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

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}

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

Answer:

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

1. f(x) = 2 – 3x, x ∈ R, x>0 (1)
2. f(x) = x² + 2, x is a real number. (1)
3. f(x) = x, x is a real number. (1)

Answer:

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

1. A × (B ∩ C) = (A × B) ∩ (A × C) (2)
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) : y² = x}
Domain of R = {9, 4, 25}
Range of R = {5, 3, 2, -2, -3, -5}

Question 3.
Find the domain of the following.

1. f(x) = \(\frac{x^{2}+2 x+1}{x^{2}-8 x+12}\) (2)
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.
x² – 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.
x² – 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) = x² 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) = x² + 2x + 1
(f – g)(x) = f(x) – g(x) = x² – 2x – 1
f(fg)(x) = f(x) × g(x)
= x²(2x +1) = 2x³ + x²

Relations And Functions Questions And Answers Pdf Question 6.
A = {1, 2}, B = {3, 4}

1. Write A × B
2. Write relation from A to B in roster form. (1)
3. 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

1. A × (B ∩ C) (1)
2. (A × B) ∩ (A × C) (2)
3. A × (B ∪ C) (1)
4. (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 – x² > 0
⇒ x² ≤ 9 ⇒ -3 ≤ x ≤ 3 ⇒ x ∈ [-3, 3]
Therefore domain of f is [-3, 3]
Put \(\sqrt{9-x^{2}}\) = y, where y ≥ 0
⇒ 9 – x² = y²⇒ x² = 9 – y²
⇒ x = \(\sqrt{9-x^{2}}\)
⇒ 9 – y² ≥ 0 ⇒ y² ≤ 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, x³): 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) = x², 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 ⇒ x² = y(1 + x²)
⇒ x² = y + yx² ⇒ x² – yx² = y
⇒ x²(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)

1. f(x) = |x – 2|
2. f(x) = x²
3. f(x) = x³
4. f(x) = \(\frac{1}{x}\)
5. f(x) = (x – 1)²
6. f(x) = 3x² – 1
7. 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) = x²

3. f(x) = x³

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

5. f(x) = (x – 1)²

6. f(x) = 3x² – 1

7. f(x) = |x| – 2

Question 9.
Consider the relation, R = {(x, 2x – 1)/x ∈ A) where A = (2, -1, 3}

1. Write R in roster form. (1)
2. 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}

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

Answer:

1. 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)}
2. Domain = {1, 2, 3, 4, 6}; Range = {1, 2, 3, 4, 6}

Question 11.
Consider the real function

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

Answer:
1. Given; f(x) = 1 ⇒ 1 = \(\frac{x^{2}+2 x+3}{x^{2}-8 x+12}\)
⇒ x² – 8x + 12 = x² + 2x + 3
⇒ 10x = 9 ⇒ x = \(\frac{9}{10}\)

2. Find the value for which denominator is zero.
⇒ x² – 8x + 12 = 0
⇒ (x – 6)(x – 2) = 0 ⇒ x = 6, 2
Therefore domain of f is R – {2, 6).

Question 12.
If f(x) = x³ + 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 = a²}

1. Write R in the roster form.
2. Find the range of R.

Answer:

1. R ={(1, 1), (2, 4)}
2. 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) = x³, x ∈ R
Answer: