Students can Download Chapter 2 Relations and Functions Questions and Answers, Plus One Maths Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

## Kerala Plus One Maths Chapter Wise Questions and Answers Chapter 2 Relations and Functions

### Plus One Maths Relations and Functions Three Mark Questions and Answers

**Plus One Maths Relations And Functions Previous Questions And Answers Question 1.**

Let A = {1, 2, 3, 4, 6}. Let R be the relation on A defined by R = {(a, b): a, b ∈ A, b is exactly divisible by a}

- Write R in roster form. (1)
- Find the domain of R. (1)
- Find the range of R. (1)

Answer:

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

**Plus One Maths Chapter Wise Questions And Answers Pdf Question 2.**

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

Answer:

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}

**Relations And Functions Class 11 Important Questions Pdf Question 3.**

A function f is defined as f(x) = 2x – 5, Write down the values of f(0), f(7), f(-3).

Answer:

Given; f(x) = 2x – 5

f(0) = -5;

f(7) = 2(7) – 5 = 14 – 5 = 9

f(-3) = 2(-3) – 5 = -6 – 5 = -11

**Hsslive Maths Textbook Answers Plus One Question 4.**

Find the range of the following functions.

- f(x) = 2 – 3x, x ∈ R, x>0 (1)
- f(x) = x
^{2}+ 2, x is a real number. (1) - f(x) = x, x is a real number. (1)

Answer:

- Given; f(x) = 2 – 3x is a first degree polynomial function, therefore the range is R.
- Given; f(x) = x
^{2}+ 2, The range of x2 is [0, ∞) , then the range of f(x) = x^{2}+ 2 is [2, ∞) - Given; f(x) = x is the identity function, therefore the range is R.

### Plus One Maths Relations and Functions Four Mark Questions and Answers

**Plus One Maths Chapter Wise Questions And Answers Question 1.**

Let A = {1, 2}, B = {1, 2, 3, 4}, C = {5, 6} and D = {5, 6, 7, 8}. Verify that

- A × (B ∩ C) = (A × B) ∩ (A × C) (2)
- A × C is a subset of B × D (2)

Answer:

1. A × (B ∩ C) ={1, 2} × Φ = Φ

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

Hence; A × (B ∩ C) = (A × B) ∩ (A × C)

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

B × D = {1, 2, 3, 4} × {5, 6, 7, 8}

= {(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.

**Relations And Functions Class 11 Important Questions Question 2.**

The arrow diagram given below shows a relation R from P to Q. Write the relation in roster form, set-builder form. Find its domain and range.

Answer:

R – {(9, -3), (9, 3), (4, -2), (4, 2), (25, -5), (25, 5)}

R = {{x, y) : y^{2} = x}

Domain of R = {9, 4, 25}

Range of R = {5, 3, 2, -2, -3, -5}

Question 3.

Find the domain of the following.

- f(x) = \(\frac{x^{2}+2 x+1}{x^{2}-8 x+12}\) (2)
- f(x) = \(\frac{x^{2}+3 x+5}{x^{2}-5 x+4}\) (2)

Answer:

1. Given; f(x) = \(\frac{x^{2}+2 x+1}{x^{2}-8 x+12}\)

The function is not defined at points where the denominator becomes zero.

x^{2} – 8x +12 = 0 ⇒ (x – 6)(x – 2) = 0 ⇒ x = 2, 6

Therefore domain of fis R – {2, 6}.

2. Given; f(x) = \(\frac{x^{2}+3 x+5}{x^{2}-5 x+4}\)

The function is not defined at points where the denominator becomes zero.

x^{2} – 5x + 4 = 0 ⇒ (x – 4)(x -1) = 0 ⇒ x = 1, 4

Therefore domain of f is R – {1, 4}.

**Plus One Maths Questions And Answers Question 4.**

Let f(x) = \(=\sqrt{x}\) and g(x) = x be two functions defined over the set of nonnegative real numbers. Find (f + g)(x), (f – g)(x), (fg)(x) and \(\left(\frac{f}{g}\right)(x)\).

Answer:

(f + g)(x) = f(x) + g(x) = \(=\sqrt{x}\) + x

(f – g)(x) = f(x) – g(x) = \(=\sqrt{x}\) – x

(fg)(x) = f(x) × g(x) = \(=\sqrt{x}\) × x = \(x^{\frac{3}{2}}\)

**Plus One Maths Relations And Functions Question 5.**

Let f(x) = x^{2} and g(x) = 2x + 1 be two functions defined over the set of nonnegative real numbers. Find (f + g)(x), (f – g)(x), (fg)(x) and \(\left(\frac{f}{g}\right)(x)\).

Answer:

(f + g)(x) = f(x) + g(x) = x^{2} + 2x + 1

(f – g)(x) = f(x) – g(x) = x^{2} – 2x – 1

f(fg)(x) = f(x) × g(x)

= x^{2}(2x +1) = 2x^{3} + x^{2}

**Relations And Functions Questions And Answers Pdf Question 6.**

A = {1, 2}, B = {3, 4}

- Write A × B
- Write relation from A to B in roster form. (1)
- Represent all possible functions from A to B (Arrow diagram may be used) (2)

Answer:

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

2. Any Subset of A × B (say R={(1, 3),(2, 4)})

3.

### Plus One Maths Relations and Functions Six Mark Questions and Answers

**Relations And Functions Class 11 Important Questions With Solutions Question 1.**

Let A = {1, 2, 3}, B = {3, 4} and C = {4, 5, 6}. Find

- A × (B ∩ C) (1)
- (A × B) ∩ (A × C) (2)
- A × (B ∪ C) (1)
- (A × B) ∪ (A × C) (2)

Answer:

1. A × (B ∩ C) = {1, 2, 3} × {4}

= {(1, 4), (2, 4), (3, 4)}

2. (A × B) ∩ (A × C)

= {(1, 3), (1, 4), (2, 3), (2, 4), (3, 3), (3, 4)} ∩ {(1, 4), (1, 5), (1, 6), (2, 4), (2, 5) , (2, 6), (3, 4), (3, 5), (3, 6)}

= {(1, 4), (2, 4), (3, 4)}

3. A × (B ∪ C) = {1, 2, 3} × {3, 4, 5, 6}

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

4. (A × B) ∪ (A × C)

= {(1, 3), (1, 4), (2, 3), (2, 4), (3, 3), (3, 4)} ∪ {(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)}

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

Question 2.

Find the domain and range of the following

Answer:

i) Given; f(x) = -|x|

D(f) = R, R(f) = (-∞, 0]

ii) Given; f(x) = \(\sqrt{9-x^{2}}\)

x can take values where 9 – x^{2} > 0

⇒ x^{2} ≤ 9 ⇒ -3 ≤ x ≤ 3 ⇒ x ∈ [-3, 3]

Therefore domain of f is [-3, 3]

Put \(\sqrt{9-x^{2}}\) = y, where y ≥ 0

⇒ 9 – x^{2} = y^{2}⇒ x^{2} = 9 – y^{2}

⇒ x = \(\sqrt{9-x^{2}}\)

⇒ 9 – y^{2} ≥ 0 ⇒ y^{2} ≤ 9 ⇒ -3 ≤ y ≤ 3

Therefore range of fis [0, 3].

iii) Given; f(x) = |x – 1|

Domain of f is R

The range of |x| is [0, ∞) , then the range of

f(x) = |x -1| is [0, ∞)

iv) Given; f(x) = \(\sqrt{x-1}\)

x can take values where x – 1 ≥ 0

⇒ x ≥ 1 ⇒ x ∈ [1, ∞]

Therefore domain of fis [1, ∞]

The range of \(\sqrt{x}\) is [0, ∞), then the range of

f(x) = \(\sqrt{x-1}\) is [0, ∞).

### Plus One Maths Relations and Functions Practice Problems Questions and Answers

Question 1.

If (x + 1, y – 2) = (3, 1), find the values of x and y.

Answer:

(x + 1, y – 2) = (3, 1) ⇒ x + 1 = 3, y – 2 = 1 ⇒ x = 2, y = 3.

Question 2.

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

Answer:

Question 3.

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

Answer:

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

Question 4.

if A = {-1, 1} find A × A × A

Answer:

A × A ={-1, 1} × {-1, 1}

= {(-1, -1), (-1, 1), (1, -1), (1, -1)}

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)}.

Question 5.

Write the relation R = {(x, x^{3}): x is a prime number less than 10} in roster form.

Answer:

2, 3, 5, 7 are the prime number less than 10.

R = {(2, 8),(3, 27),(5, 125),(7, 343)}

Question 6.

If f(x) = x^{2}, find \(\frac{f(1.1)-f(1)}{(1.1-1)}\)?

Answer:

Question 7.

Let \(\left\{\left(x, \frac{x^{2}}{1+x^{2}}\right), x \in R\right\}\) be a real function from R to R. Determine the domain and range of f.

Answer:

Domain of f is R.

Let \(\frac{x^{2}}{1+x^{2}}\) = y ⇒ x^{2} = y(1 + x^{2})

⇒ x^{2} = y + yx^{2} ⇒ x^{2} – yx^{2} = y

⇒ x^{2}(1 – y) = y

⇒ y ≥ 0, 1 – y > 0

⇒ y ≥ 0, y < 1 ⇒ 0 ≤ y ≤ 1

Therefore range of f is [0, 1).

Question 8.

Graph the following real functions. (each carries 2 scores)

- f(x) = |x – 2|
- f(x) = x
^{2} - f(x) = x
^{3} - f(x) = \(\frac{1}{x}\)
- f(x) = (x – 1)
^{2} - f(x) = 3x
^{2}– 1 - f(x) = |x| – 2

Answer:

1. f(x) = |x – 2| = \(\left\{\begin{aligned}x-2, & x \geq 2 \\-x+2, & x<2 \end{aligned}\right.\)

2. f(x) = x^{2}

3. f(x) = x^{3}

4. f(x) = \(\frac{1}{x}\)

5. f(x) = (x – 1)^{2}

6. f(x) = 3x^{2} – 1

7. f(x) = |x| – 2

Question 9.

Consider the relation, R = {(x, 2x – 1)/x ∈ A) where A = (2, -1, 3}

- Write R in roster form. (1)
- Write the range of R. (1)

Answer:

1. x = 2 ⇒ 2x – 1 = 2(2) – 1 = 3

x = -1 ⇒ 2x – 1 = 2(-1) – 1 = -3

x = 3 ⇒ 2x – 1 = 2(3) – 1 = 5

R = {(2, 3), (-1, -3), (3, 5)}

2. Range of R = {3, -3, 5}

Question 10.

Let A = {1, 2, 3, 4, 6} and R be a relation on A defined by R = {(a, b): a, b ∈ A, b is exactly divisible by a}

- Write R in the roster form. (1)
- Find the domain and range of R. (1)

Answer:

- R = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 6), (2, 2), (2, 4), (2, 6), (3, 3), (4, 4), (5, 5), (6, 6)}
- Domain = {1, 2, 3, 4, 6}; Range = {1, 2, 3, 4, 6}

Question 11.

Consider the real function

- \(f(x)=\frac{x^{2}+2 x+3}{x^{2}-8 x+12}\)
- Find the value of x if /(x) = 1
- Find the domain of f.

Answer:

1. Given; f(x) = 1 ⇒ 1 = \(\frac{x^{2}+2 x+3}{x^{2}-8 x+12}\)

⇒ x^{2} – 8x + 12 = x^{2} + 2x + 3

⇒ 10x = 9 ⇒ x = \(\frac{9}{10}\)

2. Find the value for which denominator is zero.

⇒ x^{2} – 8x + 12 = 0

⇒ (x – 6)(x – 2) = 0 ⇒ x = 6, 2

Therefore domain of f is R – {2, 6).

Question 12.

If f(x) = x^{3} + 5x and g(x) = 2x +1, find (f + g)(2) and {fg)(1).

Answer:

(f + g)(2) = f(2) + g(2) = (2)^{3} + 5(2) + 2(2) + 1

= 8 + 10 + 4 + 1 = 23

(fg)(1) = f(1)g(1) = (1 + 5)(2 + 1) = 6 × 3 = 18.

Question 13.

Let A = {1, 2, 3, 4, 5} and R be a relation on A defined by R = {(a, b):b = a^{2}}

- Write R in the roster form.
- Find the range of R.

Answer:

- R ={(1, 1), (2, 4)}
- Range = {1, 4}

Question 14.

Draw the graph of the function

f(x) – |x| + 1, x ∈ R

Answer:

Question 15.

Draw the graph of the function.

f(x) = x^{3}, x ∈ R

Answer: