the ordered pairs in the relation.

∴ Domain of R = {1, 2, 3, 4}

The whole set A is the codomain of the relation
R.

Codomain of R = A = {1, 2, 3, …, 14}

The range of R is the set of all second elements of the ordered pairs in the relation.

∴ Range of R = (3, 6, 9, 12).

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.

Sol. R = {(x, y) : y = x + 5, x is a natural number less than 4, x, y ∈ N}

The natural numbers less than 4 are 1, 2 and 3.

∴ R = {(1, 6), (2, 7), (3, 8)}

The domain of R is the set of all first elements of the ordered pairs in the relation.

∴ Domain of R = {1, 2, 3}

The range of R is the set of all second elements of the ordered pair in the relation.

∴ Range of R = {6, 7, 8}.

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 is roster form.

Sol. A = {1, 2, 3, 5} and B = {4, 6, 9}

R = {(x, y) : the difference between x and y is odd : x ∈ A, y ∈ B}

∴ The roster form of R = {(1, 4), (1, 6), (2, 9), (3, 4), (3, 6), (5, 4), (5, 6)}.

4. The figure shows the relationship between the sets P and Q. Write this relation:

(i) in set builder form.

(ii) roster form.

What is its domain and range ?

Sol. According to the figure,

P = {5, 6, 7}, Q = {3, 4, 5}

(i) The set builder form R = {(x, y) : y = x – 2; x ∈ P}

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

Domain of R = {5, 6, 7}

Range of R = {3, 4, 5}

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 is roster form.

(ii) Find the domain of R.

(iii) Find the range of R.

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

(i) The roster form 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}.

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

Sol. R = {(x, x + 5) : x ∈ (0, 1, 2, 3, 4, 5)}

The roster form 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}.

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

Sol. R = {(x, x³) : x is a prime number less than 10}

The prime number less than 10 aer 2, 3, 5 and 7.

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

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

Sol. It is given that A = (x, y, z) and B = (1, 2)

∴ A × B = {(x, 1), (x, 2), (y, 1) (y, 2), (z, 1), (z, 2)}

Since n(A × B) = 6, the number of subsets of A × B is 2⁶.

Therefore, the number of relations from A to B is 2⁶

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.

Sol. R = {(a, b) : a, b ∈ z, a – b is an integer}

It is known that the difference between any two integers is always an integer.

∴ Domain of R = Z

Range of R = Z.

Exercise 2.3

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)}.

Sol. (i) {(2, 1), (5, 1), (8, 1), (11, 1), (14, 1), (17, 1)}

As 2, 5, 8, 11, 14 and 17 are the elements of the domain of the relation corresponds to unique images.

So, this relation is a function.

Domain = {2, 5, 8, 11, 14, 17}

Range = {1}

(ii) {(2, 1), (4, 2), (6, 3), (8, 4), (10, 5), (12, 6), (14, 7)}

As 2, 4, 6, 8, 10, 12 and 14 are the elements of the domain of the relation corresponds to unique images.

So, this relation is a function.

Domain = {2, 4, 6, 8, 10, 12, 14}

Range = {1, 2, 3, 4, 5, 6, 7}

(iii) {(1, 3), (1, 5), (2, 5)}

Given, element 1 of the domain of the relation corresponds to two differents images i.e., 3 and 5.

So, this relation is not a function

2. Find the domain and range of the following real functions:

(i) f(x) = – | x |

$$\textbf{(ii) f(x) =}\sqrt{\textbf{9-x}^{2}}$$

Sol. (i) Given f(x) = – | x |

We know that

$$|\text{x}|=\begin{Bmatrix}\text{x if x}&\ge 0 \\\text{-x if x} & \lt 0\end{Bmatrix}$$

$$\text{So}\space f(\text{x})=-|\text{x}|=\begin{Bmatrix}\text{-x if x}&\ge 0 \\\text{x} & \lt 0\end{Bmatrix}$$

Since, f(x) is defined for all x ∈ R.

∴ Domain of f is R.

For all the real value of x, the value of | x | always remain positive. Therefore, the value of – | x | always be negative.

So the range if (– ∞, 0).

$$\text{(ii) Here, f(x) =}\sqrt{9-x^2}$$

For the domain of f, we have,

9 – x² ≥ 0

⇒ 9 ≥ x²

⇒ – 3 ≤ x ≤ 3

So Domain = {x : – 3 ≤ x ≤ 3, x ∈ R}

For any value of x such that – 3 ≤ x ≤ 3, the value of f(x) will lie between 0 and 3.

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).

Sol. Here f(x) = 2x – 5, so

(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.

4. The function ‘t’ which maps temperature in degree Celsius into temperature in degree Fahrenheit is defined by

$$\textbf{t}(\textbf{C})=\frac{9\textbf{C}}{5}+32.$$

Find:

(i) t(0)

(ii) t(28)

(iii) t(– 10)

(iv) The value of C, when t(C) = 212.

$$\textbf{Sol.}\space\text{Here}\space\text{t(C)}=\frac{9\text{C}}{5}+32, \text{so}\\\text{(i)\space}\text{t(0)}=\frac{9(0)}{5}+32\\= 0 + 32 = 32\\\text{(ii)\space t(28) =}\frac{9(28)}{5}+32\\=\frac{252+160}{5}=\frac{412}{5}\\\text{(iii)\space t(-10)}=\frac{9(-10)}{5}+32\\\text{= 9(– 2) + 32 = 14}\\\text{(iv)\space}\text{If}\space\text{t(C)}=212,\text{then}\\212=\frac{9\text{C}}{5}+32$$

$$\Rarr\space\frac{9\text{C}}{5}=212-32\\\Rarr\space C=\frac{180×5}{9}=100$$

So, if t(C) = 212, then C = 100.

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.

Sol. (i) Given, x > 0

⇒ 3x > 0

⇒ – 3x < 0

⇒ 2 – 3x < 2

⇒ f(x) < 2

∴ Range of f = (– ∞, 2).

(ii) Given, x is a real number.

∴ x² ≥ 0

⇒ x² + 2 ≥ 0 + 2

⇒ f(x) ≥ 2

∴ Range of f = [2, ∞).

(iii) Here, f(x) = x, x is a real number.

It is clear that the range of f is the set of all real numbers.

∴ Range of f is R.

Miscellaneous Exercise

1. The relation f is defined by

$$\textbf{f(x)}=\begin{cases}x^2,0 \le x \le3 \\3x, 3\le x\le10 \end{cases}$$

The relation g is defined by

$$\textbf{g(x)}=\begin{cases}x^2,0 \le x \le2 \\3x, 2\le x\le10 \end{cases}$$

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

Sol. Given,

$$\text{f(x)}=\begin{Bmatrix} x^2,0\le x \le3 \\3x, 3\le x \le 10 \end{Bmatrix}$$

at x = 3, f(x) = 3² = 9

f(x) = 3 × 3 = 9

So far 0 ≤ x ≤ 10 the images of f(x) are unique.

Thus the given relation is a function

$$\text{g(x)}=\begin{cases}x^2,0 \le x \le2 \\3x, 2\le x\le10 \end{cases}$$

It can observed that for x = 2, g(x) = 2² = 4

or g(x) = 3 × 2 = 6

Here element 2 of the domain of the relation of corresponds to two different images i.e., 4 and 6.

So this relation is not a function.

2. If f(x) = x²,

$$\textbf{find}\space\frac{\textbf{f(1.1)}-\textbf{f(1)}}{\textbf{1.1-1}}.$$

Sol. Here f(x) = x²

$$\text{Therefore}\space\frac{f(1.1)-f(1)}{(1.1-1)}\\=\frac{(1.1)(1)^2}{(0.1)}=2.1$$

3. Find the domain of the function

$$\textbf{f(x)}\textbf{=}\space\frac{\textbf{x}^2+\textbf{2}\textbf{x+1}}{\textbf{x}^2-\textbf{8x+12}}.$$

Sol. Here,

$$\text{f(x)}=\frac{x^2-2x+1}{x^2-8x+12}\\\Rarr\space\text{f(x)}=\frac{x^2+2x+1}{x^2-6x-2x+12}\\=\frac{(x+1)^2}{(x-2)(x-6)}$$

It can be seen that function f is defined for all real numbers except at x = 6 and x = 2.

Since, at x = 6 and x = 2 denomination becomes zero.

So, the domain of f = R – {2, 6}.

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

Sol. Given,

$$\text{f(x)}=\sqrt{(x-1)}\\\because\space \sqrt{(x-1)}\space\text{is defined for all x}\ge 1.$$

So, domain of f is [1, ∞].

If x ≥ 1, then x – 1 ≥ 0

$$\text{So, \space f(x)}=\sqrt{x-1}\ge0$$

Therefore, the range of f is [0, ∞).

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

Sol. Given, f(x) = | x – 1|

It is clear that | x – 1 | is defined for all real numbers.

∴ Domain of f is R.

Also, for x ∈ R, | x – 1 | assumes all real numbers.

So, the range of f is the set of all non-negative real numbers.

$$\textbf{6. Let f =} \begin{Bmatrix}\bigg(\textbf{x},\frac{\textbf{x}^2}{\textbf{1+x}^2}\bigg) : \textbf{x}\epsilon \textbf{R} \end{Bmatrix}$$

be a function from R into R. Determine the range of f.

Sol. Here,

$$\text{f =} \begin{Bmatrix}\bigg(\text{x},\frac{\text{x}^2}{\text{1+x}^2}\bigg) :\text{x}\epsilon\text{R} \end{Bmatrix}$$

Since, x² ≥ 0, ∀ x ∈ R

∴ x² + 1 > x²

$$\Rarr\space\frac{x^2}{x^2+1}\lt 1$$

⇒ f(x) < 1

$$\text{Also, at x = 0,}\frac{x^2}{1+x^2}=0$$

∴ Range of f = [0, 1).

7. Let f, g : R → R be defined, respectively by f(x) = x + 1, g(x) = 2x – 3

$$\textbf{Find f + g, f – g and}\space\frac{\textbf{f}}{\textbf{g}}.$$

Sol. Given, f : g R → R be defined, respectively by f(x) = x + 1, g(x) = 2x – 3

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

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

$$\frac{f}{g}=\frac{f(x)}{g(x)}\\=\frac{(x+1)}{(2x-3)}, 2x-3\neq 0\\\Rarr\space\frac{f}{g}=\frac{(x+1)}{(2x-3)},x\neq\frac{3}{2}.$$

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

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

Here, (1, 1) ∈ f

⇒ f(1) = 1

⇒ 1 = a(1) + b

⇒ b = 1 – a …(1)

Similarly, (2, 3) ∈ f

⇒ f(2) = 3

⇒ 3 = a(2) + b

On putting the value of b from equation (1), we get

2a + (1 – a) = 3

⇒ 2a + 1 – a = 3

⇒ a = 2

Now put the value of a in equation (1), we get

b = 1 – a = 1 – 2 = – 1

So, a = 2 and b = – 1.

9. Let R be the 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.

Sol. Given R = {(a, b) : a, b ∈ N and a = b²}

(i) Here, 3 ∈ N and 3² = 9 ≠ 3. So (a, a) ∈ is not true.

(ii) Here, 9 ∈ N and 3² = 9, so (9, 3) ∈ R. And 3 ∈ N but 9² = 81 ≠ 3. so (3, 9) ∉ R

Hence, the statement (a, b) ∈ R, implies (b, a) ∈ R is not true.

(iii) Here 81 ∈ N and 9² = 81, therefore (81, 9) ∈ R and 9 ∈ N and 3² = 9, therefore (9, 3) ∈ R and 3 ∈ N but 32 = 9 ≠ 81 therefore (81, 3) ∉ R.

Hence, the statement (a, b) ∈ R, (b, c) ∈ R implies (a, c) ∈ R is not true.

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.

Sol. Here A = {1, 2, 3, 4}, B = {1, 5, 9, 11, 15, 16} and f = {(1, 5), (2, 9), (3, 1), (4, 5), (2, 11)}

Therefore A × B = {(1, 1), (1, 5), (1, 8), (1, 11), (1, 15), (1, 16), (2, 1), (2, 5), (2, 9), (2, 11), (2, 15), (2, 16), (3, 1), (3, 5), (3, 9), (3, 11), (3, 15), (3, 16), (4, 1), (4, 5), (4, 9), (4, 11), (4, 15), (4, 16)}

(i) Here, the function f is a subset of (A × B), hence f is a relation from A to B.

(ii) Here element 2 of the domain of the relation f corresponds to two different images 9 and 11

So, this relation is not a function.

11. Let f be the subest of Z × Z defined by

f = {(ab, a + b) : a, b ∈ Z}.

Is f a function from Z to Z?

Justify your answer.

Sol. The relation is defined as

f = {(ab, a + b) : a, b ∈ Z}

We know that a relation f from a set A to a set B is said to be a function if every element at set A has a unique image in set B.

Since 1, 2, – 1, – 2 ∈ Z, (1 × 2, 1 + 2), (– 1 × – 2, – 1 + (– 2)) ∈ f

i.e., (2, 3), (2, – 3) ∈ f