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
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}
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}
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
Question 4.
Find the range of the following functions.
- f(x) = 2 – 3x, x ∈ R, x>0 (1)
- f(x) = x2 + 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) = x2 + 2, The range of x2 is [0, ∞) , then the range of f(x) = x2 + 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
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.
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.
- 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.
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}.
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 = \(x^{\frac{3}{2}}\)
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
Question 6.
A = {1, 2}, B = {3, 4}
- Write A × B
- Write the 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
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 – 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)
- f(x) = |x – 2|
- f(x) = x2
- f(x) = x3
- f(x) = \(\frac{1}{x}\)
- f(x) = (x – 1)2
- f(x) = 3x2 – 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) = 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}
- 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 = \(\frac{x^{2}+2 x+3}{x^{2}-8 x+12}\)
⇒ 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}
- 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) = x3, x ∈ R
Answer: